\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{mathrsfs,amssymb} \usepackage{epic} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 76, pp. 1--25.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/76\hfil A linear first-order hyperbolic equation] {A linear first-order hyperbolic equation with a discontinuous coefficient: distributional shadows and propagation of singularities} \author[H. Deguchi\hfil EJDE-2011/76\hfilneg] {Hideo Deguchi} \address{Hideo Deguchi \newline Department of Mathematics, University of Toyama \\ Toyama 930-8555, Japan} \email{hdegu@sci.u-toyama.ac.jp} \thanks{Submitted January 19, 2011. Published June 16, 2011.} \thanks{Supported by the Austrian Science Fund (FWF), Lise Meitner project M1155-N13} \subjclass[2000]{46F30, 35L03, 35A21} \keywords{First-order hyperbolic equation; discontinuous coefficient; \hfill\break\indent generalized solutions} \begin{abstract} It is well-known that distributional solutions to the Cauchy problem for $u_t + (b(t,x)u)_{x} = 0$ with $b(t,x) = 2H(x-t)$, where $H$ is the Heaviside function, are non-unique. However, it has a unique generalized solution in the sense of Colombeau. The relationship between its generalized solutions and distributional solutions is established. Moreover, the propagation of singularities is studied. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} %\newtheorem{example}[theorem]{Example} \section{Introduction}\label{intro} In this paper we study generalized solutions of the Cauchy problem for a first-order hyperbolic equation $$\label{eqn:hyperbolic1} \begin{gathered} u_t + (b(t,x)u)_{x} = 0, \quad (t,x) \in \mathbb{R}^2, \\ u|_{t = 0} = u_0, \quad x \in \mathbb{R} \end{gathered}$$ with $b(t,x) = 2H(x-t)$, where $H$ is the Heaviside function, in the framework of generalized functions introduced by Colombeau \cite{colombeau1,colombeau2}. We will seek solutions in an algebra $\mathscr{G}(\mathbb{R}^2)$ of generalized functions, which will be defined in Section \ref{sec:2} below. We mention that $\mathscr{G}(\mathbb{R})$ contains the space $\mathscr{D}'(\mathbb{R})$ of distributions so that initial data with strong singularities can be considered in our setup. The formulation of problem \eqref{eqn:hyperbolic1} in $\mathscr{G}$ will be given in Section \ref{sec:3}. Until now, the following three questions for a variety of partial differential equations in Colombeau's algebras have been addressed: (a) existence and uniqueness of generalized solutions; (b) behavior of generalized solutions in the framework of distribution theory (distributional shadow); (c) regularity of generalized solutions. For linear first-order hyperbolic systems with discontinuous coefficients, the existence and uniqueness were established in one space-dimensional case by Oberguggenberger \cite{oberguggenberger1989}, for symmetric hyperbolic systems in higher space-dimensional case by Lafon and Oberguggenberger \cite{lafon} and for hyperbolic pseudodifferential systems with generalized symbols by H\"{o}rmann \cite{hoermann2}. Almost all previous results on question (b) for differential equations have been obtained under the hypothesis of unique distributional solutions. However, it is well-known \cite{hurd} that distributional solutions of linear first-order hyperbolic equations with discontinuous coefficients may fail to exist or may be non-unique. Question (b) for the linear hyperbolic equation having no distributional solution given by Hurd and Sattinger \cite{hurd}, namely, for problem \eqref{eqn:hyperbolic1} with $b=-H$ and $u_0 \equiv 1$, where $H$ is the Heaviside function, was answered by Oberguggenberger \cite{oberguggenberger1988}. Similar equations have been studied by H\"{o}rmann and de Hoop \cite{hoermann1}. In this paper we are concerned with question (b) for linear hyperbolic equations having non-unique distributional solutions. According to Hurd and Sattinger \cite{hurd}, for $t > 0$, problem \eqref{eqn:hyperbolic1} with $u_0 \equiv 0$ has infinitely many distributional solutions $$\label{eqn:u_c} u_c(t,x):= \begin{cases} 0, & \text{if } x < 0 \text{ or } x > 2t, \\ c, & \text{if } 0 < x < t, \\ -c, & \text{if } t < x < 2t \end{cases}$$ with $c \in \mathbb{R}$. On the other hand, as stated above, it has been proved in \cite{oberguggenberger1989} that problem \eqref{eqn:hyperbolic1} has a unique generalized solution $U \in \mathscr{G}(\mathbb{R}^2)$ for any initial data. First, we will investigate how the generalized solutions are related to the distributional solutions $u_c$. Several results on behavior of generalized solutions in the framework of distribution theory of differential equations having non-unique distributional solutions have been obtained. For ordinary differential equations, see \cite{deguchi1}, and for parabolic equations, see \cite{deguchi2}. Concerning the regularity of generalized solutions of problem \eqref{eqn:hyperbolic1}, we focus on the case of initial data given by the delta function at $s \in \mathbb{R}$. As can be seen in Section \ref{sec:2}, there exist an abundant variety of elements of $\mathscr{G}(\mathbb{R})$ having the property of the delta function at $s$, which are called Dirac generalized functions at $s$. In particular, there exist Dirac generalized functions at $s$ which can be interpreted to have different strengths of singularity at $s$. Thus we have the following question: how does the strength of the singularity of a Dirac generalized function taken as initial data affect the regularity of the generalized solution of problem \eqref{eqn:hyperbolic1}? Secondly, we will give an answer to this question. The propagation of singularities for linear first-order hyperbolic equations with other particular discontinuous coefficients has been studied by H\"{o}rmann and de Hoop \cite{hoermann1}, Garetto and H\"{o}rmann \cite{garetto} and Oberguggenberger \cite{oberguggenberger2008}. The rest of this paper is organized as follows: we recall the definition and basic properties of the Colombeau algebra $\mathscr{G}$ in Section \ref{sec:2}. In Section \ref{sec:3}, we first give our formulation of problem \eqref{eqn:hyperbolic1} and describe a result on existence and uniqueness of its generalized solution $U \in \mathscr{G}(\mathbb{R}^2)$ for any initial data $U_0 \in \mathscr{G}(\mathbb{R})$ which has been obtained by Oberguggenberger \cite{oberguggenberger1989}. In Section \ref{sec:4}, we discuss how the generalized solutions are related to the distributional solutions $u_c$ given by form \eqref{eqn:u_c} (Theorem \ref{thm:1}). In Section \ref{sec:5}, we look at problem \eqref{eqn:hyperbolic1} with various Dirac generalized functions as initial data. We investigate the behavior of the generalized solutions in the framework of distribution theory, and further the regularity of the generalized solutions (Theorems \ref{thm:2}, \ref{thm:2-1}, \ref{thm:3}, \ref{thm:3-1}, \ref{thm:4} and \ref{thm:4-1}). \section{Colombeau's theory of generalized functions}\label{sec:2} We will employ the \emph{special Colombeau algebra} denoted by $\mathscr{G}^{s}$ in Grosser et al. \cite{grosser}, which was called the \emph{simplified Colombeau algebra} in Biagioni \cite{biagioni}. However, here we will simply use the letter $\mathscr{G}$ instead. Let us briefly recall the definition and basic properties of the algebra $\mathscr{G}$ of generalized functions. For more details, see Grosser et al. \cite{grosser}. Let $\Omega$ be a non-empty open subset of $\mathbb{R}^d$. Let $\mathscr{E}(\Omega)$ be the differential algebra of all maps from the interval $(0,1]$ into $C^{\infty}(\Omega)$. Thus each element of $\mathscr{E}(\Omega)$ is a family $(u^{\varepsilon})_{\varepsilon \in (0,1]}$ of real valued smooth functions on $\Omega$. The subalgebra ${\mathscr{E}}_{M}(\Omega)$ is defined by all elements $(u^{\varepsilon})_{\varepsilon \in (0,1]}$ of $\mathscr{E}(\Omega)$ with the property that, for all $K \Subset \Omega$ and $\alpha \in \mathbb{N}_0^d$, there exists $p \ge 0$ such that $\sup_{x \in K} |\partial_{x}^{\alpha} u^{\varepsilon}(x)| = O(\varepsilon^{-p}) \quad \text{as }\varepsilon \downarrow 0.$ The ideal $\mathscr{N}(\Omega)$ is defined by all elements $(u^{\varepsilon})_{\varepsilon \in (0,1]}$ of $\mathscr{E}(\Omega)$ with the property that, for all $K \Subset \Omega$, $\alpha \in \mathbb{N}_0^d$ and $q \ge 0$, $\sup_{x \in K} |\partial_{x}^{\alpha} u^{\varepsilon}(x)| = O(\varepsilon^q) \quad \text{as }\varepsilon \downarrow 0.$ \emph{The algebra} $\mathscr{G}(\Omega)$ \emph{of generalized functions} is defined by the quotient space $\mathscr{G}(\Omega) = {\mathscr{E}}_{M}(\Omega) / \mathscr{N}(\Omega).$ We use capital letters for elements of $\mathscr{G}(\Omega)$ to distinguish generalized functions from distributions and denote by $(u^{\varepsilon})_{\varepsilon \in (0,1]}$ a representative of $U \in \mathscr{G}(\Omega)$. Then for any $U$, $V \in \mathscr{G}(\Omega)$ and $\alpha \in \mathbb{N}_0^d$, we can define the partial derivative $\partial^{\alpha} U$ to be the class of $(\partial^{\alpha}u^{\varepsilon})_{\varepsilon \in (0,1]}$ and the product $UV$ to be the class of $(u^{\varepsilon}v^{\varepsilon})_{\varepsilon \in (0,1]}$. Also, for any $U =$\ class of $(u^{\varepsilon}(t,x))_{\varepsilon \in (0,1]} \in \mathscr{G}(\mathbb{R}^2)$, we can define its restriction $U|_{t = 0} \in \mathscr{G}(\mathbb{R})$ to the line $\{t = 0\}$ to be the class of $(u^{\varepsilon}(0,x))_{\varepsilon \in (0,1]}$. \begin{remark}\label{rem:imbedding} \rm The algebra $\mathscr{G}(\Omega)$ contains the space $\mathscr{E}'(\Omega)$ of compactly supported distributions. In fact, the map $f \mapsto \text{class of } (f \ast \rho_{\varepsilon}\mid_{\Omega})_{\varepsilon \in (0,1]}$ defines an imbedding of $\mathscr{E}'(\Omega)$ into $\mathscr{G}(\Omega)$, where $\rho_{\varepsilon}(x) = \frac{1}{\varepsilon^d} \rho \left(\frac{x}{\varepsilon}\right)$ and $\rho$ is a fixed element of $\mathscr{S}(\mathbb{R}^d)$ such that $\int \rho(x)\,dx = 1$ and $\int x^{\alpha}\rho(x)\,dx = 0$ for any $\alpha \in \mathbb{N}_0^d$, $|\alpha| \ge 1$. In this sense, we obtain an inclusion relation $\mathscr{E}'(\Omega) \subset \mathscr{G}(\Omega)$. This can be extended in a unique way to an imbedding of the space $\mathscr{D}'(\Omega)$ of distributions. Moreover, this imbedding turns $C^{\infty}(\Omega)$ into a subalgebra of $\mathscr{G}(\Omega)$. \end{remark} \begin{definition} \rm A generalized function $U \in \mathscr{G}(\Omega)$ is said to be \emph{associated with a distribution} $w \in \mathscr{D}'(\Omega)$ if it has a representative $(u^{\varepsilon})_{\varepsilon \in (0,1]} \in {\mathscr{E}}_{M}(\Omega)$ such that $u^{\varepsilon} \to w \quad {\rm in}\ \mathscr{D}'(\Omega) \quad {\rm as} \ \varepsilon \downarrow 0.$ We denote by $U \approx w$ and call $w$ the \emph{distributional shadow} of $U$ if $U$ is associated with $w$. \end{definition} \begin{remark} \rm A subalgebra $\mathscr{G}_{\rm log}(\Omega)$ of $\mathscr{G}(\Omega)$ is defined similarly as $\mathscr{G}(\Omega)$ by replacing the bound $\sup_{x \in K}|\partial_{x}^{\alpha} u^{\varepsilon}(x)| = O(\varepsilon^{-p})$ in ${\mathscr{E}}_{M}(\Omega)$ by the stronger bound $\sup_{x \in K}|\partial_{x}^{\alpha} u^{\varepsilon}(x)| = O((\log(1/\varepsilon))^{p})$. For any distribution $f \in \mathscr{D}'(\Omega)$, there exists a generalized function $U \in \mathscr{G}_{\rm log}(\Omega)$ which is associated with $f$, see Colombeau and Heibig \cite{colombeau and heibig}. Therefore, any distribution on $\Omega$ can be interpreted as an element of $\mathscr{G}_{\rm log}(\Omega)$ in the sense of association. \end{remark} We next define the notion of generalized functions of Dirac type. \begin{definition}\label{defn:dirac} We say that $U \in \mathscr{G}(\mathbb{R})$ is a \emph{Dirac generalized function} at $s \in \mathbb{R}$ if it has a representative $(u^{\varepsilon})_{\varepsilon \in (0,1]}$ satisfying \begin{itemize} \item[(1)] there exists $a(\varepsilon) > 0$, $a(\varepsilon) \to 0$ as $\varepsilon \downarrow 0$, such that $u^{\varepsilon}(x) = 0$ if $|x-s| \ge a(\varepsilon)$; \item[(2)] $\int_{\mathbb{R}} u^{\varepsilon}(x)\,dx = 1$ for all $\varepsilon \in (0,1]$; \item[(3)] $\sup_{\varepsilon \in (0,1]} \int_{\mathbb{R}} |u^{\varepsilon}(x)|\,dx < \infty$. \end{itemize} Then $U$ admits the delta function $\delta_s$ at $s$ as distributional shadow. \end{definition} Regularity theory for linear equations has been based on the subalgebra $\mathscr{G}^{\infty}(\Omega)$ of \emph{regular generalized functions} in $\mathscr{G}(\Omega)$ introduced by Oberguggenberger \cite{oberguggenberger1992}. It is defined by all elements which have a representative $(u^{\varepsilon})_{\varepsilon \in (0,1]}$ with the property that, for all $K \Subset \Omega$, there exists $p \ge 0$ such that, for all $\alpha \in \mathbb{N}_0^d$, $\sup_{x \in K} |\partial_{x}^{\alpha} u^{\varepsilon}(x)| = O(\varepsilon^{-p}) \quad \text{as }\varepsilon \downarrow 0.$ We observe that all derivatives of $u^{\varepsilon}$ have locally the same order of growth in $\varepsilon > 0$, unlike elements of ${\mathscr{E}}_{M}(\Omega)$. This subalgebra $\mathscr{G}^{\infty}(\Omega)$ has the property $\mathscr{G}^{\infty}(\Omega) \cap \mathscr{D}'(\Omega) = C^{\infty}(\Omega)$, see \cite[Theorem 25.2, p. 275]{oberguggenberger1992}. Hence, for the purpose of describing the regularity of generalized functions, $\mathscr{G}^{\infty}(\Omega)$ plays the same role for $\mathscr{G}(\Omega)$ as $C^{\infty}(\Omega)$ does in the setting of distributions. The $\mathscr{G}^{\infty}$-singular support (denoted by $\operatorname{sing\,supp}_{\mathscr{G}^{\infty}}$) of a generalized function is defined as the complement of the largest open set on which the generalized function is regular in the above sense. A subalgebra $\mathscr{G}_{\rm log}^{\infty}(\Omega)$ of $\mathscr{G}_{\rm log}(\Omega)$ is defined similarly as $\mathscr{G}^{\infty}(\Omega)$ by replacing the bound $\sup_{x \in K}|\partial_{x}^{\alpha} u^{\varepsilon}(x)| = O(\varepsilon^{-p})$ by the stronger bound $\sup_{x \in K}|\partial_{x}^{\alpha} u^{\varepsilon}(x)| = O((\log(1/\varepsilon))^{p})$. The $\mathscr{G}_{\rm log}^{\infty}$-singular support (denoted by $\operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}}$) can be also introduced. \begin{remark}\label{rem:strength} \rm Let $s \in \mathbb{R}$ and let $\chi$ be a fixed element of $\mathscr{D}(\mathbb{R})$ such that $\chi$ is symmetric, non-negative, with supp$\,\chi \subset [-1,1]$, $\chi(0) > 0$ and $\int_{\mathbb{R}}\chi(x)\,dx = 1$. Put $\chi_{\varepsilon}(x) = \chi(x/\varepsilon)/\varepsilon$. Then $U \in \mathscr{G}(\mathbb{R})$ defined by the class of $(\chi_{\varepsilon}(\cdot - s))_{\varepsilon \in (0,1]}$ is a Dirac generalized function at $s$ and $\operatorname{sing\,supp}_{\mathscr{G}^{\infty}} U = \{s\}$. However, if $U \in \mathscr{G}(\mathbb{R})$ is defined as the class of $(\chi_{h(\varepsilon)}(\cdot - s))_{\varepsilon \in (0,1]}$ with $h(\varepsilon) = 1/\log(1/\varepsilon)$, then it is a Dirac generalized function at $s$ again, but $\operatorname{sing\,supp}_{\mathscr{G}^{\infty}} U = \emptyset$. Hence, the speed of convergence of a representative of $U$ to the delta function at $s$ can be interpreted as the strength of the singularity of $U$ at $s$. Thus, there exist infinitely many Dirac generalized functions with different strengths of singularity at $s$ in $\mathscr{G}(\mathbb{R})$. \end{remark} \section{Existence and uniqueness of generalized solutions}\label{sec:3} We formulate problem \eqref{eqn:hyperbolic1} in Colombeau's algebra $\mathscr{G}$ in the form $$\label{eqn:generalized hyperbolic1} \begin{gathered} U_t + (BU)_{x} = 0 \quad \text{in } \mathscr{G}(\mathbb{R}^2), \\ U|_{t = 0} = U_0 \quad \text{in } \mathscr{G}(\mathbb{R}) \end{gathered}$$ with the generalized function $B \in \mathscr{G}(\mathbb{R}^2)$ having the representative $(b^{\varepsilon})_{\varepsilon \in (0,1]}$ given by $b^{\varepsilon}(t,x) := b \ast \varphi_{h(\varepsilon)} = 2\int\int_{\mathbb{R}^2} H(x-t-h(\varepsilon)y+h(\varepsilon)s) \varphi(s,y)\,dy\,ds,$ where $h(\varepsilon):= 1/\log (1/\varepsilon)$ and $\varphi$ is a fixed element of $\mathscr{D}(\mathbb{R}^2)$ such that $\varphi$ is symmetric, non-negative, with $\operatorname{supp}\varphi \subset [-1,1] \times [-1,1]$, $\varphi(0,0) > 0$ and $\int\int\varphi(t,x)\,dx\,dt = 1$. We note that $B$ belongs to $\mathscr{G}_{\rm log}(\mathbb{R}^2)$ and further admits $b(t,x) = 2H(x-t)$ as distributional shadow. \begin{definition} \rm We say that $U \in \mathscr{G}(\mathbb{R}^2)$ is a (\emph{generalized}) \emph{solution} of problem \eqref{eqn:generalized hyperbolic1} if it has a representative $(u^{\varepsilon})_{\varepsilon \in (0,1]} \in {\mathscr{E}}_{M}(\mathbb{R}^2)$ such that \begin{gather*} u^{\varepsilon}_t + (b^{\varepsilon}u^{\varepsilon})_{x} = N^{\varepsilon}, \quad (t,x) \in \mathbb{R}^2,\\ u^{\varepsilon}|_{t = 0} = u_{0}^{\varepsilon} + n^{\varepsilon}, \quad x \in \mathbb{R} \end{gather*} for some $(N^{\varepsilon})_{\varepsilon \in (0,1]} \in \mathscr{N}(\mathbb{R}^2)$ and $(n^{\varepsilon})_{\varepsilon \in (0,1]} \in \mathscr{N}(\mathbb{R})$, where $(u_{0}^{\varepsilon})_{\varepsilon \in (0,1]}$ and $(b^{\varepsilon})_{\varepsilon \in (0,1]}$ are representatives of $U_0$ and $B$, respectively. \end{definition} For any $U_0 \in \mathscr{G}(\mathbb{R})$, problem \eqref{eqn:generalized hyperbolic1} has a unique solution $U \in \mathscr{G}(\mathbb{R}^2)$, see \cite{oberguggenberger1989}, in which a more general existence and uniqueness result has been obtained. \section{Relationship to non-unique distributional solutions}\label{sec:4} In this section we establish the relationship between the generalized solutions of problem \eqref{eqn:generalized hyperbolic1} and the distributional solutions $u_c$ of problem \eqref{eqn:hyperbolic1}. For this purpose, we first prepare the following lemma. \begin{lemma}\label{lemma:1} For $\varepsilon \in (0,1)$, let \begin{align*} G^{\varepsilon}(x):= \int_{-x/h(\varepsilon)}^{2} \frac{dz}{1-\widetilde{b}^{\varepsilon}(-h(\varepsilon)z)} \quad \text{on } [-2h(\varepsilon),0), \end{align*} where $\widetilde{b}^{\varepsilon}(z):= 2\int\int_{\mathbb{R}^2} H(z-h(\varepsilon)y+h(\varepsilon)s)\varphi(s,y)\,dy\,ds$. Then $G^{\varepsilon}$ has the following four properties: \begin{itemize} \item[(i)] $G^{\varepsilon}$ is a strictly increasing continuous function on $[-2h(\varepsilon),0)$ such that $G^{\varepsilon}(-2h(\varepsilon)) = 0$ and $\lim_{x \uparrow 0}G^{\varepsilon}(x) = \infty$; \item[(ii)] there exist two constants $C_1$, $C_2 > 0$ such that, for any $x \in [-2h(\varepsilon),0)$ and $\varepsilon \in (0,1)$, $$\label{eqn:inequality for G^{varepsilon}} C_1\log\Big(\frac{2h(\varepsilon)}{-x}\Big) \le G^{\varepsilon}(x) \le C_2\log\Big(\frac{2h(\varepsilon)}{-x}\Big);$$ \item[(iii)] there exists the inverse function $(G^{\varepsilon})^{-1}$ on $[0,\infty)$, which satisfies $$\label{eqn:differentiation of (G^{varepsilon})^{-1}} \frac{d(G^{\varepsilon})^{-1}(x)}{dx} = h(\varepsilon)[1-\widetilde{b}^{\varepsilon}((G^{\varepsilon})^{-1}(x))];$$ \item[(iv)] for any $a$, $x \in [0,\infty)$, $$\label{eqn:inequality for (G^{varepsilon})^{-1}} 2 \exp \Big(-\frac{a}{C_1}\Big) h(\varepsilon)\varepsilon^{x/C_1} \le \big|(G^{\varepsilon})^{-1}\Big(\frac{x}{h(\varepsilon)} +a\Big)\big| \le 2 \exp \Big(-\frac{a}{C_2}\Big) h(\varepsilon) \varepsilon^{x/C_2},$$ where $C_1$, $C_2 > 0$ are the constants given in (ii). \end{itemize} \end{lemma} \begin{proof} Property (i) is clear. To show property (ii), rewrite $1 = 2\int\int_{s > y} \varphi(s,y)\,dy\,ds$. Then we find that \begin{align*} 1 - \widetilde{b}^{\varepsilon}(-h(\varepsilon)z) & = 1 - 2\int\int_{s > y + z} \varphi(s,y)\,dy\,ds = 2\int\int_{y < s < y + z} \varphi(s,y)\,ds\,dy. \end{align*} Hence, there exist two constants $c_1$, $c_2 > 0$ such that $c_1 z \le 1 - \widetilde{b}^{\varepsilon}(-h(\varepsilon)z) \le c_2 z$ for $0 \le z \le 2$. Putting $C_1=1/c_2$ and $C_2=1/c_1$, the reciprocal of $1 - \widetilde{b}^{\varepsilon}(-h(\varepsilon)z)$ satisfies the inequality $\frac{C_1}{z} \le \frac{1}{1 - \widetilde{b}^{\varepsilon} (-h(\varepsilon)z)} \le \frac{C_2}{z}.$ Integrating this over $[-x/h(\varepsilon),2)$ gives inequality \eqref{eqn:inequality for G^{varepsilon}}. Next, we prove property (iii). By property (i), there exists $(G^{\varepsilon})^{-1}$ on $[0,\infty)$. We differentiate $(G^{\varepsilon})^{-1}$ to get $d(G^{\varepsilon})^{-1}(x)/dx = 1/(G^{\varepsilon})'((G^{\varepsilon})^{-1}(x))$. We have $(G^{\varepsilon})'(x) = 1/h(\varepsilon)(1-\widetilde{b}^{\varepsilon}(x))$ and so get formula \eqref{eqn:differentiation of (G^{varepsilon})^{-1}}. Finally, we prove property (iv). Put $y = (G^{\varepsilon})^{-1}(x/h(\varepsilon)+a) < 0$. By property (ii), there exist two constants $C_1$, $C_2 > 0$ such that $C_1\log\Big(\frac{2h(\varepsilon)}{-y}\Big) \le G^{\varepsilon}(y) \le C_2\log\Big(\frac{2h(\varepsilon)}{-y}\Big).$ Noting that $G^{\varepsilon}(y) = x/h(\varepsilon) + a$, we have $C_1\left[\log 2h(\varepsilon) - \log(-y)\right] \le \frac{x}{h(\varepsilon)} + a \le C_2\left[\log 2h(\varepsilon) - \log(-y)\right].$ Therefore, we see that $\log 2h(\varepsilon) - \frac{a}{C_1} - \frac{x}{C_1h(\varepsilon)} \le \log(-y) \le \log 2h(\varepsilon) - \frac{a}{C_2} - \frac{x}{C_2h(\varepsilon)}.$ Since $h(\varepsilon) = 1/\log (1/\varepsilon)$, it follows that $2 \exp \Big(-\frac{a}{C_1}\Big) h(\varepsilon) \varepsilon^{x/C_1} \le -y \le 2 \exp \Big(-\frac{a}{C_2}\Big) h(\varepsilon) \varepsilon^{x/C_2}.$ Thus inequality \eqref{eqn:inequality for (G^{varepsilon})^{-1}} follows. \end{proof} We now turn to a comparison between generalized solutions of problem \eqref{eqn:generalized hyperbolic1} and the distributional solutions $u_c$ given by \eqref{eqn:u_c} of problem \eqref{eqn:hyperbolic1}. \begin{theorem}\label{thm:1} For any $c \in \mathbb{R}$ and $T > 0$, there exists initial data $U_0 \approx 0$ such that the solution $U \in \mathscr{G}(\mathbb{R}^2)$ of problem \eqref{eqn:generalized hyperbolic1} admits a distributional shadow on $(-T,T) \times \mathbb{R}$, which is given by $u(t,x) = \begin{cases} u_c(t,x), & \text{if } 0 < t < T,\; x \in \mathbb{R}, \\ 0, &\text{if } -T < t \le 0,\; x \in \mathbb{R}, \end{cases}$ where $u_c$ is the function given by \eqref{eqn:u_c}. \end{theorem} \begin{proof} We consider the Cauchy problem $$\label{eqn:generalized hyperbolic2} \begin{gathered} V_t + BV_{x} = 0 \quad \text{in } \mathscr{G}(\mathbb{R}^2), \\ V|_{t = 0} = V_0 \quad \text{in } \mathscr{G}(\mathbb{R}). \end{gathered}$$ The existence and uniqueness of solutions $V \in \mathscr{G}(\mathbb{R}^2)$ of problem \eqref{eqn:generalized hyperbolic2} are guaranteed for all initial data $V_0 \in \mathscr{G}(\mathbb{R})$ by Oberguggenberger \cite{oberguggenberger1989}. Clearly, $V_x$ satisfies problem \eqref{eqn:generalized hyperbolic1}. Define the function $v(t,x):= \int_0^x u(t,y)\,dy$. In order to prove the assertion, it suffices to show that, for any $c \in \mathbb{R}$ and $T>0$, there exists initial data $V_0$ such that $V_0' \approx 0$ on $\mathbb{R}$ and $V \approx v$ on $(-T,T) \times \mathbb{R}$. We will only prove this for the case $c > 0$. The case $c \le 0$ can be argued similarly. The proof is divided into three steps. {\bf Step 1}. Fix $c > 0$ and $\varepsilon \in (0,1)$ arbitrarily. Let $G^{\varepsilon}$ be as in Lemma \ref{lemma:1}. Recall that $G^{\varepsilon}$ is a strictly increasing continuous function on $[-2h(\varepsilon),0)$ with $G^{\varepsilon}(-2h(\varepsilon)) = 0$ and $\lim_{x \uparrow 0}G^{\varepsilon}(x) = \infty$. Hence, there exists a point $0 < \eta(\varepsilon) < 2h(\varepsilon)$ such that $G^{\varepsilon}(-\eta(\varepsilon)) = 2$. Define $$\label{eqn:w_0^{varepsilon}} w_{0}^{\varepsilon}(x):= \begin{cases} ch(\varepsilon)(G^{\varepsilon}(x)-2), & \text{if } -\eta(\varepsilon) \le x < 0,\\ ch(\varepsilon)(G^{\varepsilon}(-x)-2), & \text{if } 0 < x \le \eta(\varepsilon),\\ 0, & \text{if } |x| > \eta(\varepsilon), \end{cases}$$ and let $w^{\varepsilon}$ be a solution of the problem \begin{gather*} w^{\varepsilon}_t + b^{\varepsilon}w^{\varepsilon}_{x} = 0, \quad (t,x) \in \mathbb{R}^2,\; t \ne x \\ w^{\varepsilon}|_{t = 0} = w_0^{\varepsilon}, \quad x \in \mathbb{R},\; x \ne 0. \end{gather*} We will show below that, for $t \ge 0$, $w^{\varepsilon}(t,x) = \begin{cases} 0, & \text{if } x \le -2h(\varepsilon) \text{ or } x \ge 2t + 2h(\varepsilon),\\ cx, & \text{if } 0 \le x \le t - 2h(\varepsilon),\\ -cx + 2ct, & \text{if }t + 2h(\varepsilon) \le x \le 2t. \end{cases}$ Similarly, we can obtain that, for $t < 0$, $w^{\varepsilon}(t,x) = 0\quad \text{if }x \le t-2h(\varepsilon) \text{ or } x \ge t + 2h(\varepsilon).$ The characteristic curve $\gamma^{\varepsilon}(t,x,\tau)$ passing through $(t,x)$ at time $\tau = t$ is the solution of the problem $$\label{eqn:characteristics} \begin{gathered} \gamma^{\varepsilon}_{\tau}(t,x,\tau) = b^{\varepsilon}(\tau,\gamma^{\varepsilon}(t,x,\tau)), \\ \gamma^{\varepsilon}|_{\tau = t} = x. \end{gathered}$$ Along the characteristic curves, $w^{\varepsilon}$ is easily calculated as $$\label{eqn:w^{varepsilon}} w^{\varepsilon}(t,x) = w_0^{\varepsilon}(\gamma^{\varepsilon} (t,x,0)).$$ If $x \le -2h(\varepsilon)$ and $t > 0$, then $\gamma^{\varepsilon}(t,x,0) = x$ and $w_{0}^{\varepsilon}(x) = 0$. Hence, by \eqref{eqn:w^{varepsilon}}, we have $w^{\varepsilon}(t,x) = 0$. If $x \ge 2t + 2h(\varepsilon)$ and $t > 0$, then $\gamma^{\varepsilon}(t,x,0) = x - 2t$ and $w_{0}^{\varepsilon}(x-2t) = 0$. Hence, by \eqref{eqn:w^{varepsilon}}, we get $w^{\varepsilon}(t,x) = 0$. We next prove that $w^{\varepsilon}(t,x) = cx$ if $0 \le x \le t - 2h(\varepsilon)$. Fix $(t,x)$ arbitrarily so that $0 \le x \le t - 2h(\varepsilon)$. Let $\widetilde{b}^{\varepsilon}$ be as in Lemma \ref{lemma:1}. Then, from \eqref{eqn:characteristics}, we see that $\gamma^{\varepsilon}(t,x,\tau)$ satisfies the equation $(\gamma^{\varepsilon}(t,x,\tau) - \tau)_{\tau} = \widetilde{b}^{\varepsilon}(\gamma^{\varepsilon}(t,x,\tau) - \tau) - 1$. We divide both sides by $\widetilde{b}^{\varepsilon}(\gamma^{\varepsilon}(t,x,\tau) - \tau) - 1$ and integrate it over $[0,t]$ to get \begin{align*} \int_{0}^{t} \frac{(\gamma^{\varepsilon}(t,x,\tau) - \tau)_{\tau}}{\widetilde{b}^{\varepsilon}(\gamma^{\varepsilon} (t,x,\tau) - \tau) - 1}\,d\tau = t. \end{align*} Putting $\gamma = \gamma^{\varepsilon}(t,x,\tau) - \tau$ and noting that $\gamma^{\varepsilon}(t,x,t) = x$, we have $$\label{eqn:integral} \int_{\gamma^{\varepsilon}(t,x,0)}^{x - t} \frac{d\gamma}{\widetilde{b}^{\varepsilon}(\gamma) - 1} = t.$$ Since $x - t \le - 2h(\varepsilon)$ and $\widetilde{b}^{\varepsilon}(\gamma) = 0$ for $\gamma \le -2h(\varepsilon)$, it follows that $\int_{\gamma^{\varepsilon}(t,x,0)}^{-2h(\varepsilon)} \frac{d\gamma}{\widetilde{b}^{\varepsilon}(\gamma) - 1} = x + 2h(\varepsilon).$ Put $z = -\gamma/h(\varepsilon)$. Then \begin{align*} \int_{-\gamma^{\varepsilon}(t,x,0)/h(\varepsilon)}^{2} \frac{dz}{1 - \widetilde{b}^{\varepsilon}(-h(\varepsilon)z)} = \frac{x}{h(\varepsilon)} + 2. \end{align*} The left-hand side is rewritten as $G^{\varepsilon}(\gamma^{\varepsilon}(t,x,0))$ and so $\gamma^{\varepsilon}(t,x,0) = (G^{\varepsilon})^{-1}(x/h(\varepsilon) + 2)$. Therefore, by \eqref{eqn:w_0^{varepsilon}} and \eqref{eqn:w^{varepsilon}}, we get $w^{\varepsilon}(t,x) = cx$. We next prove that $w^{\varepsilon}(t,x) = -cx + 2ct$ if $t + 2h(\varepsilon) \le x \le 2t$. Fix $(t,x)$ arbitrarily so that $t + 2h(\varepsilon) \le x \le 2t$. The same argument as above gives \eqref{eqn:integral}. Since $x - t \ge 2h(\varepsilon)$ and $\widetilde{b}^{\varepsilon}(\gamma) = 2$ for $\gamma \ge 2h(\varepsilon)$, we have $\int_{\gamma^{\varepsilon}(t,x,0)}^{2h(\varepsilon)} \frac{d\gamma}{\widetilde{b}^{\varepsilon}(\gamma) - 1} =-x+2t+2h(\varepsilon).$ Put $z = \gamma/h(\varepsilon)$. Then $\int_{\gamma^{\varepsilon}(t,x,0)/h(\varepsilon)}^{2} \frac{dz}{\widetilde{b}^{\varepsilon}(h(\varepsilon)z)-1} = \frac{-x+2t}{h(\varepsilon)} + 2.$ The left-hand side is equal to $G^{\varepsilon}(-\gamma^{\varepsilon}(t,x,0))$, since $\widetilde{b}^{\varepsilon}(h(\varepsilon)z)-1 = 1-\widetilde{b}^{\varepsilon}(-h(\varepsilon)z)$ for $z \in \mathbb{R}$. Hence, $\gamma^{\varepsilon}(t,x,0) = -(G^{\varepsilon})^{-1}((-x+2t)/h(\varepsilon) + 2)$. Therefore, by \eqref{eqn:w_0^{varepsilon}} and \eqref{eqn:w^{varepsilon}}, we obtain that $w^{\varepsilon}(t,x) = -cx+2ct$. {\bf Step 2}. Fix $T > 0$ arbitrarily. Then $T/h(\varepsilon) > 2$ for $\varepsilon > 0$ small enough. For such $\varepsilon > 0$, we choose $0 < \lambda(\varepsilon) < \eta(\varepsilon)$ such that $G^{\varepsilon}(-\lambda(\varepsilon)) = T/h(\varepsilon)$, and put $\overline{w_0^{\varepsilon}}(x):= \begin{cases} w_0^{\varepsilon}(x), & \text{if }|x| > \lambda(\varepsilon),\\ w_0^{\varepsilon}(\lambda(\varepsilon)), & \text{if }|x| \le \lambda(\varepsilon). \end{cases}$ Let $\overline{w^{\varepsilon}}$ be a solution of the problem \begin{gather*} (\overline{w^{\varepsilon}})_t + b^{\varepsilon}(\overline{w^{\varepsilon}})_{x} = 0, \quad (t,x) \in \mathbb{R}^2, \\ \overline{w^{\varepsilon}}|_{t = 0} = \overline{w_0^{\varepsilon}}, \quad x \in \mathbb{R}. \end{gather*} Then it is easy to check that, for $t \ge 0$, $\overline{w^{\varepsilon}}(t,x) = \begin{cases} 0, & \text{if }x \le -2h(\varepsilon) \text{ or } x \ge 2t + 2h(\varepsilon), \\ cx, & \text{if }0 \le x \le \min\{t - 2h(\varepsilon), \gamma^{\varepsilon}(0,-\lambda(\varepsilon),t)\},\\ -cx + 2ct, & \text{if }\max\{t + 2h(\varepsilon), \gamma^{\varepsilon}(0,\lambda(\varepsilon),t)\} \le x \le 2t, \end{cases}$ and further that, for $t < 0$, $\overline{w^{\varepsilon}}(t,x) = 0\quad \text{if }x \le t-2h(\varepsilon) \text{ or } x \ge t + 2h(\varepsilon).$ We now prove that $\overline{w^{\varepsilon}}$ converges to $v$ in $\mathscr{D}'((-T,T) \times \mathbb{R})$ as $\varepsilon \downarrow 0$. Consider the characteristic curve $\gamma^{\varepsilon}(0,-\lambda(\varepsilon),\tau)$ passing through $(0,-\lambda(\varepsilon))$ at $\tau = 0$. There exists $t_1^{\varepsilon} > 0$ such that $t_1^{\varepsilon} = \gamma^{\varepsilon}(0,-\lambda(\varepsilon),t_1^{\varepsilon}) + 2h(\varepsilon)$. As in Step $1$, we can show that $t_1^{\varepsilon} = h(\varepsilon) \int_{\lambda(\varepsilon)/h(\varepsilon)}^{2} \frac{dz}{1 - \widetilde{b}^{\varepsilon}(-h(\varepsilon)z)} = h(\varepsilon) G^{\varepsilon}(-\lambda(\varepsilon)) = T.$ Similarly, for the characteristic curve $\gamma^{\varepsilon}(0,\lambda(\varepsilon),\tau)$ passing through $(0,\lambda(\varepsilon))$ at $\tau = 0$, there exists $t_2^{\varepsilon} > 0$ such that $t_2^{\varepsilon} = \gamma^{\varepsilon}(0,\lambda(\varepsilon), t_2^{\varepsilon}) - 2h(\varepsilon)$. Moreover, $t_2^{\varepsilon} = T$. Hence, for any $\psi \in \mathscr{D}((-T,T)\times \mathbb{R})$, we see that $$\begin{split} & \int_{-T}^{T}\int_{-\infty}^{\infty} (\overline{w^{\varepsilon}}(t,x) - v(t,x))\psi(t,x)\,dx\,dt \\ & = \int \int_{-2h(\varepsilon) < x < \min\{0,t-2h (\varepsilon)\},\, 0 < t < T} (\overline{w^{\varepsilon}}(t,x) - v(t,x))\psi(t,x)\,dx\,dt \\ & \quad + \int \int_{t-2h(\varepsilon) < x < t+2h (\varepsilon),\, -T < t < T} (\overline{w^{\varepsilon}}(t,x) - v(t,x))\psi(t,x)\,dx\,dt \\ & \quad + \int \int_{\max\{2t,t+2h(\varepsilon)\} < x < 2t+2h(\varepsilon),\, 0 < t < T} (\overline{w^{\varepsilon}}(t,x) - v(t,x))\psi(t,x)\,dx\,dt. \end{split} \label{eqn:integral2}$$ The area of $\{(t,x) \in \mathbb{R}^2 \mid -2h(\varepsilon) < x < \min\{0,t-2h(\varepsilon)\}\} \cap {\rm supp}\,\psi$ converges to $0$ as $\varepsilon \downarrow 0$. Moreover, $(\overline{w^{\varepsilon}} - v)_{\varepsilon \in (0,1]}$ is uniformly bounded on this intersection. Hence, the first integral on the right-hand side of \eqref{eqn:integral2} converges to $0$ as $\varepsilon \downarrow 0$. We can similarly show that the second and third integrals converge to $0$ as $\varepsilon \downarrow 0$. Thus, $\overline{w^{\varepsilon}}$ converges to $v$ in $\mathscr{D}'((-T,T) \times \mathbb{R})$ as $\varepsilon \downarrow 0$. {\bf Step 3}. Finally, we construct $(v_0^{\varepsilon})_{\varepsilon \in (0,1]} \in {\mathscr{E}}_{M}(\mathbb{R})$ such that $(v_0^{\varepsilon})'$ converges to $0$ in $\mathscr{D}'(\mathbb{R})$ as $\varepsilon \downarrow 0$, and further that the solution $v^{\varepsilon}$ of the problem $$\label{eqn:equation for v^{varepsilon}0} \begin{gathered} v^{\varepsilon}_t + b^{\varepsilon}v^{\varepsilon}_{x} = 0, \quad (t,x) \in \mathbb{R}^2, \\ v^{\varepsilon}|_{t = 0} = v_0^{\varepsilon}, \quad x \in \mathbb{R} \end{gathered}$$ converges to $v$ in $\mathscr{D}'((-T,T)\times\mathbb{R})$ as $\varepsilon \downarrow 0$. The existence of such $(v_0^{\varepsilon})_{\varepsilon \in (0,1]}$ implies the existence of initial data $V_0 \in \mathscr{G}(\mathbb{R})$ satisfying the desired properties that $V_0' \approx 0$ on $\mathbb{R}$ and $V \approx v$ on $(-T,T) \times \mathbb{R}$. Let $\chi \in \mathscr{D}(\mathbb{R})$ be as in Remark \ref{rem:strength}. Define the function $v_0^{\varepsilon}(x):= (\overline{w_0^{\varepsilon}} \ast \chi_{\lambda(\varepsilon)})(x)$. We have $\sup_{x \in \mathbb{R}}|\overline{w_0^{\varepsilon}}(x)| = |w_0^{\varepsilon}(\lambda(\varepsilon))| = cT - 2ch(\varepsilon)$. Furthermore, by inequality \eqref{eqn:inequality for (G^{varepsilon})^{-1}}, there exist two constants $C_1$, $C_2 > 0$ such that $2h(\varepsilon)\varepsilon^{T/C_1} \le \lambda(\varepsilon) \le 2h(\varepsilon)\varepsilon^{T/C_2}$. Hence, we see that the family of $v_0^{\varepsilon}$ defines an element of ${\mathscr{E}}_{M}(\mathbb{R})$, and further that $(v_0^{\varepsilon})'$ converges to $0$ in $\mathscr{D}'(\mathbb{R})$ as $\varepsilon \downarrow 0$. To show that the solution $v^{\varepsilon}$ of problem \eqref{eqn:equation for v^{varepsilon}0} converges to $v$ in $\mathscr{D}'((-T,T)\times\mathbb{R})$ as $\varepsilon \downarrow 0$, it suffices to prove that $v_0^{\varepsilon} - \overline{w_0^{\varepsilon}}$ converges uniformly to $0$ on any compact subset of $\mathbb{R}$ as $\varepsilon \downarrow 0$. The difference $v_0^{\varepsilon}(x) - \overline{w_0^{\varepsilon}}(x)$ satisfies the inequality $\big|v_0^{\varepsilon}(x) - \overline{w_0^{\varepsilon}}(x)\big| \le \int_{-\infty}^{\infty}|\overline{w_0^{\varepsilon}} (x-\lambda(\varepsilon) y) - \overline{w_0^{\varepsilon}}(x)|\chi(y)\,dy.$ Moreover, $|\overline{w_0^{\varepsilon}}(x-\lambda(\varepsilon) y) - \overline{w_0^{\varepsilon}}(x)| \le \sup_{-\eta(\varepsilon) \le \xi \le -\lambda(\varepsilon)}|(\overline{w_0^{\varepsilon}})'(\xi)| \lambda(\varepsilon) |y| = \frac{c}{1-\widetilde{b}^{\varepsilon}(-\lambda(\varepsilon))} \lambda(\varepsilon)|y|.$ As in the proof of Lemma \ref{lemma:1}, we have $1/(1-\widetilde{b}^{\varepsilon}(-\lambda(\varepsilon))) \le C_2h(\varepsilon)/\lambda(\varepsilon)$ for some constant $C_2 > 0$. Thus, we get $\left|v_0^{\varepsilon}(x) - \overline{w_0^{\varepsilon}}(x)\right| \le cC_2\int_{-\infty}^{\infty} |y|\chi(y)\,dy \cdot h(\varepsilon) \to 0 \quad \mbox {as}\ \varepsilon \downarrow 0.$ The proof of Theorem \ref{thm:1} is now complete. \end{proof} \begin{remark} \rm In Theorem \ref{thm:1}, for $t < 0$, the solution $U \in \mathscr{G}(\mathbb{R}^2)$ admits $0$ as distributional shadow, which is the unique distributional solution of problem \eqref{eqn:hyperbolic1} for negative time with $0$ initial data, see Hurd and Sattinger \cite{hurd}. \end{remark} \begin{remark} \rm Theorem \ref{thm:1} means that, in the setting of Colombeau's theory, all distributional solutions $u_c$ with initial data $0$ can be regarded as generalized solutions with different initial data. \end{remark} \begin{remark} \rm A similar result to Theorem \ref{thm:1} does not necessarily hold for other differential equations having non-unique distributional solutions. In fact, there exists an ordinary differential equation having a classical solution with which none of its generalized solutions is associated. For details, see \cite{deguchi1}. \end{remark} \section{Propagation of singularities}\label{sec:5} In this section we study the propagation of singularities for problem \eqref{eqn:generalized hyperbolic1}. The coefficient $B$ in problem \eqref{eqn:generalized hyperbolic1} is $\mathscr{G}^{\infty}$-regular, since $B$ is an element of $\mathscr{G}_{\rm log}(\mathbb{R}^2) \subset \mathscr{G}^{\infty}(\mathbb{R}^2)$. Hence, the subalgebra $\mathscr{G}_{\rm log}^{\infty} \subset \mathscr{G}_{\rm log}$ is suitable to study the propagation of singularities for problem \eqref{eqn:generalized hyperbolic1}. However, we are also interested in the propagation of singularities in $\mathscr{G}^{\infty}$, since $U_0 \in \mathscr{G}^{\infty}(\mathbb{R})$ does not necessarily imply that $U \in \mathscr{G}^{\infty}(\mathbb{R}^2)$. Thus, we discuss the propagation of singularities in both $\mathscr{G}_{\rm log}^{\infty}$ and $\mathscr{G}^{\infty}$ for problem \eqref{eqn:generalized hyperbolic1}. Let $\chi \in \mathscr{D}(\mathbb{R})$ be as in Remark \ref{rem:strength}. Assume that $U_0 \in \mathscr{G}(\mathbb{R})$ is given by the class of $(\chi_{h(\varepsilon)})_{\varepsilon \in (0,1]}$. Then $U_0$ is a Dirac generalized function at $0$ and belongs to $\mathscr{G}^{\infty}(\mathbb{R}) \setminus \mathscr{G}_{\rm log}^{\infty}(\mathbb{R})$. As may be seen in the following theorem, the singularity in $\mathscr{G}_{\rm log}^{\infty}$ of the initial data $U_0$ splits in two directions at the origin due to the discontinuity of the coefficient. \begin{figure}[ht] \begin{center} \setlength{\unitlength}{1mm} \begin{picture}(70,60)(0,0) \put(0,30){\line(1,0){60}} \put(59.8,29.2){$\rightarrow$} \put(28.1,26){$0$} \put(61,26){$x$} \put(30,0){\line(0,1){60}} \put(29.2,60){$\uparrow$} \put(26,60){$t$} \put(15,40){$\delta(x)/2$} \put(24,45){$\nearrow$} \put(10,50){$0$} \put(42,50){$0$} \put(10,17){$0$} \put(42,17){$0$} \put(0,0){\line(1,1){30}} \put(13,5){$\delta(x-t)$} \put(14,9){$\nwarrow$} \dottedline{1}(30,30)(60,60) \put(60,55){$t=x$} \put(30,30){\line(2,1){30}} \put(60,41){$t=x/2$} \put(50,33){$\delta(x-2t)/2$} \put(52,37){$\nwarrow$} \end{picture} \end{center} \caption{Distributional shadow}\label{fig1} \end{figure} \begin{theorem}\label{thm:2} Let $U_0 \in \mathscr{G}(\mathbb{R})$ be as above. Then the solution $U \in \mathscr{G}(\mathbb{R}^2)$ of problem \eqref{eqn:generalized hyperbolic1} admits a distributional shadow, which is given by $u(t,x) = \begin{cases} \frac{\delta(x) + \delta(x-2t)}{2}, & \text{if }t \ge 0,\; x \in \mathbb{R},\\ \delta(x-t), & \text{if }t < 0,\; x \in \mathbb{R}. \end{cases}$ Furthermore, $\operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U = \{(t,0) \mid t \ge 0\} \cup \{(t,2t) \mid t \ge 0\} \cup \{(t,t) \mid t \le 0\} \ (= \operatorname{sing\,supp}\, u).$ \end{theorem} \begin{proof} Let $v_0^{\varepsilon} = H \ast \chi_{h(\varepsilon)}$ and let $V_0 \in \mathscr{G}(\mathbb{R})$ be given by the class of $(v_0^{\varepsilon})_{\varepsilon \in (0,1]}$. In order to prove the first assertion, it suffices to show that the solution $V \in \mathscr{G}(\mathbb{R}^2)$ of problem \eqref{eqn:generalized hyperbolic2} admits a distributional shadow, which is given by $v(t,x) = \begin{cases} \frac{H(x) + H(x-2t)}{2}, & \text{if }t \ge 0,\; x \in \mathbb{R}, \\ H(x-t), & \text{if }t < 0,\; x \in \mathbb{R}. \end{cases}$ Let $(v^{\varepsilon})_{\varepsilon \in (0,1]}$ be a representative of $V \in \mathscr{G}(\mathbb{R}^2)$ satisfying $$\label{eqn:equation for v^{varepsilon}} \begin{gathered} v^{\varepsilon}_t + b^{\varepsilon}v^{\varepsilon}_{x} = 0, \quad (t,x) \in \mathbb{R}^2, \\ v^{\varepsilon}|_{t = 0} = v_0^{\varepsilon}, \quad x \in \mathbb{R}. \end{gathered}$$ Consider the characteristic curve $\gamma^{\varepsilon}(0,x,t)$ passing through $(0,x)$ at time $t = 0$. Along the characteristic curves, we have $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,x,t)) = v_0^{\varepsilon}(x)$. We can easily check that $v^{\varepsilon}$ converges to $H(x-t)$ a.e. in $(-\infty,0) \times \mathbb{R}$ as $\varepsilon \downarrow 0$. We now fix $0 < a \le 1$ arbitrarily, and put $t_1^{\varepsilon}:= \gamma^{\varepsilon}(0,-ah(\varepsilon),t_1^{\varepsilon}) + 2h(\varepsilon)$. As in Step $1$ of the proof of Theorem \ref{thm:1}, we have $t_1^{\varepsilon} = h(\varepsilon)\int_{a}^{2} \frac{dz}{1 - \widetilde{b}^{\varepsilon}(-h(\varepsilon)z)} = h(\varepsilon)\int_{a}^{2} \frac{dz}{1 - 2\int\int_{s > y + z}\varphi(s,y)\,dy\,ds} \to 0 \quad \text{as}\ \varepsilon \downarrow 0.$ Note that, for any $t \ge t_1^{\varepsilon}$, $\gamma^{\varepsilon}(0,-ah(\varepsilon),t) = \gamma^{\varepsilon}(0,-ah(\varepsilon),t_1^{\varepsilon})$. Hence, for any $t \ge 0$, we have $\gamma^{\varepsilon}(0,-ah(\varepsilon),t) \to 0$ as $\varepsilon \downarrow 0$. Moreover, $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-ah(\varepsilon),t)) = v_0^{\varepsilon}(-ah(\varepsilon)) = \int_{-\infty}^{-a}\chi(y)\,dy,$ so that $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-h(\varepsilon),t)) = 0$ and $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-ah(\varepsilon),t)) \uparrow 1/2$ as $a \downarrow 0$. Similarly, we take $t_2^{\varepsilon}:= \gamma^{\varepsilon}(0,ah(\varepsilon), t_2^{\varepsilon}) - 2h(\varepsilon)$ to get $t_2^{\varepsilon} = h(\varepsilon)\int_{a}^{2} \frac{dz}{\widetilde{b}^{\varepsilon}(h(\varepsilon)z) - 1} = h(\varepsilon)\int_{a}^{2} \frac{dz}{2\int\int_{s > y - z}\varphi(s,y)\,dy\,ds - 1} \to 0 \quad \text{as}\ \varepsilon \downarrow 0.$ Note that, for any $t \ge t_2^{\varepsilon}$, $\gamma^{\varepsilon}(0,ah(\varepsilon),t) = 2t - \gamma^{\varepsilon}(0,ah(\varepsilon),t_2^{\varepsilon}) +4h(\varepsilon)$. Therefore, for any $t \ge 0$, $\gamma^{\varepsilon}(0,ah(\varepsilon),t) \to 2t$ as $\varepsilon \downarrow 0$. Moreover, \begin{align*} v^{\varepsilon}(t,\gamma^{\varepsilon}(0,ah(\varepsilon),t)) = v_0^{\varepsilon}(ah(\varepsilon)) = \int_{-\infty}^{a}\chi(y)\,dy, \end{align*} so that $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,h(\varepsilon),t)) = 1$ and $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,ah(\varepsilon),t)) \downarrow 1/2$ as $a \downarrow 0$. Hence, taking into account the fact that $v^{\varepsilon}(t,x)$ is non-decreasing in $x$, we obtain that $v^{\varepsilon}$ converges to $(H(x) + H(x-2t))/2$ a.e. in $(0,\infty) \times \mathbb{R}$ as $\varepsilon \downarrow 0$. Thus, the first assertion follows. Next, we prove the second assertion. The proof is divided into four steps. {\bf Step 1}. First, we prove that $U \in \mathscr{G}(\mathbb{R}^2)$ is $\mathscr{G}_{\rm log}^{\infty}$-regular on $\{(t,x) \in \mathbb{R}^2 \mid x < \min\{0,t\}\} \cup \{(t,x) \in \mathbb{R}^2 \mid x > \max\{t,2t\}\}$. It is easy to check that $V \in \mathscr{G}(\mathbb{R}^2)$ equals $0$ on $\{(t,x) \in \mathbb{R}^2 \mid x < \min\{0,t\}\}$ and further equals $1$ on $\{(t,x) \in \mathbb{R}^2 \mid x > \max\{t,2t\}\}$. Hence, $U = V_x \in \mathscr{G}(\mathbb{R}^2)$ is $\mathscr{G}_{\rm log}^{\infty}$-regular on the union of these two sets. {\bf Step 2}. Secondly, we prove that $\{(t,t) \mid t \le 0\}$ is contained in $\operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U$. Fix $t < 0$ arbitrarily. We see that $v^{\varepsilon}(t,x) = v_0^{\varepsilon}(\gamma^{\varepsilon}(t,x,0))$. Since $v_0^{\varepsilon} = H \ast \chi_{h(\varepsilon)}$, we get $$\label{eqn:step 2-1} v_x^{\varepsilon}(t,x) = (v_0^{\varepsilon})'(\gamma^{\varepsilon}(t,x,0)) \gamma^{\varepsilon}_x(t,x,0) = \frac{1}{h(\varepsilon)} \chi\left(\frac{\gamma^{\varepsilon}(t,x,0)}{h(\varepsilon)}\right) \gamma^{\varepsilon}_x(t,x,0).$$ From problem \eqref{eqn:characteristics} and the definition of $\widetilde{b}^{\varepsilon}$, we find that $\gamma^{\varepsilon}(t,x,\tau)$ satisfies the problem $$\label{eqn:characteristics'} \begin{gathered} \gamma_{\tau}^{\varepsilon}(t,x,\tau) = \widetilde{b}^{\varepsilon}(\gamma^{\varepsilon}(t,x,\tau)-\tau), \\ \gamma^{\varepsilon}|_{\tau = t} = x. \end{gathered}$$ We differentiate these equations in $x$ to get \begin{gather*} \gamma_{\tau x}^{\varepsilon}(t,x,\tau) = (\widetilde{b}^{\varepsilon})'(\gamma^{\varepsilon}(t,x,\tau)-\tau) \gamma_{x}^{\varepsilon}(t,x,\tau), \\ \gamma_x^{\varepsilon}|_{\tau = t} = 1. \end{gather*} We divide the first equation by $\gamma_x^{\varepsilon}(t,x,\tau)$ and integrate it over $[t,0]$ to see that $\int_t^0 \frac{\gamma_{\tau x}^{\varepsilon}(t,x,\tau)}{\gamma_{x}^{\varepsilon}(t,x,\tau)}\,d\tau = \int_t^0 (\widetilde{b}^{\varepsilon})'(\gamma^{\varepsilon}(t,x,\tau)-\tau)\,d\tau.$ A simple calculation shows that $\gamma_x^{\varepsilon}(t,x,0) = \exp\Big(\int_t^0 (\widetilde{b}^{\varepsilon})' (\gamma^{\varepsilon}(t,x,\tau)-\tau)\,d\tau\Big).$ Since $\gamma^{\varepsilon}(t,t,\tau) = \tau$, we see that $\gamma_x^{\varepsilon}(t,t,0) = \exp\Big(\int_t^0 (\widetilde{b}^{\varepsilon})'(0)\,d\tau\Big) = \exp\Big(-(\widetilde{b}^{\varepsilon})'(0)t\Big).$ By the definition of $\widetilde{b}^{\varepsilon}$, we have $(\widetilde{b}^{\varepsilon})'(0) = 2\int_{-1}^{1} \varphi(s,s)\,ds/h(\varepsilon)$. Hence, noting that $h(\varepsilon) = 1/\log (1/\varepsilon)$, we see that $$\label{eqn:step 2-2} \gamma_x^{\varepsilon}(t,t,0) = \exp\Big(\frac{2}{h(\varepsilon)}\int_{-1}^{1} \varphi(s,s)\,ds \cdot (-t)\Big) = \big(\frac{1}{\varepsilon}\big)^{2\int_{-1}^{1} \varphi(s,s)\,ds \cdot (-t)}.$$ Combining equations \eqref{eqn:step 2-1} and \eqref{eqn:step 2-2}, we obtain that, for $\varepsilon > 0$ small enough, $v_x^{\varepsilon}(t,t) = \frac{1}{h(\varepsilon)} \chi(0) \big(\frac{1}{\varepsilon}\big)^{2\int_{-1}^{1} \varphi(s,s)\,ds \cdot (-t)} \ge \chi(0) \big(\frac{1}{\varepsilon}\big)^{2\int_{-1}^{1} \varphi(s,s)\,ds \cdot (-t)}.$ Since $U = V_x \in \mathscr{G}(\mathbb{R}^2)$, this shows that $\{(t,t) \mid t \le 0\} \subset \operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U$. {\bf Step 3}. Thirdly, we prove that $\{(t,0) \mid t \ge 0\}$ and $\{(t,2t) \mid t \ge 0\}$ are contained in $\operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U$. Put $t_1^{\varepsilon} = \gamma^{\varepsilon} (0,-ah(\varepsilon),t_1^{\varepsilon}) + 2h(\varepsilon)$. Then as shown above, $t_1^{\varepsilon} \downarrow 0$ as $\varepsilon \downarrow 0$. For $t \ge t_1^{\varepsilon}$, consider $\frac{v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-ah(\varepsilon),t)) - v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-2h(\varepsilon),t))}{\gamma^{\varepsilon}(0,-ah(\varepsilon),t) - \gamma^{\varepsilon}(0,-2h(\varepsilon),t)},$ where $0 < a < 1$ is a constant such that $\int_{-\infty}^{-a} \chi(y)\,dy > 0$. As shown above, we have $\gamma^{\varepsilon}(0,-ah(\varepsilon),t) = \gamma^{\varepsilon}(0,-ah(\varepsilon),t_1^{\varepsilon}) = h(\varepsilon)\int_{a}^{2} \frac{dz}{1 - \widetilde{b}^{\varepsilon} (-h(\varepsilon)z)} - 2h(\varepsilon).$ Since $\gamma^{\varepsilon}(0,-2h(\varepsilon),t) = -2h(\varepsilon)$, we get $0 < \gamma^{\varepsilon}(0,-ah(\varepsilon),t) - \gamma^{\varepsilon}(0,-2h(\varepsilon),t) = h(\varepsilon)\int_{a}^{2} \frac{dz}{1 - \widetilde{b}^{\varepsilon}(-h(\varepsilon)z)}.$ Furthermore, \begin{gather*} v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-ah(\varepsilon),t)) = \int_{-\infty}^{-a} \chi(y)\,dy > 0,\\ v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-2h(\varepsilon),t)) = 0. \end{gather*} Therefore, $\frac{v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-ah(\varepsilon),t)) - v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-2h(\varepsilon),t))} {\gamma^{\varepsilon}(0,-ah(\varepsilon),t) - \gamma^{\varepsilon}(0,-2h(\varepsilon),t)} = \frac{\int_{-\infty}^{-a} \chi(y)\,dy}{\int_{a}^{2} \frac{dz}{1 - \widetilde{b}^{\varepsilon}(-h(\varepsilon)z)}} \cdot \frac{1}{h(\varepsilon)}.$ By the mean value theorem, there exists $x_1^{\varepsilon} \in (\gamma^{\varepsilon}(0,-2h(\varepsilon),t), \gamma^{\varepsilon}(0,-ah(\varepsilon),t))$ such that $v^{\varepsilon}_x(t,x_1^{\varepsilon}) = \frac{\int_{-\infty}^{-a} \chi(y)\,dy}{\int_{a}^{2} \frac{dz}{1 - \widetilde{b}^{\varepsilon}(-h(\varepsilon)z)}} \cdot \frac{1}{h(\varepsilon)}.$ Note that $\partial_x^{\alpha} v^{\varepsilon}(t,\gamma^{\varepsilon} (0,-2h(\varepsilon),t)) = 0$ for $\alpha \in \mathbb{N}$. Then, repeating the above process gives us $(x^{\varepsilon}_{\alpha})_{\alpha \ge 2}$ such that $x^{\varepsilon}_{\alpha} \in (\gamma^{\varepsilon}(0,-2h (\varepsilon),t), x^{\varepsilon}_{\alpha-1})$ and \begin{align*} \partial^{\alpha}_x v^{\varepsilon}(t,x_{\alpha}^{\varepsilon}) & = \frac{\partial^{\alpha-1}_x v^{\varepsilon} (t,x^{\varepsilon}_{\alpha-1}) - \partial^{\alpha-1}_x v^{\varepsilon}(t,\gamma^{\varepsilon}(0, -2h(\varepsilon),t))}{x_{\alpha - 1}^{\varepsilon} - \gamma^{\varepsilon}(0,-2h(\varepsilon),t)} \\ & \ge \frac{\int_{-\infty}^{-a} \chi(y)\,dy} {\big(\int_{a}^{2} \frac{dz}{1 - \widetilde{b}^{\varepsilon} (-h(\varepsilon)z)}\big)^{\alpha}} \cdot \frac{1}{h(\varepsilon)^{\alpha}}. \end{align*} Since $U = V_x \in \mathscr{G}(\mathbb{R}^2)$, this shows that $\{(t,0) \mid t \ge 0\} \subset \operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U$. In a similar way, we can show that $\{(t,2t) \mid t \ge 0\} \subset \operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U$. {\bf Step 4}. Fourthly, we prove that $U \in \mathscr{G}(\mathbb{R}^2)$ is $\mathscr{G}_{\rm log}^{\infty}$-regular on $\{(t,x) \in \mathbb{R}^2 \mid 0 < x < 2t\}$. {\bf Step 4-1}. To do so, we first estimate $\gamma^{\varepsilon}(t,x,0)$ for all $(t,x)$ such that $0 \le x \le 2t$. When $0 \le x \le t-2h(\varepsilon)$, as seen in Step 1 of the proof of Theorem \ref{thm:1}, $\gamma^{\varepsilon}(t,x,0) = (G^{\varepsilon})^{-1}(x/h(\varepsilon) + 2)$. By inequality \eqref{eqn:inequality for (G^{varepsilon})^{-1}}, there exists a constant $C_2 > 0$ such that $$\label{eqn:step 4-1} 0 < -\gamma^{\varepsilon}(t,x,0) \le 2\exp\big(-\frac{2}{C_2}\big)h(\varepsilon) \varepsilon^{x/C_2}.$$ When $t + 2h(\varepsilon) \le x \le 2t$, we have $\gamma^{\varepsilon}(t,x,0) = -(G^{\varepsilon})^{-1}((2t-x)/h(\varepsilon) + 2)$. Hence, by \eqref{eqn:inequality for (G^{varepsilon})^{-1}}, we have $0 < \gamma^{\varepsilon}(t,x,0) \le 2\exp\big(-\frac{2}{C_2}\big)h(\varepsilon) \varepsilon^{(2t-x)/C_2}.$ When $t-2h(\varepsilon) \le x < t$, we get $\int_{-\gamma^{\varepsilon}(t,x,0)/h(\varepsilon)}^{(t-x) /h(\varepsilon)} \frac{dz}{1-\widetilde{b}^{\varepsilon} (-h(\varepsilon)z)} = \frac{t}{h(\varepsilon)}.$ As seen from the proof of Lemma \ref{lemma:1}, we have $1/(1-\widetilde{b}^{\varepsilon}(-h(\varepsilon)z)) \le C_2/z$ for $0 \le z \le 2$ and so $\frac{t}{h(\varepsilon)} \le \int_{-\gamma^{\varepsilon}(t,x,0)/h(\varepsilon)}^{(t-x) /h(\varepsilon)} \frac{C_2}{z}\,dz = C_2 \log \frac{t-x}{-\gamma^{\varepsilon}(t,x,0)}.$ Since $h(\varepsilon)=1/\log(1/\varepsilon)$, it follows that $$\label{eqn:step 4-3} 0 < -\gamma^{\varepsilon}(t,x,0) \le (t-x)\varepsilon^{t/C_2}.$$ When $t < x \le t + 2h(\varepsilon)$, we get $\int_{\gamma^{\varepsilon}(t,x,0)/h(\varepsilon)}^{(x-t)/h(\varepsilon)} \frac{dz}{1-\widetilde{b}^{\varepsilon}(-h(\varepsilon)z)} = \frac{t}{h(\varepsilon)},$ and so $0 < \gamma^{\varepsilon}(t,x,0) \le (x-t)\varepsilon^{t/C_2}.$ When $t = x$, we have $\gamma^{\varepsilon}(t,t,0) = 0$. {\bf Step 4-2}. We next estimate $\gamma^{\varepsilon}_x(t,x,0)$. When $0 \le x \le t-2h(\varepsilon)$, we have $\gamma^{\varepsilon}(t,x,0) = (G^{\varepsilon})^{-1}(x/h(\varepsilon) + 2)$. Hence, as in the proof of Lemma \ref{lemma:1}, we get, for some constant $c_2 > 0$, $\gamma^{\varepsilon}_x(t,x,0) = 1 - \widetilde{b}^{\varepsilon}(\gamma^{\varepsilon}(t,x,0)) \le c_2 \frac{|\gamma^{\varepsilon}(t,x,0)|}{h(\varepsilon)} \le 2c_2\exp\big(-\frac{2}{C_2}\big) \varepsilon^{x/C_2},$ where we used formula \eqref{eqn:differentiation of (G^{varepsilon})^{-1}} in the first step and inequality \eqref{eqn:step 4-1} in the last step. When $t + 2h(\varepsilon) \le x \le 2t$, $\gamma^{\varepsilon}(t,x,0) = -(G^{\varepsilon})^{-1}((2t-x)/h(\varepsilon) + 2)$. Similarly, we get $\gamma^{\varepsilon}_x(t,x,0) = 1 - \widetilde{b}^{\varepsilon}(-\gamma^{\varepsilon}(t,x,0)) \le 2c_2\exp\big(-\frac{2}{C_2}\big) \varepsilon^{(2t-x)/C_2}.$ When $t-2h(\varepsilon) \le x < t$, we have $\gamma^{\varepsilon}(t,x,0) = (G^{\varepsilon})^{-1}(t/h(\varepsilon) + G^{\varepsilon}(x-t))$. Differentiating this in $x$ gives $$\label{eqn:differentiation} \gamma_x^{\varepsilon}(t,x,0) = \frac{1 - \widetilde{b}^{\varepsilon} (\gamma^{\varepsilon}(t,x,0))}{1 - \widetilde{b}^{\varepsilon}(x-t)}.$$ The numerator of \eqref{eqn:differentiation} can be estimated as follows: $1 - \widetilde{b}^{\varepsilon}(\gamma^{\varepsilon}(t,x,0)) \le c_2 \frac{|\gamma^{\varepsilon}(t,x,0)|}{h(\varepsilon)} \le c_2 \frac{(t-x) \varepsilon^{t/C_2}}{h(\varepsilon)},$ where we used inequality \eqref{eqn:step 4-3} in the last step. Similarly, the denominator of \eqref{eqn:differentiation} is estimated as follows: for some constant $c_1 > 0$, we have $1 - \widetilde{b}^{\varepsilon}(x-t) \ge c_1(t-x)/h(\varepsilon)$. Hence, $0 < \gamma_x^{\varepsilon}(t,x,0) \le \frac{c_2}{c_1}\varepsilon^{t/C_2}.$ When $t < x \le t + 2h(\varepsilon)$, we have $\gamma^{\varepsilon}(t,x,0) = -(G^{\varepsilon})^{-1}(t/h(\varepsilon) + G^{\varepsilon}(t-x))$ and so get $0 < \gamma_x^{\varepsilon}(t,x,0) \le \frac{c_2}{c_1}\varepsilon^{t/C_2}.$ To estimate $\gamma_x^{\varepsilon}(t,t,0)$, we consider problem \eqref{eqn:characteristics'}. As in Step 2, we can derive that $$\label{eqn:x-derivative of gamma} \gamma_x^{\varepsilon}(t,x,s) = \exp\Big(-\int_s^t (\widetilde{b}^{\varepsilon})' (\gamma^{\varepsilon}(t,x,\tau)-\tau)\,d\tau\Big).$$ Note that $\gamma^{\varepsilon}(t,t,\tau) = \tau$ and $(\widetilde{b}^{\varepsilon})'(0) = 2\int_{-1}^{1} \varphi(s,s)\,ds/h(\varepsilon)$. Hence, $\gamma_x^{\varepsilon}(t,t,0) = \varepsilon^{(2\int_{-1}^{1} \varphi(s,s)\,ds)t}.$ {\bf Step 4-3}. Finally, we prove that, for all $K \Subset \{(t,x) \in \mathbb{R}^2 \mid 0 < x < 2t\}$ and $\alpha \in \mathbb{N}_0^2$, $$\label{eqn:estimate for v} \|\partial^{\alpha}v_x^{\varepsilon}(t,x)\|_{L^{\infty}(K)} \to 0 \quad \text{as }\varepsilon \downarrow 0.$$ This implies that $U = V_x \in \mathscr{G}(\mathbb{R}^2)$ is $\mathscr{G}_{\rm log}^{\infty}$-regular on $\{(t,x) \in \mathbb{R}^2 \mid 0 < x < 2t\}$. Note that $v_x^{\varepsilon}(t,x) = \chi\Big(\frac{\gamma^{\varepsilon}(t,x,0)}{h(\varepsilon)}\Big) \frac{\gamma_x^{\varepsilon}(t,x,0)}{h(\varepsilon)}.$ Hence, to prove \eqref{eqn:estimate for v}, it suffices to show that, for all $K \Subset \{(t,x) \in \mathbb{R}^2 \mid 0 < x < 2t\}$ and $\alpha \in \mathbb{N}^2$, $$\label{eqn:estimate for gamma} \frac{\|\partial^{\alpha}\gamma^{\varepsilon}(t,x,0) \|_{L^{\infty}(K)}}{h(\varepsilon)} \to 0 \quad \text{as } \varepsilon \downarrow 0.$$ Since $(\widetilde{b}^{\varepsilon})'(z) \ge 0$ for $z \in \mathbb{R}$, we see from \eqref{eqn:x-derivative of gamma} that, for $0 \le s \le t$, $$\label{eqn:x-derivative of gamma1} 0 < \gamma_x^{\varepsilon}(t,x,s) \le 1.$$ From \eqref{eqn:x-derivative of gamma} again, we find that \begin{align*} \gamma_{xx}^{\varepsilon}(t,x,s) & = -\gamma_x^{\varepsilon}(t,x,s) \int_s^t (\widetilde{b}^{\varepsilon})''(\gamma^{\varepsilon}(t,x,\tau)-\tau) \gamma_x^{\varepsilon}(t,x,\tau)\,d\tau. \end{align*} We see that $\|(\widetilde{b}^{\varepsilon})''\|_{L^{\infty}(\mathbb{R})} \le C''/h(\varepsilon)^2$ for some constant $C'' > 0$. In view of this inequality and \eqref{eqn:x-derivative of gamma1}, we get $|\gamma_{xx}^{\varepsilon}(t,x,s)| \le C''(t-s)/h(\varepsilon)^2$ for $0 \le s \le t$. Furthermore, if $s = 0$, then from Step 4-2, we see that $\frac{|\gamma_{xx}^{\varepsilon}(t,x,0)|}{h(\varepsilon)} \le C''\frac{|\gamma_x^{\varepsilon}(t,x,0)|t}{h(\varepsilon)^{3}} \to 0 \quad {\rm on}\ K \quad \text{as }\varepsilon \downarrow 0.$ By repeating this process, we obtain that a similar estimate holds for any derivative of $\gamma^{\varepsilon}(t,x,0)$ in $x$. We also find from problem \eqref{eqn:characteristics'} that $$\label{eqn:t-derivative of gamma} \gamma_t^{\varepsilon}(t,x,s) = -\widetilde{b}^{\varepsilon}(x-t) \exp\Big(-\int_s^t (\widetilde{b}^{\varepsilon})' (\gamma^{\varepsilon}(t,x,\tau)-\tau)\,d\tau\Big).$$ In view of \eqref{eqn:t-derivative of gamma}, we can similarly show inequality \eqref{eqn:estimate for gamma}. The proof of Theorem \ref{thm:2} is now complete. \end{proof} \begin{remark} \rm We assumed that $\chi$ is symmetric. Hence, $\int_{-\infty}^{0} \chi(y)\,dy = 1/2$. If, instead of the symmetry of $\chi$, we assume that $\int_{-\infty}^{0} \chi(y)\,dy = a$ for $0 \le a \le 1$, then the solution $U \in \mathscr{G}(\mathbb{R}^2)$ of problem \eqref{eqn:generalized hyperbolic1} with the initial data $U_0$ given by the class of $(\chi_{h(\varepsilon)})_{\varepsilon \in (0,1]}$ possesses the distributional shadow $u(t,x) = \begin{cases} a\delta(x) + (1-a)\delta(x-2t), & \text{if }t \ge 0,\; x \in \mathbb{R}, \\ \delta(x-t), & \text{if }t < 0,\; x \in \mathbb{R}. \end{cases}$ \end{remark} Next, we calculate the $\mathscr{G}^{\infty}$-singular support of the solution $U \in \mathscr{G}(\mathbb{R}^2)$ with the same initial data $U_0$ as in Theorem \ref{thm:2}. The following theorem shows that the splitting of the singularity at the origin does not occur in the sense of $\mathscr{G}^{\infty}$. \begin{theorem}\label{thm:2-1} Under the same assumption as in Theorem \ref{thm:2}, it holds that $\operatorname{sing\,supp}_{\mathscr{G}^{\infty}} U = \{(t,t) \mid t \le 0\}.$ \end{theorem} \begin{proof} The proof is divided into two steps. {\bf Step 1}. First, we prove that $U \in \mathscr{G}(\mathbb{R}^2)$ is $\mathscr{G}^{\infty}$-regular on $\mathbb{R}^2 \setminus \{(t,t) \mid t \le 0\}$. As can be seen in Step 1 of the proof of Theorem \ref{thm:2}, the solution $U \in \mathscr{G}(\mathbb{R}^2)$ is $\mathscr{G}^{\infty}$-regular on $\{(t,x) \in \mathbb{R}^2 \mid x < \min\{0,t\}\} \cup \{(t,x) \in \mathbb{R}^2 \mid x > \max\{t,2t\}\}$. Hence, it suffices to prove that $U \in \mathscr{G}(\mathbb{R}^2)$ is $\mathscr{G}^{\infty}$-regular on $(0,\infty) \times \mathbb{R}$. Let $(t,x) \in (0,\infty) \times \mathbb{R}$ and $0 \le s \le t$. As in Step 4-3 of the proof of Theorem \ref{thm:2}, we get $0 < \gamma_x^{\varepsilon}(t,x,s) \le 1$ and $|\gamma_{xx}^{\varepsilon}(t,x,s)| \le C''(t-s)/h(\varepsilon)^2$. Similarly, we can prove that all derivatives of $\gamma^{\varepsilon}(t,x,s)$ in $x$ are dominated by a finite sum of terms in the form of $\kappa_i(t-s)^j/h(\varepsilon)^k$ with a constant $\kappa_i > 0$. Note by \eqref{eqn:t-derivative of gamma} that $\gamma_t^{\varepsilon}(t,x,s) = -\widetilde{b}^{\varepsilon}(x-t) \gamma_x^{\varepsilon}(t,x,s)$. Hence, we see that, all derivatives of $\gamma^{\varepsilon}(t,x,s)$ in $t$ and $x$ are also dominated by a finite sum of terms in the form of $\kappa_i(t-s)^j/h(\varepsilon)^k$. Let us recall that the solution $v^{\varepsilon}$ of problem \eqref{eqn:equation for v^{varepsilon}} satisfies \eqref{eqn:step 2-1}. Then, we see that, for all $K \Subset (0,\infty) \times \mathbb{R}$ and $\alpha \in \mathbb{N}_0^2$, $\|\partial^{\alpha} v_x^{\varepsilon}(t,x)\|_{L^{\infty}(K)} = O(\varepsilon^{-1}) \quad \text{as }\varepsilon \downarrow 0.$ Since $U = V_x \in \mathscr{G}(\mathbb{R}^2)$, this shows that $U$ is $\mathscr{G}^{\infty}$-regular on $(0,\infty) \times \mathbb{R}$. {\bf Step 2}. Secondly, we prove that $\{(t,t) \mid t \le 0\}$ is contained in $\operatorname{sing\,supp}_{\mathscr{G}^{\infty}}U$. For $t < 0$, consider $\frac{v^{\varepsilon}(t,\gamma^{\varepsilon}(0,0,t)) - v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-2h(\varepsilon),t))}{\gamma^{\varepsilon}(0,0,t) - \gamma^{\varepsilon}(0,-2h(\varepsilon),t)}.$ Clearly, $\gamma^{\varepsilon}(0,0,t) = t$. As in Step 1 of the proof of Theorem \ref{thm:1}, we see that $t - \gamma^{\varepsilon}(0,-2h(\varepsilon),t) = -(G^{\varepsilon})^{-1}(-t/h(\varepsilon))$. Hence, by inequality \eqref{eqn:inequality for (G^{varepsilon})^{-1}}, we get, for some constant $C_2 > 0$, $0 < t - \gamma^{\varepsilon}(0,-2h(\varepsilon),t) \le 2h(\varepsilon)\varepsilon^{-t/C_2}.$ Furthermore, \begin{gather*} v^{\varepsilon}(t,\gamma^{\varepsilon}(0,0,t)) = \frac{1}{2}, \\ v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-2h(\varepsilon),t)) = 0. \end{gather*} Therefore, $\frac{v^{\varepsilon}(t,\gamma^{\varepsilon}(0,0,t)) - v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-2h (\varepsilon),t))}{\gamma^{\varepsilon}(0,0,t) - \gamma^{\varepsilon}(0,-2h(\varepsilon),t)} \ge \frac{1}{4h(\varepsilon)} \cdot \frac{1}{\varepsilon^{-t/C_2}}.$ By the mean value theorem, there exists $x_1^{\varepsilon} \in (\gamma^{\varepsilon}(0,-2h(\varepsilon),t),t)$ such that $v_x^{\varepsilon}(t,x_1^{\varepsilon}) \ge \frac{1}{4h(\varepsilon)} \cdot \frac{1}{\varepsilon^{-t/C_2}}.$ Note that $\partial_x^{\alpha} v^{\varepsilon} (t,\gamma^{\varepsilon}(0,-2h(\varepsilon),t)) = 0$ for $\alpha \in \mathbb{N}$. Then we repeat this process to find $(x_{\alpha}^{\varepsilon})_{\alpha \ge 2}$ such that $x_{\alpha}^{\varepsilon} \in (\gamma^{\varepsilon} (0,-2h(\varepsilon),t),x_{\alpha-1}^{\varepsilon})$ and $\partial_x^{\alpha} v^{\varepsilon}(t,x_\alpha^{\varepsilon}) = \frac{\partial_x^{\alpha-1}v^{\varepsilon}(t,x_{\alpha-1}^{\varepsilon}) - \partial_x^{\alpha-1} v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-2h(\varepsilon),t))}{x_{\alpha-1}^{\varepsilon} - \gamma^{\varepsilon}(0,-2h(\varepsilon),t)} \ge \frac{1}{2^{\alpha+1}h(\varepsilon)^{\alpha}} \cdot \frac{1}{\varepsilon^{-\alpha t/C_2}}.$ Since $U = V_x \in \mathscr{G}(\mathbb{R}^2)$, this shows that $\{(t,t) \mid t \le 0\} \subset \operatorname{sing\,supp}_{\mathscr{G}^{\infty}} U$. The proof of Theorem \ref{thm:2-1} is now complete. \end{proof} Next, we discuss the case of initial data $U_0$ given by other Dirac generalized functions at $0$. As stated in Remark \ref{rem:strength}, there exist infinitely many Dirac generalized functions with different strengths of singularity at $0$. For any constant $t_0 > 0$, we define $c(\varepsilon):= -(G^{\varepsilon})^{-1}(t_0/h(\varepsilon))/2h(\varepsilon)$. We find from inequality \eqref{eqn:inequality for (G^{varepsilon})^{-1}} that there exist two constants $C_1$, $C_2 > 0$ independent of $t_0$ such that $\varepsilon^{t_0/C_1} \le c(\varepsilon) \le \varepsilon^{t_0/C_2}$ for $\varepsilon \in (0,1]$. Hence, we have $(\chi_{c(\varepsilon)})_{\varepsilon \in (0,1]} \in {\mathscr{E}}_{M}(\mathbb{R})$, which allows us to define $U_0 \in \mathscr{G}(\mathbb{R})$ as the class of $(\chi_{c(\varepsilon)})_{\varepsilon \in (0,1]}$. Then $U_0$ is a Dirac generalized function at $0$ and does not belong to $\mathscr{G}^{\infty}(\mathbb{R})$. Furthermore, the singularity of $U_0$ at $0$ can be interpreted to become stronger as $t_0$ becomes large. As may be seen in the following theorem, the stronger the singularity of the initial data $U_0$ at $0$ becomes, the longer the singularity in $\mathscr{G}_{\rm log}^{\infty}$ propagates along the line $\{t = x\}$, and it splits at time $t_0$. \begin{figure}[ht] \begin{center} \setlength{\unitlength}{1mm} \begin{picture}(70,60)(0,0) \put(0,30){\line(1,0){60}} \put(59.8,29.2){$\rightarrow$} \put(28.1,26){$0$} \put(61,26){$x$} \put(30,0){\line(0,1){60}} \put(29.2,60){$\uparrow$} \put(26,60){$t$} %\put(15,40){$\delta(x)/2$} %\put(24,45){$\nearrow$} \put(10,50){$0$} \put(44,54){$0$} \put(10,17){$0$} \put(44,17){$0$} \put(0,0){\line(1,1){40}} \put(13,5){$\delta(x-t)$} \put(14,9){$\nwarrow$} \dottedline{1}(40,40)(63,63) \put(60,56){$t=x$} \put(40,40){\line(5,2){23}} \put(60,45){$t=x/2+t_0/2$} \put(51,37){$\delta(x-2t+t_0)/2$} \put(51,41){$\nwarrow$} \put(40,40){\line(0,1){20}} \put(18,47){$\delta(x-t_0)/2$} \put(35,51){$\nearrow$} \dottedline{1}(30,40)(40,40) \put(25,39){$t_0$} \end{picture} \end{center} \caption{Distributional shadow}\label{fig2} \end{figure} \begin{theorem}\label{thm:3} Let $t_0$ and $U_0 \in \mathscr{G}(\mathbb{R})$ be as above. Then the solution $U \in \mathscr{G}(\mathbb{R}^2)$ of problem \eqref{eqn:generalized hyperbolic1} admits a distributional shadow, which is given by $u(t,x) = \begin{cases} \frac{\delta(x-t_0) + \delta(x-2t+t_0)}{2}, & \text{if }t \ge t_0,\; x \in \mathbb{R}, \\ \delta(x-t), & \text{if }t < t_0,\; x \in \mathbb{R}. \end{cases}$ Furthermore, \begin{align*} & \{(t,t_0) \mid t \ge t_0\} \cup \{(t,2t-t_0) \mid t \ge t_0\} \cup \{(t,t) \mid t \le t_0\} \\ & \subset \operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U \subset \{(t,t_0) \mid t \ge t_0\} \cup \{(t,2t-t_0) \mid t \ge t_0\} \cup \{(t,t) \mid t \in \mathbb{R}\}. \end{align*} \end{theorem} \begin{proof} Let $v_0^{\varepsilon} = H \ast \chi_{c(\varepsilon)}$ and let $V_0 \in \mathscr{G}(\mathbb{R})$ be given by the class of $(v_0^{\varepsilon})_{\varepsilon \in (0,1]}$. In order to prove the first assertion, it suffices to show that the solution $V \in \mathscr{G}(\mathbb{R}^2)$ of problem \eqref{eqn:generalized hyperbolic2} admits a distributional shadow, which is given by $v(t,x) = \begin{cases} \frac{H(x-t_0) + H(x-2t+t_0)}{2}, & \text{if }t \ge t_0,\; x \in \mathbb{R}, \\ H(x-t), & \text{if }t < t_0, \; x \in \mathbb{R}. \end{cases}$ Let $(v^{\varepsilon})_{\varepsilon \in (0,1]}$ be a representative of $V \in \mathscr{G}(\mathbb{R}^2)$ satisfying problem \eqref{eqn:equation for v^{varepsilon}}. Then we have $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,x,t)) = v_0^{\varepsilon}(x)$ and see that $v^{\varepsilon}$ converges to $H(x-t)$ a.e. in $(-\infty,0) \times \mathbb{R}$ as $\varepsilon \downarrow 0$. We now fix $0 < a \le 1$ arbitrarily, and put $t_1^{\varepsilon}:= \gamma^{\varepsilon}(0,-ac(\varepsilon),t_1^{\varepsilon}) + 2h(\varepsilon)$. As in Step 1 of the proof of Theorem \ref{thm:1}, we have $t_1^{\varepsilon} = h(\varepsilon)\int_{ac(\varepsilon)/h(\varepsilon)}^{2} \frac{dz}{1 - \widetilde{b}^{\varepsilon}(-h(\varepsilon)z)}.$ By the definition of $c(\varepsilon)$, we get $t_1^{\varepsilon} = t_0 - h(\varepsilon) \int^{ac(\varepsilon)/h(\varepsilon)}_{2c(\varepsilon)} \frac{dz}{1 - \widetilde{b}^{\varepsilon}(-h(\varepsilon)z)}.$ As in the proof of Lemma \ref{lemma:1}, we get, for some constant $C_2 > 0$ and $\varepsilon > 0$ small enough, $h(\varepsilon)\int^{ac(\varepsilon) /h(\varepsilon)}_{2c(\varepsilon)} \frac{dz} {1 - \widetilde{b}^{\varepsilon}(-h(\varepsilon)z)} \le h(\varepsilon)\int^{ac(\varepsilon) /h(\varepsilon)}_{2c(\varepsilon)} \frac{C_2}{z}\,dz.$ We have $h(\varepsilon)\int^{ac(\varepsilon)/h(\varepsilon)}_{2c(\varepsilon)} \frac{C_2}{z}\,dz = -C_2h(\varepsilon) \log \frac{2h(\varepsilon)}{a} \to 0 \quad \text{as } \varepsilon \downarrow 0,$ so that $t_1^{\varepsilon} \to t_0$ as $\varepsilon \downarrow 0$. Note that for any $t \ge t_1^{\varepsilon}$, $\gamma^{\varepsilon}(0,-ac(\varepsilon),t) = \gamma^{\varepsilon}(0,-ac(\varepsilon),t_1^{\varepsilon})$. Therefore, for any $0 \le t \le t_0$, $\gamma^{\varepsilon}(0,-ac(\varepsilon),t) \to t$ as $\varepsilon \downarrow 0$, and for any $t \ge t_0$, $\gamma^{\varepsilon}(0,-ac(\varepsilon),t) \to t_0$ as $\varepsilon \downarrow 0$. Furthermore, $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-ac(\varepsilon),t)) = v_0^{\varepsilon}(-ac(\varepsilon)) = \int_{-\infty}^{-a}\chi(y)\,dy,$ and so $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-c(\varepsilon),t)) = 0$ and $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-ac(\varepsilon),t)) \uparrow 1/2$ as $a \downarrow 0$. Similarly, we take $t_2^{\varepsilon}:= \gamma^{\varepsilon} (0,ac(\varepsilon),t_2^{\varepsilon}) - 2h(\varepsilon)$ to get $t_2^{\varepsilon} = h(\varepsilon)\int_{ac(\varepsilon)/h(\varepsilon)}^{2} \frac{dz}{\widetilde{b}^{\varepsilon}(h(\varepsilon)z) - 1} \to t_0 \quad \text{as}\ \varepsilon \downarrow 0.$ Note that, for any $t \ge t_2^{\varepsilon}$, $\gamma^{\varepsilon}(0,ac(\varepsilon),t) = 2t - \gamma^{\varepsilon}(0,ac(\varepsilon),t_2^{\varepsilon}) + 4h(\varepsilon)$. Hence, for any $0 \le t \le t_0$, $\gamma^{\varepsilon}(0,ac(\varepsilon),t) \to t$ as $\varepsilon \downarrow 0$, and for any $t \ge t_0$, $\gamma^{\varepsilon}(0,ac(\varepsilon),t) \to 2t - t_0$ as $\varepsilon \downarrow 0$. Furthermore, $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,ac(\varepsilon),t)) = v_0^{\varepsilon}(ac(\varepsilon)) = \int_{-\infty}^{a}\chi(y)\,dy,$ and so $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,c(\varepsilon),t)) = 1$ and $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,ac(\varepsilon),t)) \downarrow 1/2$ as $a \downarrow 0$. Therefore, in view of the fact that $v^{\varepsilon}(t,x)$ is non-decreasing in $x$, we obtain that $v^{\varepsilon}$ converges to $H(x-t)$ a.e. in $(0,t_0) \times \mathbb{R}$ and to $(H(x-t_0) + H(x-2t+t_0))/2$ a.e. in $(t_0,\infty) \times \mathbb{R}$ as $\varepsilon \downarrow 0$. Thus, the first assertion follows. Next, we prove the second assertion. We will do so in four steps. {\bf Step 1}. First, we prove that $U \in \mathscr{G}(\mathbb{R}^2)$ is $\mathscr{G}_{\rm log}^{\infty}$-regular on $\{(t,x) \in \mathbb{R}^2 \mid x < \min\{t_0,t\}\} \cup \{(t,x) \in \mathbb{R}^2 \mid x > \max\{t,2t-t_0\}\}$. It is easy to check that $V \in \mathscr{G}(\mathbb{R}^2)$ equals $0$ on $\{(t,x) \in \mathbb{R}^2 \mid x < \min\{t_0,t\}\}$ and further equals $1$ on $\{(t,x) \in \mathbb{R}^2 \mid x > \max\{t,2t-t_0\}\}$. Hence, $U = V_x \in \mathscr{G}(\mathbb{R}^2)$ is $\mathscr{G}_{\rm log}^{\infty}$-regular on the union of these two sets. {\bf Step 2}. Secondly, we prove that $\{(t,t) \mid t \le t_0\}$ is contained in $\operatorname{sing\,supp}_{\mathscr{G} _{\rm log}^{\infty}} U$. Put $t_1^{\varepsilon} = \gamma^{\varepsilon}(0,-c(\varepsilon), t_1^{\varepsilon}) + 2h(\varepsilon)$. Then, as shown above, $t_1^{\varepsilon} \uparrow t_0$ as $\varepsilon \downarrow 0$. For $t < t_1^{\varepsilon}$, consider $\frac{v^{\varepsilon}(t,\gamma^{\varepsilon}(0,0,t)) - v^{\varepsilon}(t,\gamma^{\varepsilon}(0, -c(\varepsilon),t))}{\gamma^{\varepsilon}(0,0,t) -\gamma^{\varepsilon}(0,-c(\varepsilon),t)}.$ As in Step 1 of the proof of Theorem \ref{thm:1}, we get $G^{\varepsilon}(\gamma^{\varepsilon}(0,-c(\varepsilon),t)-t) = \frac{t_1^{\varepsilon}-t}{h(\varepsilon)}.$ Using inequality \eqref{eqn:inequality for (G^{varepsilon})^{-1}}, we have, for some constant $C_2 > 0$, $0 < t - \gamma^{\varepsilon}(0,-c(\varepsilon),t) \le 2h(\varepsilon)\varepsilon^{(t_1^{\varepsilon}-t)/C_2}.$ Since $\gamma^{\varepsilon}(0,0,t) = t$, it follows that $$\label{eqn:difference} 0 < \gamma^{\varepsilon}(0,0,t) - \gamma^{\varepsilon}(0,-c(\varepsilon),t) \le 2h(\varepsilon)\varepsilon^{(t_1^{\varepsilon}-t)/C_2}.$$ We use $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,0,t)) = 1/2$, $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-c(\varepsilon),t)) = 0$ and \eqref{eqn:difference} to see that $\frac{v^{\varepsilon}(t,\gamma^{\varepsilon}(0,0,t)) - v^{\varepsilon}(t,\gamma^{\varepsilon} (0,-c(\varepsilon),t))}{\gamma^{\varepsilon}(0,0,t) -\gamma^{\varepsilon}(0,-c(\varepsilon),t)} \ge \frac{1}{4h(\varepsilon)} \cdot \frac{1}{\varepsilon^{(t_1^{\varepsilon}-t)/C_2}}.$ By the mean value theorem, there exists $x^{\varepsilon} \in (\gamma^{\varepsilon}(0,-c(\varepsilon),t), \gamma^{\varepsilon}(0,0,t))$ such that $v_x^{\varepsilon}(t,x^{\varepsilon}) \ge \frac{1}{4h(\varepsilon)} \cdot \frac{1}{\varepsilon^{(t_1^{\varepsilon}-t)/C_2}}.$ Since $U = V_x \in \mathscr{G}(\mathbb{R}^2)$, this means that $\{(t,t) \mid t \le t_0\} \subset \operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U$. {\bf Step 3}. Thirdly, we prove that $\{(t,t_0) \mid t \ge t_0\}$ and $\{(t,2t-t_0) \mid t \ge t_0\}$ are contained in $\operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U$. For $t > t_0$, consider $\frac{v^{\varepsilon}(t,\gamma^{\varepsilon} (0,-ac(\varepsilon),t)) - v^{\varepsilon} (t,\gamma^{\varepsilon}(0,-c(\varepsilon),t))}{\gamma^{\varepsilon} (0,-ac(\varepsilon),t) - \gamma^{\varepsilon}(0,-c(\varepsilon),t)}$ for $0 < a \le 1$. As shown above, if we put $t_1^{\varepsilon} = \gamma^{\varepsilon}(0,-ac(\varepsilon),t_1^{\varepsilon}) + 2h(\varepsilon)$, then we have $t_1^{\varepsilon} \uparrow t_0$ as $\varepsilon \downarrow 0$ and $\gamma^{\varepsilon}(0,-ac(\varepsilon),t) = \gamma^{\varepsilon}(0,-ac(\varepsilon),t_1^{\varepsilon}) = h(\varepsilon)\int_{ac(\varepsilon)/h(\varepsilon)}^{2} \frac{dz}{1 - \widetilde{b}^{\varepsilon}(-h(\varepsilon)z)} - 2h(\varepsilon).$ Hence, $\gamma^{\varepsilon}(0,-ac(\varepsilon),t) - \gamma^{\varepsilon}(0,-c(\varepsilon),t) = h(\varepsilon)\int^{c(\varepsilon)/h(\varepsilon)}_{ac (\varepsilon)/h(\varepsilon)} \frac{dz}{1 - \widetilde{b}^{\varepsilon} (-h(\varepsilon)z)}.$ We now take $0 < a < 1$ so that $\int_{-\infty}^{-a} \chi(y)\,dy > 0$. Then, as in the proof of Lemma \ref{lemma:1}, we get, for some constant $C_2 > 0$, \begin{align*} 0 < \gamma^{\varepsilon}(0,-ac(\varepsilon),t) - \gamma^{\varepsilon}(0,-c(\varepsilon),t) \le h(\varepsilon)\int^{c(\varepsilon)/h(\varepsilon)}_{ac (\varepsilon)/h(\varepsilon)} \frac{C_2}{z}\,dz = C_2 h(\varepsilon) \log \frac{1}{a}. \end{align*} Furthermore, \begin{gather*} v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-ac(\varepsilon),t)) = \int_{-\infty}^{-a}\chi(y)\,dy > 0, \\ v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-c(\varepsilon),t)) = 0. \end{gather*} Hence, $\frac{v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-ac(\varepsilon),t)) - v^{\varepsilon}(t,\gamma^{\varepsilon} (0,-c(\varepsilon),t))}{\gamma^{\varepsilon}(0,-ac(\varepsilon),t) - \gamma^{\varepsilon}(0,-c(\varepsilon),t)} \ge \frac{\int_{-\infty}^{-a} \chi(y)\,dy}{C_2 \log 1/a} \cdot \frac{1}{h(\varepsilon)}.$ By the mean value theorem, there exists $x_1^{\varepsilon} \in (\gamma^{\varepsilon}(0,-c(\varepsilon),t), \gamma^{\varepsilon}(0,-ac(\varepsilon),t))$ such that $v^{\varepsilon}_x(t,x_1^{\varepsilon}) \ge \frac{\int_{-\infty}^{-a} \chi(y)\,dy}{C_2 \log 1/a} \cdot \frac{1}{h(\varepsilon)}.$ Note that $\partial_x^{\alpha} v^{\varepsilon}(t, \gamma^{\varepsilon}(0,-c(\varepsilon),t)) = 0$ for $\alpha \in \mathbb{N}$. Hence, we repeat this process to get $(x^{\varepsilon}_{\alpha})_{\alpha \ge 2}$ such that $x^{\varepsilon}_{\alpha} \in (\gamma^{\varepsilon}(0,-c(\varepsilon),t), x^{\varepsilon}_{\alpha-1})$ and $\partial^{\alpha}_x v^{\varepsilon}(t,x_{\alpha}^{\varepsilon}) = \frac{\partial_x^{\alpha-1} v^{\varepsilon}(t,x_{\alpha-1}^{\varepsilon}) - \partial_x^{\alpha-1} v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-c(\varepsilon),t))}{x_{\alpha-1}^{\varepsilon} - \gamma^{\varepsilon}(0,-c(\varepsilon),t)} \ge \frac{\int_{-\infty}^{-a} \chi(y)\,dy}{\left(C_2 \log 1/a\right)^{\alpha}} \cdot \frac{1}{h(\varepsilon)^{\alpha}}.$ Since $U = V_x \in \mathscr{G}(\mathbb{R}^2)$, this shows that $\{(t,t_0) \mid t \ge t_0\} \subset \operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U$. In a similar way, we can show that $\{(t,2t-t_0) \mid t \ge t_0\} \subset \operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U$. {\bf Step 4}. Fourthly, we prove that $U \in \mathscr{G}(\mathbb{R}^2)$ is $\mathscr{G}_{\rm log}^{\infty}$-regular on $\{(t,x) \in \mathbb{R}^2 \mid t_0 < x < t\}$ and $\{(t,x) \in \mathbb{R}^2 \mid t < x < 2t-t_0\}$. {\bf Step 4-1}. To do so, we first estimate $\gamma^{\varepsilon}(t,x,0)$ for all $(t,x)$ such that $t_0 \le x \le t-2h(\varepsilon)$ or $t+2h(\varepsilon) \le x \le 2t-t_0$. When $t_0 \le x \le t-2h(\varepsilon)$, as seen in Step 1 of the proof of Theorem \ref{thm:1}, $\gamma^{\varepsilon}(t,x,0) = (G^{\varepsilon})^{-1}(x/h(\varepsilon) + 2)$. Hence, $G^{\varepsilon}(\gamma^{\varepsilon}(t,x,0)) = x/h(\varepsilon) + 2$. By the definition of $c(\varepsilon)$, we have $G^{\varepsilon}(-2h(\varepsilon)c(\varepsilon)) = t_0/h(\varepsilon)$. Taking the difference gives $G^{\varepsilon}(\gamma^{\varepsilon}(t,x,0)) - G^{\varepsilon}(-2h(\varepsilon)c(\varepsilon)) = \frac{x - t_0}{h(\varepsilon)} + 2.$ By the definition of $G^{\varepsilon}$, we have $\int_{-\gamma^{\varepsilon}(t,x,0) /h(\varepsilon)}^{2c(\varepsilon)} \frac{dz}{1-\widetilde{b}^{\varepsilon}(-h(\varepsilon)z)} = \frac{x - t_0}{h(\varepsilon)} + 2.$ As seen from the proof of Lemma \ref{lemma:1}, we have $1/(1-\widetilde{b}^{\varepsilon}(-h(\varepsilon)z)) \le C_2/z$ for $0 \le z \le 2$ and so $\frac{x-t_0}{h(\varepsilon)} + 2 \le \int_{-\gamma^{\varepsilon}(t,x,0) /h(\varepsilon)}^{2c(\varepsilon)} \frac{C_2}{z}\,dz = C_2 \log \frac{2h(\varepsilon)c(\varepsilon)}{-\gamma^{\varepsilon} (t,x,0)}.$ Since $h(\varepsilon)=1/\log(1/\varepsilon)$, we have $\big(\frac{1}{\varepsilon}\big)^{x-t_0}e^2 \le \Big( \frac{2h(\varepsilon)c(\varepsilon)} {-\gamma^{\varepsilon}(t,x,0)}\Big)^{C_2}.$ Hence, $$\label{eqn:step 4-1'} 0 < -\gamma^{\varepsilon}(t,x,0) \le 2\exp\big(-\frac{2}{C_2}\big)h(\varepsilon)c(\varepsilon) \varepsilon^{(x-t_0)/C_2}.$$ When $t + 2h(\varepsilon) \le x \le 2t-t_0$, we have $\gamma^{\varepsilon}(t,x,0) = -(G^{\varepsilon})^{-1}((2t-x)/h(\varepsilon) + 2)$. A similar argument to the one above gives $0 < \gamma^{\varepsilon}(t,x,0) \le 2\exp\big(-\frac{2}{C_2}\big)h(\varepsilon)c(\varepsilon) \varepsilon^{(2t-x-t_0)/C_2}.$ {\bf Step 4-2}. We next estimate $\gamma^{\varepsilon}_x(t,x,0)$. When $t_0 \le x \le t-2h(\varepsilon)$, we have $\gamma^{\varepsilon}(t,x,0) = (G^{\varepsilon})^{-1}(x/h(\varepsilon) + 2)$. Hence, as in the proof of Lemma \ref{lemma:1}, we get, for some constant $c_2 > 0$, $\gamma^{\varepsilon}_x(t,x,0) = 1 - \widetilde{b}^{\varepsilon}(\gamma^{\varepsilon}(t,x,0)) \le c_2 \frac{|\gamma^{\varepsilon}(t,x,0)|}{h(\varepsilon)} \le 2c_2\exp\big(-\frac{2}{C_2}\big)c(\varepsilon) \varepsilon^{(x-t_0)/C_2},$ where we used formula \eqref{eqn:differentiation of (G^{varepsilon})^{-1}} in the first step and inequality \eqref{eqn:step 4-1'} in the last step. When $t + 2h(\varepsilon) \le x \le 2t-t_0$, $\gamma^{\varepsilon}(t,x,0) = -(G^{\varepsilon})^{-1}((2t-x)/h(\varepsilon) + 2)$. Similarly, we get $\gamma^{\varepsilon}_x(t,x,0) = 1 - \widetilde{b}^{\varepsilon}(-\gamma^{\varepsilon}(t,x,0)) \le 2c_2\exp\big(-\frac{2}{C_2}\big)c(\varepsilon) \varepsilon^{(2t-x-t_0)/C_2}.$ {\bf Step 4-3}. Finally, we prove that, for all $K \Subset \{(t,x) \in \mathbb{R}^2 \mid t_0 < x < t\} \cup \{(t,x) \in \mathbb{R}^2 \mid t < x < 2t-t_0\}$ and $\alpha \in \mathbb{N}_0^2$, $$\label{eqn:estimate for v'} \|\partial^{\alpha}v_x^{\varepsilon}(t,x)\|_{L^{\infty}(K)} \to 0 \quad \text{as }\varepsilon \downarrow 0.$$ This implies that $U = V_x \in \mathscr{G}(\mathbb{R}^2)$ is $\mathscr{G}_{\rm log}^{\infty}$-regular on $\{(t,x) \in \mathbb{R}^2 \mid t_0 < x < t\} \cup \{(t,x) \in \mathbb{R}^2 \mid t < x < 2t-t_0\}$. Note that $v_x^{\varepsilon}(t,x) = \chi\Big(\frac{\gamma^{\varepsilon}(t,x,0)}{c(\varepsilon)}\Big) \frac{\gamma_x^{\varepsilon}(t,x,0)}{c(\varepsilon)}.$ Hence, to prove \eqref{eqn:estimate for v'}, it suffices to show that, for all $K \Subset \{(t,x) \in \mathbb{R}^2 \mid t_0 < x < t\} \cup \{(t,x) \in \mathbb{R}^2 \mid t < x < 2t-t_0\}$ and $\alpha \in \mathbb{N}^2$, $\frac{\|\partial^{\alpha}\gamma^{\varepsilon} (t,x,0)\|_{L^{\infty}(K)}}{c(\varepsilon)} \to 0 \quad \text{as }\varepsilon \downarrow 0.$ This can be done similarly to Step 4-3 of the proof of Theorem \ref{thm:2}. The proof of Theorem \ref{thm:3} is now complete. \end{proof} \begin{remark} \rm It can be conjectured in Theorem \ref{thm:3} that \begin{align*} &\operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U \\ &= \{(t,t_0) \mid t \ge t_0\} \cup \{(t,2t-t_0) \mid t \ge t_0\} \cup \{(t,t) \mid t \le t_0\} \quad (= \operatorname{sing\,supp}\, u), \end{align*} but this is open. \end{remark} As in Step 2 of the proof of Theorem \ref{thm:2-1}, we can apply the mean value theorem repeatedly in Step 2 of the proof of Theorem \ref{thm:3} to show the following inclusion relation on the $\mathscr{G}^{\infty}$-singular support of the solution $U$. However, it is open whether equality holds. \begin{theorem}\label{thm:3-1} Under the same assumption as in Theorem \ref{thm:3}, it holds that $\operatorname{sing\,supp}_{\mathscr{G}^{\infty}} U \supset \{(t,t) \mid t \le t_0\}.$ \end{theorem} Finally, we discuss the case that $U_0 \in \mathscr{G}(\mathbb{R})$ is defined as the class of $(\kappa_1\chi_{a_1(\varepsilon)} (\cdot+s_1) + \kappa_2\chi_{a_2(\varepsilon)}(\cdot-s_2))_{\varepsilon \in (0,1]}$, where $\kappa_1$, $\kappa_2 \in \mathbb{R}$, $s_1$, $s_2 > 0$, $a_1(\varepsilon)$, $a_2(\varepsilon) \le h(\varepsilon)$. Then $U_0 \approx \kappa_1\delta_{-s_1} + \kappa_2\delta_{s_2}$, where $\delta_{-s_1}$ and $\delta_{s_2}$ are the delta functions at $-s_1$ and $s_2$, respectively. As may be seen in the following theorem, the $\mathscr{G}_{\rm log}^{\infty}$-singular support of the corresponding solution $U \in \mathscr{G}(\mathbb{R}^2)$ and the singular support of its distributional shadow do not necessarily coincide. \begin{figure}[ht] \begin{center} \setlength{\unitlength}{1mm} \begin{picture}(70,60)(0,0) \put(0,30){\line(1,0){60}} \put(59.8,29.2){$\rightarrow$} \put(28.1,26){$0$} \put(61,31.5){$x$} \put(30,0){\line(0,1){60}} \put(29.2,60){$\uparrow$} \put(26,60){$t$} \put(8,27){$-s_1$} \put(39,27){$s_2$} \put(8,50){$0$} \put(40,50){$0$} \put(8,18){$0$} \put(40,18){$0$} \put(0,0){\line(1,1){20}} \put(14,11){$\kappa_2\delta(x-t)$} \put(18.5,15){$\nwarrow$} \put(1,-2){$(\kappa_1+\kappa_2)\delta(x-t)$} \put(5.5,1.5){$\nwarrow$} \put(15,15){\line(0,1){45}} \put(-4,40){$\kappa_1\delta(x+s_1)$} \put(10.5,44){$\nearrow$} \dottedline{1}(20,20)(63,63) \put(60,55){$t=x$} \put(20,20){\line(2,1){43}} \put(60,37){$t=x/2-s_2/2$} \put(47,25){$\kappa_2\delta(x-2t-s_2)$} \put(49,29.2){$\nwarrow$} \end{picture} \end{center} \caption{Distributional shadow for the case $s_1 > s_2$} \label{fig3} \end{figure} \begin{theorem}\label{thm:4} Let $U_0 \in \mathscr{G}(\mathbb{R})$ be as above. Then the solution $U \in \mathscr{G}(\mathbb{R}^2)$ of problem \eqref{eqn:generalized hyperbolic1} admits a distributional shadow, which is given by $u(t,x) = \begin{cases} \kappa_1\delta(x+s_1) + \kappa_2\delta(x-2t-s_2), \\ \quad \text{if }t \ge \max\{-s_1,-s_2\},\\[3pt] \kappa_1\delta(x+s_1) + \kappa_2\delta(x-t), \\ \quad \text{if }\min\{-s_1,-s_2\} < t < \max\{-s_1,-s_2\} \text{ and }s_1 > s_2,\\[3pt] \kappa_1\delta(x-t) + \kappa_2\delta(x-2t-s_2), \\ \quad \text{if }\min\{-s_1,-s_2\} < t < \max\{-s_1,-s_2\} \text{ and }s_1 < s_2,\\[3pt] (\kappa_1+\kappa_2)\delta(x-t), \\ \quad \text{if }t \le \min\{-s_1,-s_2\}. \end{cases}$ Furthermore, \begin{aligned} &\operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U\\ & = \{(t,-s_1) \mid t \ge -s_1\} \cup \{(t,2t+s_2) \mid t \ge -s_2\} \cup \{(t,t) \mid t \le \max\{-s_1,-s_2\}\}. \end{aligned}\label{eqn:singsupp3} Thus, if $\kappa_1 = -\kappa_2 \ne 0$, then $\operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}}U \ne \operatorname{sing\,supp} u.$ \end{theorem} \begin{proof} The first assertion can be proved similarly to the proof of Theorem \ref{thm:2}. We will only prove the second assertion for the case $\kappa_1\kappa_2 \ne 0$. The case $\kappa_1\kappa_2 = 0$ can be argued similarly. Let $v_0^{\varepsilon} = H \ast (\kappa_1\chi_{a_1(\varepsilon)} (\cdot+s_1) + \kappa_2\chi_{a_2(\varepsilon)}(\cdot-s_2))$ and let $V_0 \in \mathscr{G}(\mathbb{R})$ be given by the class of $(v_0^{\varepsilon})_{\varepsilon \in (0,1]}$. In order to prove the second assertion, it suffices to investigate the behavior of the solution $V \in \mathscr{G}(\mathbb{R}^2)$ of problem \eqref{eqn:generalized hyperbolic2}. Let $(v^{\varepsilon})_{\varepsilon \in (0,1]}$ be a representative of $V \in \mathscr{G}(\mathbb{R}^2)$ satisfying problem \eqref{eqn:equation for v^{varepsilon}}. As in the proof of Theorems \ref{thm:2} and \ref{thm:3}, we can apply the method of characteristic curves to see that $V_x$ identically equals $0$ on the complement of the set given by \eqref{eqn:singsupp3}. Since $U = V_x \in \mathscr{G}(\mathbb{R}^2)$, it follows that $U$ is $\mathscr{G}_{\rm log}^{\infty}$-regular on the complement of the set given by \eqref{eqn:singsupp3}. Now, we show that $\{(t,-s_1) \mid t \ge -s_1\} \cup \{(t,2t+s_2) \mid t \ge -s_2\} \subset \operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U$. We have $v^{\varepsilon}(t,x) = v_0^{\varepsilon}(\gamma^{\varepsilon}(t,x,0))$. If $-s_1-a_1(\varepsilon) \le x \le -s_1 + a_1(\varepsilon)$ and $t \ge -s_1 + a_1(\varepsilon) + 2h(\varepsilon)$, then $\gamma^{\varepsilon}(t,x,0) = x$, so that $v^{\varepsilon}(t,x) = v_0^{\varepsilon}(x)$. We see that $\partial_x^{\alpha}v^{\varepsilon}(t,-s_1) = \kappa_1 \chi^{(\alpha-1)}(0)/a_1(\varepsilon)^{\alpha}$ for $\alpha \in \mathbb{N}$. Hence, $\{(t,-s_1) \mid t \ge -s_1\} \subset \operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U$. Similarly, if $2t+s_2-a_2(\varepsilon) \le x \le 2t+s_2 + a_2(\varepsilon)$ and $t \ge -s_2 + a_2(\varepsilon) + 2h(\varepsilon)$, then $\gamma^{\varepsilon}(t,x,0) = x - 2t$, so that $v^{\varepsilon}(t,x) = v_0^{\varepsilon}(x-2t)$. We see that $\partial_x^{\alpha}v^{\varepsilon}(t,2t+s_2) = \kappa_2 \chi^{(\alpha-1)}(0)/a_2(\varepsilon)^{\alpha}$ for $\alpha \in \mathbb{N}$. Hence, $\{(t,2t+s_2) \mid t \ge -s_2\} \subset \operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U$. Finally, we prove that $\{(t,t) \mid t \le \max\{-s_1,-s_2\}\} \subset \operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U$. For $t < -s_1$, consider $\frac{v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-s_1,t)) - v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-s_1 -a_1(\varepsilon),t))}{\gamma^{\varepsilon}(0,-s_1,t) -\gamma^{\varepsilon}(0,-s_1-a_1(\varepsilon),t)}.$ As in Step 1 of the proof of Theorem \ref{thm:1}, we get, for $\varepsilon > 0$ small enough, \begin{gather} G^{\varepsilon}(\gamma^{\varepsilon}(0,-s_1,t)-t) = -\frac{t+s_1}{h(\varepsilon)} + 2, \label{eqn:G^{varepsilon}1} \\ G^{\varepsilon}(\gamma^{\varepsilon}(0,-s_1-a_1(\varepsilon),t)-t) = -\frac{t+s_1+a_1(\varepsilon)}{h(\varepsilon)} + 2. \label{eqn:G^{varepsilon}2} \end{gather} By \eqref{eqn:inequality for (G^{varepsilon})^{-1}} and \eqref{eqn:G^{varepsilon}2}, we have, for some constant $C_2 > 0$, $$\label{eqn:inequality for gamma2} t - \gamma^{\varepsilon}(0,-s_1-a_1(\varepsilon),t) \le 2 \exp\big(-\frac{2}{C_2}\big)h(\varepsilon)\varepsilon^{-(t +s_1+a_1(\varepsilon))/C_2}.$$ Subtracting \eqref{eqn:G^{varepsilon}2} from \eqref{eqn:G^{varepsilon}1} gives $$\label{eqn:difference of G^{varepsilon}} G^{\varepsilon}(\gamma^{\varepsilon}(0,-s_1,t)-t) - G^{\varepsilon}(\gamma^{\varepsilon}(0,-s_1-a_1(\varepsilon),t)-t) = \frac{a_1(\varepsilon)}{h(\varepsilon)}.$$ On the other hand, by the definition of $G^{\varepsilon}$, we get \begin{align*} 0 &< G^{\varepsilon}(\gamma^{\varepsilon}(0,-s_1,t)-t) - G^{\varepsilon}(\gamma^{\varepsilon}(0,-s_1-a_1(\varepsilon),t)-t) \\ & \hspace{0.5cm} = \int_{(t - \gamma^{\varepsilon}(0,-s_1,t)) /h(\varepsilon)}^{(t - \gamma^{\varepsilon} (0,-s_1-a_1(\varepsilon),t))/h(\varepsilon)} \frac{dz}{1 - \widetilde{b}^{\varepsilon}(-h(\varepsilon)z)}. \end{align*} As in the proof of Lemma \ref{lemma:1}, we have $1/(1 - \widetilde{b}^{\varepsilon}(-h(\varepsilon)z)) \ge C_1/z$ for some constant $C_1 > 0$ and so \begin{align*} & G^{\varepsilon}(\gamma^{\varepsilon}(0,-s_1,t)-t) - G^{\varepsilon}(\gamma^{\varepsilon}(0,-s_1-a_1(\varepsilon),t)-t)\\ & \ge \int_{(t - \gamma^{\varepsilon}(0,-s_1,t))/h(\varepsilon)}^{(t - \gamma^{\varepsilon}(0,-s_1-a_1(\varepsilon),t))/h(\varepsilon)} \frac{C_1}{z}\,dz\\ & \ge C_1\big[\frac{t - \gamma^{\varepsilon}(0,-s_1-a_1 (\varepsilon),t)}{h(\varepsilon)} - \frac{t - \gamma^{\varepsilon}(0,-s_1,t)}{h(\varepsilon)}\big] \big[\frac{h(\varepsilon)}{t - \gamma^{\varepsilon} (0,-s_1-a_1(\varepsilon),t)}\big]\\ & = \frac{C_1[\gamma^{\varepsilon}(0,-s_1,t) - \gamma^{\varepsilon}(0,-s_1-a_1(\varepsilon),t)]}{t - \gamma^{\varepsilon}(0,-s_1-a_1(\varepsilon),t)}. \end{align*} Hence, by \eqref{eqn:difference of G^{varepsilon}}, we have $$\label{eqn:difference of G^{varepsilon}2} \gamma^{\varepsilon}(0,-s_1,t)-\gamma^{\varepsilon} (0,-s_1-a_1(\varepsilon),t) \le \frac{t - \gamma^{\varepsilon}(0,-s_1-a_1(\varepsilon),t)}{C_1} \cdot \frac{a_1(\varepsilon)}{h(\varepsilon)}.$$ We combine \eqref{eqn:inequality for gamma2} and \eqref{eqn:difference of G^{varepsilon}2} to see that $$\label{eqn:difference of G^{varepsilon}3} \gamma^{\varepsilon}(0,-s_1,t)-\gamma^{\varepsilon} (0,-s_1-a_1(\varepsilon),t) \le \frac{2}{C_1} \exp\big(-\frac{2}{C_2}\big) a_1(\varepsilon) \varepsilon^{-(t+s_1+a_1(\varepsilon))/C_2}.$$ We use $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-s_1,t)) = \kappa_1/2$, $v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-s_1-a_1(\varepsilon),t)) = 0$ and \eqref{eqn:difference of G^{varepsilon}3} to get \begin{align*} & \frac{v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-s_1,t)) - v^{\varepsilon}(t,\gamma^{\varepsilon}(0,-s_1-a_1(\varepsilon),t))}{\gamma^{\varepsilon}(0,-s_1,t)-\gamma^{\varepsilon}(0,-s_1-a_1(\varepsilon),t)} \\ & \ge \frac{\kappa_1C_1}{4 \exp(-2/C_2) a_1(\varepsilon)} \cdot \frac{1}{\varepsilon^{-(t+s_1+a_1(\varepsilon))/C_2}}. \end{align*} Then by the mean value theorem, we find $x^{\varepsilon} \in (\gamma^{\varepsilon}(0,-s_1-a_1(\varepsilon),t), \gamma^{\varepsilon}(0,-s_1,t))$ such that $v_x^{\varepsilon}(t,x^{\varepsilon}) \ge \frac{\kappa_1C_1}{4 \exp(-2/C_2) a_1(\varepsilon)} \cdot \frac{1}{\varepsilon^{-(t+s_1+a_1(\varepsilon))/C_2}}.$ Since $U = V_x \in \mathscr{G}(\mathbb{R}^2)$, this means that $\{(t,t) \mid t \le -s_1\} \subset \operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U$. Similarly, we can see that $\{(t,t) \mid t \le -s_2\} \subset \operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U$. Therefore, $\{(t,t) \mid t \le \max\{-s_1,-s_2\}\} \subset \operatorname{sing\,supp}_{\mathscr{G}_{\rm log}^{\infty}} U$. Thus, \eqref{eqn:singsupp3} follows. \end{proof} \begin{remark} \rm In Theorem \ref{thm:4}, when $s_1 = s_2$ and $\kappa_1 = -\kappa_2 \ne 0$, the solution $U \in \mathscr{G}(\mathbb{R}^2)$ of problem \eqref{eqn:generalized hyperbolic1} admits a distributional shadow on $(-s_1,\infty) \times \mathbb{R}$, which is given by $u(t,x) = \kappa_1\delta(x+s_1) + \kappa_2\delta(x-2t-s_1)$. Since, for any $\kappa_1$, $\kappa_2$ such that $\kappa_1 = -\kappa_2 \ne 0$, this distribution $u$ satisfies problem \eqref{eqn:hyperbolic1} with initial data $0$ at $t = -s_1$, it follows that there exist infinitely many different distributional solutions with initial data $0$ at $t = -s_1$. Thus, Theorem \ref{thm:4} means that, in the setting of Colombeau's theory, these distributional solutions with initial data $0$ at $t=-s_1$ can be regarded as generalized solutions with different initial data, as in Theorem \ref{thm:1}. \end{remark} As in Step 2 of the proof of Theorem \ref{thm:2-1}, we can use the mean value theorem repeatedly in the last part of the proof of Theorem \ref{thm:4} to get the following equality on the $\mathscr{G}^{\infty}$-singular support of the solution $U$. Hence, we see that, even if $U_0 \in \mathscr{G}^{\infty}(\mathbb{R})$, the singularity in $\mathscr{G}^{\infty}$ occurs suddenly when the propagation of singularities is observed backward in time. \begin{theorem}\label{thm:4-1} Under the same assumption as in Theorem \ref{thm:4}, if $U_0 \in \mathscr{G}^{\infty}(\mathbb{R})$, then $\operatorname{sing\,supp}_{\mathscr{G}^{\infty}} U = \{(t,t) \mid t \le \max\{-s_1,-s_2\}\}.$ \end{theorem} \subsection*{Acknowledgments} The author expresses his most heartfelt thanks to G\"{u}nther H\"{o}rmann for the warm hospitality and valuable discussions during his visit to the Fakult\"{a}t f\"{u}r Mathematik, Universit\"{a}t Wien from October 2, 2009 to March 31, 2011. \begin{thebibliography}{00} \bibitem{biagioni} H. A. Biagioni. \emph{A nonlinear theory of generalized functions}. Lect. Notes Math. 1421. Springer-Verlag, Berlin, 1990. \bibitem{colombeau1} J. F. Colombeau. \emph{New generalized functions and multiplication of distributions}. North-Holland Math. Stud. 84. North-Holland, Amsterdam, 1984. \bibitem{colombeau2} J. F. Colombeau. \emph{Elementary introduction to new generalized functions}. North-Holland Math. Stud. 113. North-Holland, Amsterdam, 1985. \bibitem{colombeau and heibig} J. F. Colombeau and A. Heibig. Generalized solutions to Cauchy problems. \emph{Monatsh. Math.}, 117:33--49, 1994. \bibitem{deguchi1} H. Deguchi. Generalized solutions of semilinear parabolic equations. \emph{Monatsh. Math.}, 146:279--294, 2005. \bibitem{deguchi2} H. Deguchi. Generalized solutions of parabolic initial value problems with discontinuous nonlinearities. \emph{J. Math. Anal. Appl.}, 323:583--596, 2006. \bibitem{garetto} C. Garetto and G. H\"{o}rmann. On duality theory and pseudodifferential techniques for Colombeau algebras: generalized delta functionals, kernels and wave front sets. \emph{Bull. Cl. Sci. Math. Nat. Sci. Math.} 31:115--136, 2006. \bibitem{grosser} M. Grosser, M. Kunzinger, M. Oberguggenberger, and R. Steinbauer. \emph{Geometric theory of generalized functions with applications to general relativity}. Mathematics and its Applications 537. Kluwer Acad. Publ., Dordrecht, 2001. \bibitem{hoermann1} G. H\"{o}rmann and M. V. de Hoop. Microlocal analysis and global solutions of some hyperbolic equations with discontinuous coefficients. \emph{Acta Appl. Math.}, 67:173--224, 2001. \bibitem{hoermann2} G. H\"{o}rmann. First-order hyperbolic pseudodifferential equations with generalized symbols. \emph{J. Math. Anal. Appl.}, 293:40--56, 2004. \bibitem{hurd} A. E. Hurd and D. H. Sattinger. Questions of existence and uniqueness for hyperbolic equations with discontinuous coefficients. \emph{Trans. Amer. Math. Soc.}, 132:159--174, 1968. \bibitem{lafon} F. Lafon and M. Oberguggenberger. Generalized solutions to symmetric hyperbolic systems with discontinuous coefficients: the multidimensional case. \emph{J. Math. Anal. Appl.}, 160:93--106, 1991. \bibitem{oberguggenberger1988} M. Oberguggenberger. Hyperbolic systems with discontinuous coefficients: examples. In: B. Stankovi\'{c}, E. Pap, S. Pilipovi\'{c}, and V. S. Vladimirov (Eds.), Generalized functions, convergence structures, and their applications, pp. 257--266, Plenum, New York, 1988. \bibitem{oberguggenberger1989} M. Oberguggenberger. Hyperbolic systems with discontinuous coefficients: generalized solutions and a transmission problem in acoustics. \emph{J. Math. Anal. Appl.}, 142:452--467, 1989. \bibitem{oberguggenberger1992} M. Oberguggenberger. \emph{Multiplication of distributions and applications to partial differential equations}. Pitman Research Notes Math. 259. Longman Scientific \& Technical, Harlow 1992. \bibitem{oberguggenberger2008} M. Oberguggenberger. Hyperbolic systems with discontinuous coefficients: generalized wavefront set. In: L. Rodino and M. W. Wong (Eds.), New developments in pseudo--differential operators, pp. 117--136, Birkh\"{a}user, Basel, 2009. \end{thebibliography} \end{document}