%% R.E. Showalter: Chapter 4 \chapter{First Order Evolution Equations} \section{Introduction} \setcounter{equation}{0} \setcounter{theorem}{0} We consider first an initial-boundary value problem for the equation of heat conduction. That is, we seek a function $u:[0,\pi]\times [0,\infty]\to\RR$ which satisfies the partial differential equation \label{eq411} u_t = u_{xx}\ ,\qquad 00 with the boundary conditions $$\label{eq412} u(0,t)=0\ ,\quad u(\pi,t)=0\ ,\qquad t>0$$ and the initial condition \label{eq413} u(x,0)= u_0(x)\ ,\qquad 00 with the boundary conditions \eqn{412} and initial condition \eqn{413}. Suppose that for each $t>0$, $f(\cdot,t)\in L^2[0,\pi]$ and, hence, has the eigenfunction expansion $$\label{eq416} f(x,t) = \sum_{n=1}^\infty f_n(t) \sin (nx)\quad ,\quad f_n(t) \equiv {2\over\pi} \int_0^\pi f(\xi,t)\sin (n\xi)\,d\xi\ .$$ We look for the solution in the form $u(x,t) = \sum_{n=1}^\infty u_n(t) \sin (nx)$ and find from \eqn{415} and \eqn{413} that the coefficients must satisfy \begin{array}{ll} u'_n (t) + n^2 u_n(t) = f_n(t)\ ,&t\ge 0\ ,\\ \noalign{\vskip6pt} u_n(0) = u_n^0\ ,&n\ge 1\ . \end{array} Hence we have $$u_n(t) = u_n^0 e^{-n^2t} + \int_0^t e^{-n^2(t-\tau)} f_n(\tau)\,d\tau$$ and the solution is given by $$u(x,t)= S(t)u_0(x) + \int_0^t \int_0^\pi \biggl\{ {2\over\pi} \sum_{n=1}^\infty e^{-n^2(t -\tau)} \sin (nx) \sin (n\xi)\biggr\} f(\xi,\tau)\, d\xi\, d\tau\ .$$ But from \eqn{416} it follows that we have the representation $$\label{eq417} u(\cdot,t) = S(t) u_0 (\cdot) + \int_0^t S(t-\tau)f(\cdot,\tau)\,d\tau$$ for the solution of \eqn{415}, \eqn{412}, \eqn{413}. The preceding computations will be made precise in this chapter and \eqn{417} will be used to prove existence and uniqueness of a solution. \section{The Cauchy Problem} % 2 \setcounter{equation}{0} Let $H$ be a Hilbert space, $D(A)$ a subspace of $H$, and $A\in L(D(A),H)$. We shall consider the evolution equation $$\label{eq421} u'(t) + Au(t)=0\ .$$ The {\it Cauchy problem\/} is to find a function $u\in C([0,\infty],H)\cap C^1((0,\infty),H)$ such that, for $t>0$, $u(t) \in D(A)$ and \eqn{421} holds, and $u(0) = u_0$, where the initial value $u_0\in H$ is prescribed. Assume that for every $u_0\in D(A)$ there exists a unique solution of the Cauchy problem. Define $S(t) u_0=u(t)$ for $t\ge0$, $u_0\in D(A)$, where $u(\cdot)$ denotes that solution of \eqn{421} with $u(0) = u_0$. If $u_0,v_0\in D(A)$ and if $a,b\in \RR$, then the function $t\mapsto aS(t) u_0 + bS(t)v_0$ is a solution of \eqn{421}, since $A$ is linear, and the uniqueness of solutions then implies $$S(t) (au_0 + bv_0) = aS (t) u_0 + bS(t)v_0\ .$$ Thus, $S(t) \in L(D(A))$ for all $t\ge0$. If $u_0 \in D(A)$ and $\tau \ge0$, then the function $t\mapsto S(t+\tau)u_0$ satisfies \eqn{421} and takes the initial value $S(\tau)u_0$. The uniqueness of solutions implies that $$S(t+\tau) u_0 = S(t) S(\tau)u_0\ ,\qquad u_0\in D(A)\ .$$ Clearly, $S(0)=I$. We define the operator $A$ to be {\it accretive\/} if $$\Re (Ax,x)_H\ge0\ ,\qquad x\in D(A)\ .$$ If $A$ is accretive and if $u$ is a solution of the Cauchy problem for \eqn{421}, then \begin{eqnarray*} D_t (\|u(t)\|^2) & = & 2\Re (u'(t),u(t))H\\ \noalign{\vskip6pt} & = & -2\Re (Au(t),u(t))_H \le 0\ ,\qquad t>0\ , \end{eqnarray*} so it follows that $\|u(t)\| \le \|u(0)\|$, $t\ge 0$. This shows that $$\|S(t) u_0\| \le \|u_0\|\ ,\qquad u_0\in D(A)\ ,\ t\ge0\ ,$$ so each $S(t)$ is a contraction in the $H$-norm and hence has a unique extension to the closure of $D(A)$. When $D(A)$ is dense, we thereby obtain a contraction semigroup on $H$. \definition A {\it contraction semigroup\/} on $H$ is a set $\{S(t):t\ge0\}$ of linear operators on $H$ which are contractions and satisfy \begin{eqnarray} &S(t+\tau) = S(t)\cdot S(\tau)\ ,\quad S(0)=I\ , &t,\tau \ge0\ , \label{eq422}\\ \noalign{\vskip6pt} &S(\cdot)x\in C([0,\infty),H)\ ,\quad \hphantom{\ S(0)=I\ ,} &x\in H\ .\label{eq423} \end{eqnarray} The {\it generator\/} of the contraction semigroup $\{S(t):t\ge0\}$ is the operator with domain $$D(B) = \Bigl\{ x\in H: \lim_{h\to0^+} h^{-1} (S(h)-I)x=D^+ (S(0)x) \hbox{ exists in } H\Bigr\}$$ and value $Bx=\lim_{h\to0^+} h^{-1}(S(h)-I)x=D^+ (S(0)x)$. Note that $Bx$ is the right-derivative at $0$ of $S(t)x$. \qed The equation \eqn{422} is the {\it semigroup identity\/}. The definition of solution for the Cauchy problem shows that \eqn{423} holds for $x\in D(A)$, and an elementary argument using the uniform boundedness of the (contraction) operators $\{S(t):t\ge0\}$ shows that \eqn{423} holds for all $x\in H$. The property \eqn{423} is the {\it strong continuity\/} of the semigroup. \begin{theorem}\label{thm4-2A} Let $A\in L(D(A),H)$ be accretive with $D(A)$ dense in $H$. Suppose that for every $u_0\in D(A)$ there is a unique solution $u\in C^1([0,\infty),H)$ of \eqn{421} with $u(0)=u_0$. Then the family of operators $\{S(t) :t\ge0\}$ defined as above is a contraction semigroup on $H$ whose generator is an extension of $-A$. \end{theorem} \proof Note that uniqueness of solutions is implied by $A$ being accretive, so the semigroup is defined as above. We need only to verify that $-A$ is a restriction of the generator. Let $B$ denote the generator of $\{S(t):t\ge0\}$ and $u_0\in D(A)$. Since the corresponding solution $u(t) = S(t)u_0$ is right-differentiable at $0$, we have $$S(h) u_0-u_0 = \int_0^h u' (t)\,dt = -\int_0^h Au (t)\,dt \ ,\qquad h>0 \ .$$ Hence, we have $D^+ (S(0)u_0) = -Au_0$, so $u_0\in D(B)$ and $Bu_0=-Au_0$. \qed We shall see later that if $-A$ is the generator of a contraction semigroup, then $A$ is accretive, $D(A)$ is dense, and for every $u_0\in D(A)$ there is a unique solution $u\in C^1([0,\infty),H)$ of \eqn{421} with $u(0)=u_0$. But first, we consider a simple example. \begin{theorem}\label{thm4-2B} For each $B\in \L(H)$, the series $\sum_{n=0}^\infty (B^n/n!)$ converges in $\L(H)$; denote its sum by $\exp (B)$. The function $t\mapsto \exp (tB) :\RR\to \L(H)$ is infinitely differentiable and satisfies $$\label{eq424} D\bigl[ \exp (tB)\bigr] = B\cdot\exp (tB) = \exp (tB) \cdot B\ ,\qquad t\in \RR\ .$$ If $B_1,B_2\in \L(H)$ and if $B_1\cdot B_2 = B_2\cdot B_1$, then $$\label{eq425} \exp (B_1+B_2) = \exp (B_1)\cdot \exp (B_2)\ .$$ \end{theorem} \proof The convergence of the series in $\L(H)$ follows from that of\newline $\sum_{n=0}^\infty \|B\|_{\L(H)}^n /n! = \exp (\|B\|)$ in $\RR$. To verify the differentiability of $\exp (tB)$ at $t=0$, we note that $$\Bigl[ \bigl( \exp (tB)-I\bigr)/t\Bigr] -B = (1/t) \sum_{n=2}^\infty (tB)^n/n!\ ,\qquad t\ne 0\ ,$$ and this gives the estimate $$\|\Bigl[\bigl(\exp (tB)-I\bigr)/t\Bigr] -B\| \le (1/|t|) \Bigl[ \exp (|t|\cdot\|B\|) - 1- |t|\, \|B\|\Bigr]\ .$$ Since $t\mapsto \exp (t\|B\|)$ is (right) differentiable at $0$ with (right) derivative $\|B\|$, it follows that \eqn{424} holds at $t=0$. The semigroup property shows that \eqn{424} holds at every $t\in\RR$. (We leave \eqn{425} as an exercise.) \section{Generation of Semigroups} % 3 \setcounter{equation}{0} Our objective here is to characterize those operators which generate contraction semigroups. To first obtain necessary conditions, we assume that $B:D(B)\to H$ is the generator of a contraction semigroup $\{S(t):t\ge0\}$. If $t\ge0$ and $x\in D(B)$, then the last term in the identity $$h^{-1}(S(t+h)x-S(t)x) = h^{-1}(S(h)-I)S(t)x = h^{-1}S(t)(S(h)x-x)\ ,\qquad h>0\ ,$$ has a limit as $h\to 0^+$, hence, so also does each term and we obtain $$D^+ S(t) x = BS (t) x= S(t) Bx\ ,\qquad x\in D(B)\ ,\ t\ge0\ .$$ Similarly, using the uniform boundedness of the semigroup we may take the limit as $h\to 0^+$ in the identity $$h^{-1} (S(t) x- S(t-h) x) = S(t-h) h^{-1} (S(h) x-x)\ ,\qquad 00\ .$$ We summarize the above. \plainlemma {For each $x\in D(B)$, $S(\cdot) x\in C^1 (\RR_0^+,H)$, $S(t)x\in D(B)$, and $$\label{eq431} S(t) x-x = \int_0^t BS(s) x\,ds = \int_0^t S(s) Bx\,dx \ ,\qquad t\ge0\ .$$ } \plaincor {$B$ is closed.} \proof Let $x_n\in D(B)$ with $x_n\to x$ and $Bx_n\to y$ in $H$. For each $h>0$ we have from \eqn{431} $$h^{-1} (S(h)x_n-x_n) = h^{-1} \int_0^h S(s)Bx_n\,ds\ ,\qquad n\ge1\ .$$ Letting $n\to\infty$ and then $h\to 0^+$ gives $D^+S(0)x=y$, hence, $Bx=y$. \plainlemma {$D(B)$ is dense in $H$; for each $t\ge0$ and $x\in H$, $\int_0^t S(s)x\,ds\in D(B)$ and $$\label{eq432} S(t) x-x=B\int_0^t S(s)x\,ds\ ,\qquad x\in H\ ,\ t\ge0\ .$$ } \proof Define $x_t = \int_0^t S(s)x\,ds$. Then for $h>0$ \begin{eqnarray*} h^{-1}(S(h)x_t-x_t) & = & h^{-1}\biggl\{ \int_0^t S(h+s)x\,ds - \int_0^t S(s)x\,ds\biggr\}\\ \noalign{\vskip6pt} & = & h^{-1} \biggl\{ \int_h^{t+h} S(s)x\,ds - \int_0^t S(s)x\,ds\biggr\}\ . \end{eqnarray*} Adding and subtracting $\int_t^h S(s)x\,ds$ gives the equation $$h^{-1}(S(h)x_t-x_t) = h^{-1} \int_t^{t+h} S(s)x\,ds - h^{-1} \int_0^h S(s)x\,ds\ ,$$ and letting $h\to0$ shows that $x_t \in D(B)$ and $Bx_t = S(t)x-x$. Finally, from $t^{-1} x_t \to x$ as $t\to 0^+$, it follows that $D(B)$ is dense in $H$. Let $\lambda> 0$. Then it is easy to check that $\{ e^{-\lambda t} S(t):t\ge0\}$ is a contraction semigroup whose generator is $B-\lambda$ with domain $D(B)$. From \eqn{431} and \eqn{432} applied to this semigroup we obtain \begin{array}{rclll} e^{-\lambda t} S(t) x-x &=&\displaystyle \int_0^t e^{-\lambda s} S(s) (B-\lambda)x\,ds\ , &\qquad x\in D(B)\ ,&t\ge 0\ ,\\ \noalign{\vskip6pt} e^{-\lambda t} S(t) y-y &=&\displaystyle (B-\lambda) \int_0^t e^{-\lambda s} S(s)y\,ds\ , &\qquad y\in H\ ,&t\ge 0\ .\end{array} Letting $t\to\infty$ (and using the fact that $B$ is closed to evaluate the limit of the last term) we find that \begin{array}{rcll} x&=&\displaystyle \int_0^\infty e^{-\lambda s} S(s)(\lambda-B)x\,ds\ ,&\qquad x\in D(B)\ ,\\ \noalign{\vskip6pt} y&=&\displaystyle (\lambda-B) \int_0^\infty e^{-\lambda s} S(s)y\,ds\ ,&\qquad y\in H\ . \end{array} These identities show that $\lambda-B$ is injective and surjective, respectively, with $$\|(\lambda-B)^{-1}y\|\le \int_0^\infty e^{-\lambda s} \,ds \|y\| = \lambda^{-1}\|y\|\ ,\qquad y\in H\ .$$ This proves the necessity part of the following fundamental result. \begin{theorem}\label{thm4-3A} Necessary and sufficient conditions that the operator \newline $B:D(B)\to H$ be the generator of a contraction semigroup on $H$ are that \begin{description} \item[] $D(B)$ is dense in $H$ and $\lambda-B:D(B)\to H$ is a bijection with $\|\lambda(\lambda-B)^{-1} \|_{\L(H)} \le 1$ for all $\lambda >0$. \end{description} \end{theorem} \proof (Continued) It remains to show that the indicated conditions on $B$ imply that it is the generator of a semigroup. We shall achieve this as follows: (a)~approximate $B$ by bounded operators, $B_\lambda$, (b)~obtain corresponding semigroups $\{S_\lambda (t):t\ge0\}$ by exponentiating $B_\lambda$, then (c)~show that $S(t)\equiv \lim_{\lambda\to\infty} S_\lambda (t)$ exists and is the desired semigroup. Since $\lambda-B :D(B)\to H$ is a bijection for each $\lambda >0$, we may define $B_\lambda = \lambda B(\lambda-B)^{-1}$, $\lambda>0$. \plainlemma {For each $\lambda >0$, $B_\lambda \in \L(H)$ and satisfies $$\label{eq433} B_\lambda = -\lambda +\lambda^2 (\lambda-B)^{-1}\ .$$ For $x\in D(B)$, $\|B_\lambda (x)\| \le \|Bx\|$ and $\lim_{\lambda\to\infty} B_\lambda (x) = Bx$.} \proof Equation \eqn{433} follows from $(B_\lambda+\lambda)(\lambda-B) x= \lambda^2 x$, $x\in D(B)$. The estimate is obtained from $B_\lambda = \lambda (\lambda-B)^{-1}B$ and the fact that $\lambda (\lambda-B)^{-1}$ is a contraction. Finally, we have from \eqn{433} $$\|\lambda (\lambda-B)^{-1} x-x\| = \|\lambda^{-1} B_\lambda x \| \le \lambda^{-1} \|Bx\|\ ,\qquad \lambda>0\ ,\ x\in D(B)\ ,$$ hence, $\lambda (\lambda-B)^{-1} x\mapsto x$ for all $x\in D(B)$. But $D(B)$ dense and $\{\lambda(\lambda-B)^{-1}\}$ uniformly bounded imply $\lambda (\lambda-B)^{-1} x\to x$ for all $x\in H$, and this shows $B_\lambda x = \lambda (\lambda-B)^{-1} Bx\to Bx$ for $x\in D(B)$. Since $B_\lambda$ is bounded for each $\lambda >0$, we may define by Theorem \ref{thm4-2B} $$S_\lambda (t) = \exp (tB_\lambda)\ ,\qquad \lambda >0\ ,\ t\ge0\ .$$ \plainlemma {For each $\lambda >0$, $\{S_\lambda (t):t\ge 0\}$ is a contraction semigroup on $H$ with generator $B_\lambda$. For each $x\in D(B)$, $\{S_\lambda (t)x\}$ converges in $H$ as $\lambda\to \infty$, and the convergence is uniform for $t\in [0,T]$, $T>0$.} \proof The first statement follows from $$\|S_\lambda (t)\| = e^{-\lambda t} \|\exp (\lambda^2 (\lambda-B)^{-1}t)\| \le e^{-\lambda t} e^{\lambda t} = 1\ ,$$ and $D(S_\lambda (t)) = B_\lambda S_\lambda (t)$. Furthermore, \begin{eqnarray*} S_\lambda (t)-S_\mu (t) & = &\int_0^t D_sS_\mu (t-s) S_\lambda (s)\,ds\\ \noalign{\vskip6pt} &=& \int_0^t S_\mu (t-s)S_\lambda (s)(B_\lambda -B_\mu) \,ds\ ,\qquad \mu,\lambda >0\ , \end{eqnarray*} in $\L(H)$, so we obtain $$\|S_\lambda (t) x-S_\mu(t) s\| \le t\|B_\lambda x-B_\mu x\|\ ,\qquad \lambda,\mu>0\ ,\ t\ge 0\ ,\ x\in D(B)\ .$$ This shows $\{S_\lambda (t)x\}$ is uniformly Cauchy for $t$ on bounded intervals, so the Lemma follows. \qed Since each $S_\lambda (t)$ is a contraction and $D(B)$ is dense, the indicated limit holds for all $x\in H$, and uniformly on bounded intervals. We define $S(t) x=\lim_{\lambda\to\infty} S_\lambda (t)x$, $x\in H$, $t\ge0$, and it is clear that each $S(t)$ is a linear contraction. The uniform convergence on bounded intervals implies $t\mapsto S(t)x$ is continuous for each $x\in H$ and the semigroup identity is easily verified. Thus $\{S(t):t\ge0\}$ is a contraction semigroup on $H$. If $x\in D(B)$ the functions $S_\lambda (\cdot)B_\lambda x$ converge uniformly to $S(\cdot) Bx$ and, hence, for $h>0$ we may take the limit in the identity $$S_\lambda (h)x-x = \int_0^h S_\lambda (t) B_\lambda x\,dt$$ to obtain $$S(h) x - x= \int_0^h S(t)Bx\,dt\ ,\qquad x\in D(B)\ ,\ h>0\ .$$ This implies that $D^+ (S(0)x)= Bx$ for $x\in D(B)$. If $C$ denotes the generator of $\{S(t) :t\ge0\}$, we have shown that $D(B)\subset D(C)$ and $Bx=Cx$ for all $x\in D(B)$. That is, $C$ is an extension of $B$. But $I-B$ is surjective and $I-C$ is injective, so it follows that $D(B) = D(C)$. \begin{corollary}\label{cor4-3B} If $-A$ is the generator of a contraction semigroup, then for each $u_0\in D(A)$ there is a unique solution $u\in C^1([0,\infty),H)$ of \eqn{421} with $u(0)=u_0$. \end{corollary} \proof This follows immediately from \eqn{431}. \begin{theorem}\label{thm4-3C} If $-A$ is the generator of a contraction semigroup, then for each $u_0\in D(A)$ and each $f\in C^1([0,\infty),H)$ there is a unique $u\in C^1 ([0,\infty),H)$ such that $u(0)=u_0$, $u(t)\in D(A)$ for $t\ge0$, and $$\label{eq434} u'(t) + Au(t) = f(t)\ ,\qquad t\ge0\ .$$ \end{theorem} \proof It suffices to show that the function $$g(t) = \int_0^t S(t-\tau )f(\tau)\,d\tau\ ,\qquad t\ge0\ ,$$ satisfies \eqn{434} and to note that $g(0) =0$. Letting $z=t-\tau$ we have \begin{eqnarray*} (g(t+h)-g(t))/h &=& \int_0^t S(z)(f(t+h-z) -f(t-z))h^{-1}\,dz \\ \noalign{\vskip6pt} &&\qquad + h^{-1} \int_t^{t+h} S(z) f(t+h-z)\,dz \end{eqnarray*} so it follows that $g'(t)$ exists and $$g'(t) = \int_0^t S(z) f'(t-z)\,dz + S(t)f(0)\ .$$ Furthermore we have \begin{eqnarray} (g(t+h)-g(t))/h &=& h^{-1} \biggl\{ \int_0^{t+h} S(t+h-\tau)f(\tau)\,d\tau - \int_0^t S(t-\tau ) f(\tau)\,d\tau\biggr\} \nonumber\\ \noalign{\vskip6pt} &=&(S(h)-I)h^{-1} \int_0^t S(t-\tau) f(\tau)\,d\tau\nonumber\\ \noalign{\vskip6pt} &&\qquad + h^{-1} \int_t^{t+h} S(t+h-\tau) f(\tau)\,d\tau\ .\label{eq435} \end{eqnarray} Since $g'(t)$ exists and since the last term in \eqn{435} has a limit as $h\to 0^+$, it follows from \eqn{435} that $$\int_0^t S(t-\tau) f(\tau)\,d\tau \in D(A)$$ and that $g$ satisfies \eqn{434}. \section{Accretive Operators; two examples} % 4 \setcounter{equation}{0} We shall characterize the generators of contraction semigroups among the negatives of accretive operators. In our applications to boundary value problems, the conditions of this characterization will be more easily verified than those of Theorem \ref{thm4-3A}. These applications will be illustrated by two examples; the first contains a first order partial differential equation and the second is the second order equation of heat conduction in one dimension. Much more general examples of the latter type will be given in Section~7. The two following results are elementary and will be used below and later. \begin{lemma}\label{lem4-4A} Let $B\in \L(H)$ with $\|B\|<1$. Then $(I-B)^{-1}\in \L(H)$ and is given by the power series $\sum_{n=0}^\infty B^n$ in $\L(H)$. \end{lemma} \begin{lemma}\label{lem4-4B} Let $A\in L(D(A),H)$ where $D(A)\le H$, and assume ${(\mu-A)^{-1}}\in \L(H)$, with $\mu \in \CC$. Then $(\lambda-A)^{-1}\in \L(H)$ for $\lambda \in \CC$, if and only if $[I-(\mu-\lambda)(\mu-A)^{-1}]^{-1} \in \L(H)$, and in that case we have $$(\lambda-A)^{-1} = (\mu-A)^{-1} \bigl[ I- (\mu-\lambda)(\mu-A)^{-1}\bigr]^{-1}\ .$$ \end{lemma} \proof Let $B\equiv I-(\mu-\lambda) (\mu-A)^{-1}$ and assume $B^{-1}\in\L(H)$. Then we have \begin{eqnarray*} (\lambda-A)(\mu-A)^{-1}B^{-1} &=& [(\lambda-\mu)+(\mu-A)] (\mu-A)^{-1} B^{-1}\\ &=& [(\lambda-\mu)(\mu-A)^{-1} +I] B^{-1} = I\ , \end{eqnarray*} and \begin{eqnarray*} (\mu-A)^{-1}B^{-1}(\lambda-A) &=& (\mu-A)^{-1} B^{-1} [(\lambda-\mu)+(\mu-A)]\\ &=& (\mu-A)^{-1} B^{-1} [B(\mu-A)]=I\ ,\ \hbox{ on }\ D(A)\ . \end{eqnarray*} The converse is proved similarly. \qed Suppose now that $-A$ generates a contraction semigroup on $H$. From Theorem \ref{thm4-3A} it follows that $$\label{eq441} \|(\lambda+A)x\| \ge \lambda \|x\|\ ,\qquad \lambda >0\ ,\ x\in D(A)\ ,$$ and this is equivalent to $$2\Re (Ax,x)_H \ge -\|Ax\|^2 /\lambda\ ,\qquad \lambda >0\ ,\ x\in D(A)\ .$$ But this shows $A$ is accretive and, hence, that Theorem \ref{thm4-3A} implies the necessity part of the following. \begin{theorem}\label{thm4-4C} The linear operator $-A:D(A)\to H$ is the generator of a contraction semigroup on $H$ if and only if $D(A)$ is dense in $H$, $A$ is accretive, and $\lambda+A$ is surjective {\em for some} $\lambda >0$. \end{theorem} \proof (Continued) It remains to verify that the above conditions on the operator $A$ imply that $-A$ satisfies the conditions of Theorem \ref{thm4-3A}. Since $A$ is accretive, the estimate \eqn{441} follows, and it remains to show that $\lambda+A$ is surjective for every $\lambda >0$. We are given $(\mu+A)^{-1}\in \L(H)$ for some $\mu>0$ and $\|\mu(\mu+A)^{-1} \|\le 1$. For any $\lambda \in C$ we have $\|(\lambda-\mu)(\mu+A)^{-1}\| \le |\lambda-\mu|/\mu$, hence Lemma \ref{lem4-4A} shows that $I-(\lambda-\mu)(\lambda+A)^{-1}$ has an inverse which belongs to $\L(H)$ if $|\lambda-\mu| <\mu$. But then Lemma \ref{lem4-4B} implies that $(\lambda +A)^{-1}\in \L(H)$. Thus, $(\mu+A)^{-1}\in \L(H)$ with $\mu>0$ implies that $(\lambda+A)^{-1} \in \L(H)$ for all $\lambda >0$ such that $|\lambda-\mu|<\mu$, i.e., $0<\lambda <2\mu$. The result then follows by induction. \qed \exam{1} Let $H=L^2(0,1)$, $c\in \CC$, $D(A) = \{u\in H^1(0,1):u(0)=cu(1)\}$, and $A=\partial$. Then we have for $u\in H^1(0,1)$ $$2\Re (Au,u)_H = \int_0^1 (\partial u\cdot\bar u + \overline{\partial u} \cdot u) = |u(1)|^2 - |u(0)|^2\ .$$ Thus, $A$ is accretive if (and only if) $|c|\le1$, and we assume this hereafter. Theorem \ref{thm4-4C} implies that $-A$ generates a contraction semigroup on $L^2(0,1)$ if (and only if) $I+A$ is surjective. But this follows from the solvability of the problem $$u+\partial u = f\ ,\qquad u(0) = cu(1)$$ for each $f\in L^2(0,1)$; the solution is given by \begin{eqnarray*} u(x) & = & \int_0^1 G(x,s)f(s)\,ds\ ,\\ \noalign{\vskip6pt} G(x,s) & = &\cases{[e/(e-c)] e^{-(x-s)}\ ,&$0\le s0$, and one can thereby show that the solution is infinitely differentiable in the open cylinder $(0,1)\times (0,\infty)$. Finally, the series will in general not converge if $t<0$. This occurs because of the exponential terms, and severe conditions must be placed on the initial data $u_0$ in order to obtain convergence at a point where $t<0$. Even when a solution exists on an interval $[-T,0]$ for some $T>0$, it will not depend continuously on the initial data (cf., Exercise 1.3). The preceding situation is typical of Cauchy problems which are resolved by {\it analytic semigroups\/}. Such Cauchy problems are (appropriately) called {\it parabolic\/} and we shall discuss these notions in Sections~6 and 7 and again in Chapters~V and VI. \section{Generation of Groups; a wave equation} % 5 \setcounter{equation}{0} We are concerned here with a situation in which the evolution equation can be solved on the whole real line $\RR$, not just on the half-line $\RR^+$. This is the case when $-A$ generates a {\it group\/} of operators on $H$. \definition A {\it unitary group\/} on $H$ is a set $\{ G(t):t\in\RR\}$ of linear operators on $H$ which satisfy \begin{eqnarray} &&G(t+\tau) = G(t)\cdot G(\tau)\ ,\quad G(0)=I\ ,\qquad t,\tau\in\RR\ , \label{eq451}\\ \noalign{\vskip4pt} &&G(\cdot)x \in C(\RR,H)\ ,\qquad x\in H\ ,\label{eq452}\\ \noalign{\vskip4pt} &&\|G(t)\|_{\L(H)} = 1\ ,\qquad t\in \RR\ .\label{eq453} \end{eqnarray} The {\it generator\/} of this unitary group is the operator $B$ with domain $$D(B) = \Bigl\{ x\in H: \lim_{h\to0} h^{-1}(G(h)-I) x\ \hbox{ exists in }\ H\Bigr\}$$ with values given by $Bx=\lim_{h\to0} h^{-1} (G(h)-I)x= D(G(0)x)$, the (two-sided) derivative at $0$ of $G(t)x$. Equation \eqn{451} is the group condition, \eqn{452} is the condition of strong continuity of the group, and \eqn{453} shows that each operator $G(t)$, $t\in \RR$, is an isometry. Note that \eqn{451} implies $$G(t)\cdot G(-t)=I\ ,\qquad t\in \RR\ ,$$ so each $G(t)$ is a bijection of $H$ onto $H$ whose inverse is given by $$G^{-1} (t) = G(-t)\ ,\qquad t\in \RR\ .$$ If $B\in \L(H)$, then \eqn{451} and \eqn{452} are satisfied by $G(t)\equiv \exp (tB)$, $t\in \RR$ (cf., Theorem \ref{thm4-2B}). Also, it follows from \eqn{424} that $B$ is the generator of $\{G(t):t\in\RR\}$ and $$D(\|G(t)x\|^2) = 2\Re (BG (t)x,G(t)x)_H \ ,\qquad x\in H\ ,\ t\in \RR\ ,$$ hence, \eqn{453} is satisfied if and only if $\Re (Bx,x)_H =0$ for all $x\in H$. These remarks lead to the following. \begin{theorem}\label{thm4-5A} The linear operator $B:D(B)\to H$ is the generator of a unitary group on $H$ if and only if $D(B)$ is dense in $H$ and $\lambda-B$ is a bijection with $\|\lambda(\lambda-B)^{-1}\|_{\L(H)} \le1$ for all $\lambda\in\RR$, $\lambda \ne 0$. \end{theorem} \proof If $B$ is the generator of the unitary group $\{G(t):t\in\RR\}$, then $B$ is the generator of the contraction semigroup $\{G(t):t\ge0\}$ and $-B$ is the generator of the contraction semigroup $\{G(-t):t\ge0\}$. Thus, both $B$ and $-B$ satisfy the necessary conditions of Theorem \ref{thm4-3A}, and this implies the stated conditions on $B$. Conversely, if $B$ generates the contraction semigroup $\{S_+(t):t\ge0\}$ and $-B$ generates the contraction semigroup $\{S_-(t):t\ge0\}$, then these operators commute. For each $x_0\in D(B)$ we have $$D[S_+ (t)S_-(-t)x_0] =0\ ,\qquad t\ge0\ ,$$ so $S_+ (t)S_-(-t) = I$, $t\ge0$. This shows that the family of operators defined by G(t) = \cases{S_+(t)\ ,&t\ge0\cr \noalign{\vskip6pt} S_- (-t)\ ,&t<0\cr} satisfies \eqn{451}. The condition \eqn{452} is easy to check and \eqn{453} follows from $$1= \|G(t) \cdot G(-t)\| \le \|G(t)\|\cdot \|G(-t)\| \le \|G(t)\|\le 1\ .$$ Finally, it suffices to check that $B$ is the generator of $\{G(t):t\in\RR\}$ and then the result follows. \begin{corollary}\label{cor4-5B} The operator $A$ is the generator of a unitary group on $H$ if and only if for each $u_0\in D(A)$ there is a unique solution $u\in C^1(\RR,H)$ of \eqn{421} with $u(0) = u_0$ and $\|u(t)\| = \|u_0\|$, $t\in \RR$. \end{corollary} \proof This is immediate from the proof of Theorem \ref{thm4-5A} and the results of Theorem \ref{thm4-2A} and Corollary \ref{cor4-3B}. \begin{corollary}\label{cor4-5C} If $A$ generates a unitary group on $H$, then for each $u_0\in D(A)$ and each $f\in C^1(\RR,H)$ there is a unique solution $u\in C^1(\RR,H)$ of \eqn{433} and $u(0)=u_0$. This solution is given by $$u(t) = G(t) u_0 + \int_0^t G(t-\tau) f(\tau)\,d\tau\ ,\qquad t\in\RR\ .$$ \end{corollary} Finally, we obtain an analogue of Theorem \ref{thm4-4C} by noting that both $+A$ and $-A$ are accretive exactly when $A$ satisfies the following. \definition The linear operator $A\in L(D(A),H)$ is said to be {\it conservative\/} if $$\Re (Ax,x)_H=0\ ,\qquad x\in D(A)\ .$$ \begin{corollary}\label{cor4-5D} The linear operator $A:D(A)\to H$ is the generator of a unitary group on $H$ if and only if $D(A)$ is dense in $H$, $A$ is conservative, and $\lambda+A$ is surjective for some $\lambda>0$ and for some $\lambda<0$. \end{corollary} \example Take $H= L^2(0,1)\times L^2(0,1)$, $D(A) = H_0^1 (0,1)\times H^1(0,1)$, and define $$A[u,v] = [-i\partial v,i\partial u]\ ,\qquad [u,v]\in D(A)\ .$$ Then we have $$(A[u,v],[u,v])_H = i\int_0^1(\partial v\cdot\bar u- \partial u\cdot\bar v)\ , \qquad [u,v]\in D(A)$$ and an integration-by-parts gives $$\label{eq454} 2\Re (A[u,v],[u,v])_H = i(\,\bar u(x)v(x)-u(x)\bar v(x))\Big|_{x=0}^{x=1} =0\ ,$$ since $u(0)=u(1)=0$. Thus, $A$ is a conservative operator. If $\lambda\ne0$ and $[f_1,f_2]\in H$, then $$\lambda [u,v] + A[u,v] = [f_1,f_2]\ ,\qquad [u,v]\in D(A)$$ is equivalent to the system \begin{eqnarray} -\partial^2u+\lambda^2 u &=& \lambda f_1 -i\partial f_2\ , \qquad u\in H_0^1(0,1)\ ,\label{eq455}\\ \noalign{\vskip6pt} -i\partial u+\lambda v &=& f_2\ ,\hphantom{\lambda-i\partial f_2} \qquad v\in H^1(0,1)\ .\label{eq456} \end{eqnarray} But \eqn{455} has a unique solution $u\in H_0^1 (0,1)$ by Theorem III.\ref{thm3-2B} since $\lambda f_1-i\partial f_2\in (H_0^1)'$ from Theorem II.\ref{thm2-2B}. Then \eqn{456} has a solution $v\in L^2(0,1)$ and it follows from \eqn{456} that $$(i\lambda) \partial v = \lambda f_1-\lambda^2 u\in L^2(0,1)\ ,$$ so $v\in H^1(0,1)$. Thus $\lambda+A$ is surjective for $\lambda\ne0$. Corollaries \ref{cor4-5C} and \ref{cor4-5D} imply that the Cauchy problem \label{eq457} \begin{array}{rcll} D\bu (t) + A\bu (t) &=&[0,f(t)]\ ,&\qquad t\in \RR\ ,\\ \noalign{\vskip6pt} \bu (0) &=& [u_0,v_0]& \end{array} is well-posed for $u_0\in H_0^1(0,1)$, $v_0\in H^1(0,1)$, and $f\in C^1(\RR,H)$. Denoting by $u(t)$, $v(t)$, the components of $\bu (t)$, i.e., $\bu (t)\equiv [u(t),v(t)]$, it follows that $u\in C^2(\RR,L^2(0,1))$ satisfies the wave equation $$\partial_t^2 u(x,t) - \partial_x^2 u(x,t) = f(x,t)\ ,\qquad 00, u(t)\in D(A^p) for every p\ge1. \end{corollary} There are some important differences between Corollary \ref{cor4-6D} and its counterpart, Corollary \ref{cor4-3B}. In particular we note that Corollary \ref{cor4-6D} solves the Cauchy problem for all initial data in H, while Corollary \ref{cor4-3B} is appropriate only for initial data in D(A). Also, the infinite differentiability of the solution from Corollary \ref{cor4-6D} and the consequential inclusion in the domain of every power of A at each t>0 are properties not generally true in the situation of Corollary \ref{cor4-3B}. These regularity properties are typical of parabolic problems (cf., Section~7). \begin{theorem}\label{thm4-6E} If A is the operator of Theorem \ref{thm4-6A}, then for each u_0\in H and each {\em H\"older continuous} f:[0,\infty)\to H:$$\|f(t)-f(\tau)\| \le K (t-\tau)^\alpha\ ,\qquad 0\le \tau\le t\ ,$$where K and \alpha are constant, 0<\alpha\le1, there is a unique u\in C([0,\infty),H) \cap C^1 ((0,\infty),H) such that u(0)=u_0, u(t) \in D(A) for t>0, and$$u'(t)+Au (t) = f(t)\ ,\qquad t>0\ .$$\end{theorem} \proof It suffices to show that the function$$g(t) = \int_0^t T(t-\tau) f(\tau)\,d\tau\ ,\qquad t\ge0\ ,$$is a solution of the above with u_0=0. Note first that for t>0$$g(t) = \int_0^t T(t-\tau) (f(\tau)-f(t))\,d\tau + \int_0^t T(t-\tau) \,d\tau\cdot f(t)\ .$$from Theorem \ref{thm4-6B}(c) and the H\"older continuity of f we have$$\|A\cdot T(t-\tau) (f(\tau)-f(t))\| \le C(\theta_0) K|t-\tau|^{\alpha-1}\ ,$$and since A is closed we have g(t)\in D(A) and$$Ag(t) = A\int_0^t T(t-\tau)(f(\tau)-f(t))\,d\tau + (I-T(t))\cdot f(t)\ .$$The result now follows from the computation \eqn{435} in the proof of Theorem \ref{thm4-3C}. \section{Parabolic Equations} % 7 \setcounter{equation}{0} We were led to consider the abstract Cauchy problem in a Hilbert space H $$\label{eq471} u'(t) + Au(t) = f(t)\ ,\qquad t>0\ ;\ u(0)=u_0$$ by an initial-boundary value problem for the parabolic partial differential equation of heat conduction. Some examples of \eqn{471} will be given in which A is an operator constructed from an abstract boundary value problem. In these examples A will be a linear unbounded operator in the Hilbert space L^2(G) of equivalence classes of functions on the domain G, so the construction of a representative U(\cdot,t) of u(t) is non-trivial. In particular, if such a representative is chosen arbitrarily, the functions t\mapsto U(x,t) need not even be measurable for a given x\in G. We begin by constructing a measurable representative U(\cdot,\cdot) of a solution u(\cdot) of \eqn{471} and then make precise the correspondence between the vector-valued derivative u'(t) and the partial derivative \partial_tU(\cdot,t). \begin{theorem}\label{thm4-7A} Let I= [a,b], a closed interval in \RR and G be an open (or measurable) set in \RR^n. \begin{description} \item[{\rm(a)}] If u\in C(I,L^2(G)), then there is a measurable function U:I\to\RR such that $$\label{eq472} u(t) = U(\cdot,t)\ ,\qquad t\in I\ .$$ \item[{\rm(b)}] If u\in C^1(I,L^2(G)), U and V are measurable real-valued functions on G\times I for which \eqn{472} holds for a.e.\ t\in I and$$u'(t)=V(\cdot,t)\ ,\qquad \hbox{a.e. }\ t\in I\ ,$$then V= \partial_tU in \D^* (G\times I). \end{description} \end{theorem} \proof (a) For each t\in I, let U_0(\cdot,t) be a representative of u(t). For each integer n\ge1, let a=t_00$$\Re \Big\{ a_1(v,v) + a_2 \bigl(\gamma(v),\gamma(v)\bigr)\Bigr\} \ge c\|v\|_V^2\ ,\qquad v\in V\ .$$Let U_0\in L^2(G) and a measurable F: G\times [0,T]\to \KK be given for which F(\cdot,t)\in L^2(G) for all t\in [0,T] and for some K\in L^2(G) and \alpha, 0<\alpha\le1, we have$$|F(x,t)-F(x,\tau)| \le K(x) | t-\tau|^\alpha\ ,\ \hbox{ a.e. }\ x\in G\ ,\ t\in [0,T]\ .Then there exists a U\in L^2(G\times [0,T]) such that for all t>0 \label{eq473} \left.\begin{array}{l} U(\cdot,t)\in V\ ,\ \partial_tU(\cdot,t)+A_1U(\cdot,t)=F(\cdot,t)\ \hbox{ in } \ L^2(G)\ ,\\ \noalign{\vskip6pt} \hbox{and }\ \partial_1 U(\cdot,t) + \A_2 (\gamma U(\cdot,t))=0\ \hbox{ in }\ B'\ ,\end{array} \right\} and\lim_{t\to0} \int_G |U(x,t) -U_0(x)|^2\,dx = 0\ .$$\end{theorem} We shall give some examples which illustrate particular cases of Theorem \ref{thm4-7C}. Each of the following corresponds to an elliptic boundary value problem in Section~III.4, and we refer to that section for details on the computations. \subsection{} % 7.1 Let the open set G in \RR^n, coefficients a_{ij}, a_j\in L^\infty(G), and sesquilinear form a(\cdot,\cdot) = a_1(\cdot,\cdot), and spaces H and B be given as in Section III.4.1. Let U_0\in L^2(G) be given together with a function F:G\times[0,T]\to\KK as in Theorem \ref{thm4-7C}. If we choose$$V= \{v\in H^1(G) :\gamma_0 v(s)=0\ ,\ \hbox{ a.e. }\ s\in\Gamma\}where \Gamma is a prescribed subset of \partial G, then a solution U of \eqn{473} satisfies \label{eq474} \left.\begin{array}{l} \displaystyle \partial_t U- \sum_{i,j=1}^n \partial_j (a_{ij} \partial_iU) + \sum_{j=0}^n a_j \partial_j U=F\ \hbox{ in }\ L^2(G\times [0,T])\ ,\\ \noalign{\vskip6pt} U(s,t) = 0\ ,\qquad t>0\ ,\ \hbox{ a.e. }\ s\in \Gamma\ ,\ \hbox{ and}\\ \noalign{\vskip6pt} \displaystyle {\partial U(s,t) \over \partial\nu_A} = 0\ ,\qquad t>0\ ,\ \hbox{ a.e. }\ s\in\partial G\sim\Gamma\ , \end{array} \right\} where{\partial U\over\partial\nu_A } \equiv \sum_{i=1}^n \partial_i U \biggl( \sum_{j=1}^n a_{ij}\nu_j\biggr)$$denotes the derivative in the direction determined by \{a_{ij}\} and the unit outward normal \nu on \partial G. The second equation in \eqn{474} is called the boundary condition of {\it first type\/} and the third equation is known as the boundary condition of {\it second type\/}. \subsection{} % 7.2 Let V be a closed subspace of H^1(G) to be chosen below, H=L^2(G), V_0=H_0^1 (G) and define$$a_1 (u,v) = \int_G \nabla u\cdot \overline{\nabla v}\ ,\qquad u,v\in V\ .$$Then A_1= -\Delta_n and \partial_1 is an extension of the normal derivative \partial/\partial\nu on \partial G. Let \alpha \in L^\infty (\partial G) and define$$a_2(\varphi,\psi) = \int_{\partial G} \alpha (s) \varphi (s) \overline{\psi (s)}\,ds \ ,\qquad \varphi,\psi \in L^2 (\partial G)\ .(Note that B\subset L^2(\partial G)\subset B' and \A_2\varphi = \alpha\cdot\varphi.) Let U_0 \in L^2(G) and F be given as in Theorem \ref{thm4-7C}. Then (exercise) Theorem \ref{thm4-7C} asserts the existence of a solution of \eqn{473}. If we choose V= H^1(G), this solution satisfies \label{eq475} \left.\begin{array}{l} \partial_t U-\Delta_n U= F\ \hbox{ in }\ L^2(G\times [0,T])\ ,\\ \noalign{\vskip6pt} \displaystyle {\partial U(s,t)\over\partial \nu} +\alpha (s)U(s,t)=0\ , \qquad t>0\ ,\ \hbox{ a.e. }\ s\in \partial G \end{array}\right\} If we choose V= \{v\in H^1(G) : \gamma v= constant\}, then U satisfies \label{eq476} \left.\begin{array}{l} \partial_tU-\Delta_n U=F\ \hbox{ in }\ L^2(G\times [0,T])\ ,\\ \noalign{\vskip6pt} U(s,t) = u_0(t)\ ,\qquad t>0\ ,\ \hbox{ a.e. }\ s\in \partial G\ ,\\ \noalign{\vskip6pt} \displaystyle \int_{\partial G} {\partial U(s,t)\over\partial\nu}\,ds + \int_{\partial G} \alpha (s)\,ds \cdot u_0(t) =0\ ,\qquad t>0\ . \end{array}\right\} The boundary conditions in \eqn{475} and \eqn{476} are known as the {\it third type\/} and {\it fourth type\/}, respectively. Other types of problems can be solved similarly, and we leave these as exercises. In particular, each of the examples from Section~III.4 has a counterpart here. Our final objective of this chapter is to demonstrate that the weak solutions of certain of the preceding mixed initial-boundary value problems are necessarily strong or classical solutions. Specifically, we shall show that the weak solution is smooth for problems with smooth or regular data. Consider the problem \eqn{474} above with F\equiv 0. The solution u(\cdot) of the abstract problem is given by the semigroup constructed in Theorem \ref{thm4-6B} as u(t)=T(t)u_0. (We are assuming that a(\cdot,\cdot) is V-elliptic.) Since T(t) \in L(H,D(A)) and AT(t)\in\L(H) for all t>0, we obtain from the identity (T(t/m))^m=T(t) that T(t)\in L(H,D(A^m)) for integer m\ge1. This is an abstract regularity result; generally, for parabolic problems D(A^m) consists of increasingly smooth functions as m gets large. Assume also that a(\cdot,\cdot) is k-regular over V (cf. Section~6.4) for some integer k\ge0. Then A^{-1} maps H^s(G) into H^{2+s}(G) for 0\le s\le k, so D(A^m) \subset H^{2+k} whenever 2m\ge2+k. Thus, we have the spatial regularity result that u(t)\in H^{2+k}(G) for all t>0 when a(\cdot,\cdot) is V-elliptic and k-regular. One can clearly use the imbedding results of Section II.4 to show U(\cdot,t)\in C_u^p(G) when 2(2+k)>2p+n. We consider the regularity in time of the solution of the abstract problem corresponding to \eqn{474}. First note that A^m:D(A^m)\to H defines a scalar product on D(A^m) for which D(A^m) is a Hilbert space. Fix t>\tau>0 and consider the identity(1/h)(u(t+h) - u(t)) = A^{-m}\bigl[ (1/h)(T(t+h-\tau)- T(t-\tau))A^m u(t) \bigr]$$for 0<|h| 0 and integer m. This is an abstract temporal regularity result. Assume now that a(\cdot,\cdot) is k-regular over V. The preceding remarks show that the above difference quotients converge to u'(t) = \partial_t U(\cdot,t) in the space H^{2+k}(G). The convergence holds in C_u^p (G) if 2(2+k)>2p+n as before, and the solution U is a classical solution for p\ge2. Thus, \eqn{474} has a classical solution when the above hypotheses hold for some k>n/2. \exercises \begin{description} \item[1.1.] Supply all details in Section 1. \item[1.2.] Develop analogous series representations for the solution of \eqn{415} and \eqn{413} with the boundary conditions \begin{description} \item[(a)] u_x(0,t)=u_x (\pi,t)=0 of Neumann type (cf.\ Section III.7.7), \item[(b)] u(0,t)= u(\pi,t), u_x(0,t)=u_x(\pi,t) of periodic type (cf.\ Section III.7.8). \end{description} \item[1.3.] Find the solution of the {\it backward heat equation}$$u_t + u_{xx}=0\ ,\qquad 00$$subject to u(0,t) = u(\pi,t) =0 and u(x,0) = \sin (nx)/n. Discuss the dependence of the solution on the initial data u(x,0). \medskip \item[2.1.] If A is given as in Section III.7.C, obtain the eigenfunction series representation for the solution of \eqn{421}. \item[2.2.] Show that if u,v\in C^1((0,T),H), then$$D_t(u(t),v(t))_H = (u'(t),v(t))_H + (u(t),v'(t))_H\ ,\qquad 00$and that its generator is$B-\lambda$. \item[3.2.] Verify the limits as$t\to\infty$in the two identities leading up to Theorem \ref{thm4-3A}. \item[3.3.] Show$B(\lambda-B)^{-1} = (\lambda-B)^{-1}B$for$B$as in Theorem \ref{thm4-3A}. \item[3.4.] Show that Theorem \ref{thm4-3C} holds if we replace the given hypothesis on$f$by$f:\RR^+\to D(A)$and$Af(\cdot) \in C([0,\infty),H)$. \medskip \item[4.1.] Prove Lemma \ref{lem4-4A}. \item[4.2.] Show that the hypothesis in Theorem \ref{thm4-4C} that$D(A)$is dense in$H$is unnecessary. Hint: If$x\in D(A)^\bot$, then$x= (\lambda+A)z$for some$z\in D(A)$and$z=\theta$. \item[4.3.] Show that \eqn{441} follows from$A$being accretive. \item[4.4.] For the operator$A$in Example 4(a), find the kernel and range of$\lambda+A$for each$\lambda\ge0$and$c$with$|c|\le1$. \item[4.5.] Solve \eqn{445} by the methods of Chapter III. \item[4.6.] Solve \eqn{445} and \eqn{446} when the Dirichlet conditions are replaced by Neumann conditions. Repeat for other boundary conditions. \medskip \item[5.1.] Show that operators$\{S_+(t)\}$and$\{S_-(t)\}$commute in the proof of Theorem \ref{thm4-5A}. \item[5.2.] Verify all details in the Example of Section 5. \item[5.3.] If$A$is self-adjoint on the complex Hilbert space$H$, show$iA$generates a unitary group. Discuss the Cauchy problem for the Schrodinger equation$u_t = i\Delta_n u$on$\RR^n\times \RR$. \item[5.4.] Formulate and discuss some well-posed problems for the equation$\partial_t u+\partial_x^3u =0$for$00$. \medskip \item[6.1.] Verify all the estimates which lead to the convergence of the integral \eqn{468}. \item[6.2.] Finish the proof of Theorem \ref{thm4-6E}. \item[6.3.] Show that$f(t)\equiv \int_0^t F(s)\,ds$is H\"older continuous if$F(\cdot) \in L^p(0,T;H)$for some$p>1$. \item[6.4.] Show that for$0<\varep<\theta_0$and integer$n\ge1$, there is a constant$c_{\varep,n}$for which$\|t^n A^nT(t)\| \le c_{\varep,n}$for$t\in S(\theta_0-\varep)$in the situation of Theorem \ref{thm4-6B}. \medskip \item[7.1.] In the proof of Theorem \ref{thm4-7A}(a), verify$\lim_{n\to\infty} \|U_n(\cdot,t)-u(t)\| =0$. For Theorem \ref{thm4-7A}(b), show$\varphi \in C_0^\infty (I,L^2(G))$. \item[7.2.] Give a proof of Theorem \ref{thm4-7B} without appealing to the results of Corollary III.\ref{cor3-3B}. \item[7.3.] Show that the change of variable$u(t) = e^{\lambda t} v(t)$in \eqn{471} gives a corresponding equation with$A$replaced by$A+\lambda$. Verify that \eqn{474} is well-posed if$a_1(\cdot,\cdot)$is strongly elliptic. \item[7.4.] Show that \eqn{473} is equivalent to \eqn{475} for an appropriate choice of$V$. Show how to solve \eqn{475} with a non-homogeneous boundary condition. \item[7.5.] Show that \eqn{473} is equivalent to \eqn{476} for an appropriate choice of$V$. Show how to solve \eqn{476} with a non-homogeneous boundary condition. If$G$is an interval, show {\it periodic boundary conditions\/} are obtained. \item[7.6.] Solve initial-boundary value problems corresponding to each of examples in Sections~4.3, 4.4, and 4.5 of Chapter~III. \item[7.7.] Show that$u(t) = T(t)u_0$converges to$u_0$in$D(A^m)$if and only if$u_0\in D(A^m)$. Discuss the corresponding limit$\lim_{t\to0^+} U(\cdot,t)\$ in \eqn{474}. \end{description}