%% 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
\begin{equation}\label{eq411}
u_t = u_{xx}\ ,\qquad 00
\end{equation}
with the boundary conditions
\begin{equation}\label{eq412}
u(0,t)=0\ ,\quad u(\pi,t)=0\ ,\qquad t>0
\end{equation}
and the initial condition
\begin{equation}\label{eq413}
u(x,0)= u_0(x)\ ,\qquad 00
\end{equation}
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
\begin{equation}\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\ .
\end{equation}
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
\begin{equation}\label{eq417}
u(\cdot,t) = S(t) u_0 (\cdot) + \int_0^t S(t-\tau)f(\cdot,\tau)\,d\tau
\end{equation}
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
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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
\begin{equation}\label{eq421}
u'(t) + Au(t)=0\ .
\end{equation}
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
\begin{equation}\label{eq424}
D\bigl[ \exp (tB)\bigr] = B\cdot\exp (tB) = \exp (tB) \cdot B\ ,\qquad
t\in \RR\ .
\end{equation}
If $B_1,B_2\in \L(H)$ and if $B_1\cdot B_2 = B_2\cdot B_1$, then
\begin{equation}\label{eq425}
\exp (B_1+B_2) = \exp (B_1)\cdot \exp (B_2)\ .
\end{equation}
\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
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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
\begin{equation}\label{eq431}
S(t) x-x = \int_0^t BS(s) x\,ds = \int_0^t S(s) Bx\,dx \ ,\qquad t\ge0\ .
\end{equation} }
\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
\begin{equation}\label{eq432}
S(t) x-x=B\int_0^t S(s)x\,ds\ ,\qquad x\in H\ ,\ t\ge0\ .
\end{equation} }
\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
\begin{equation}\label{eq433}
B_\lambda = -\lambda +\lambda^2 (\lambda-B)^{-1}\ .
\end{equation}
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
\begin{equation}\label{eq434}
u'(t) + Au(t) = f(t)\ ,\qquad t\ge0\ .
\end{equation}
\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
\begin{equation}\label{eq441}
\|(\lambda+A)x\| \ge \lambda \|x\|\ ,\qquad \lambda >0\ ,\ x\in D(A)\ ,
\end{equation}
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
\begin{equation}\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\ ,
\end{equation}
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
\begin{equation}\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}
\end{equation}
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)\ .$$
The result now follows from the computation \eqn{435} in the
proof of Theorem \ref{thm4-3C}.
\section{Parabolic Equations} % 7
\setcounter{equation}{0}
We were led to consider the abstract Cauchy problem in a Hilbert space $H$
\begin{equation}\label{eq471}
u'(t) + Au(t) = f(t)\ ,\qquad t>0\ ;\ u(0)=u_0
\end{equation}
by an initial-boundary value problem for the parabolic partial differential
equation of heat conduction.
Some examples of \eqn{471} will be given in which $A$ is an operator
constructed from an abstract boundary value problem.
In these examples $A$ will be a linear unbounded operator in the Hilbert
space $L^2(G)$ of equivalence classes of functions on the domain $G$, so
the construction of a representative $U(\cdot,t)$ of $u(t)$ is non-trivial.
In particular, if such a representative is chosen arbitrarily, the functions
$t\mapsto U(x,t)$ need not even be measurable for a given $x\in G$.
We begin by constructing a measurable representative $U(\cdot,\cdot)$ of a
solution $u(\cdot)$ of \eqn{471} and then make precise the correspondence
between the vector-valued derivative $u'(t)$ and the partial derivative
$\partial_tU(\cdot,t)$.
\begin{theorem}\label{thm4-7A}
Let $I= [a,b]$, a closed interval in $\RR$ and $G$ be an open (or measurable)
set in $\RR^n$.
\begin{description}
\item[{\rm(a)}] If $u\in C(I,L^2(G))$, then there is a measurable function
$U:I\to\RR$ such that
\begin{equation}\label{eq472}
u(t) = U(\cdot,t)\ ,\qquad t\in I\ .
\end{equation}
\item[{\rm(b)}] If $u\in C^1(I,L^2(G))$, $U$ and $V$ are measurable
real-valued functions on $G\times I$ for which \eqn{472} holds for
a.e.\ $t\in I$ and
$$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]\ .$$
Then there exists a $U\in L^2(G\times [0,T])$ such that for all $t>0$
\begin{equation}\label{eq473}
\left.\begin{array}{l}
U(\cdot,t)\in V\ ,\ \partial_tU(\cdot,t)+A_1U(\cdot,t)=F(\cdot,t)\ \hbox{ in }
\ L^2(G)\ ,\\
\noalign{\vskip6pt}
\hbox{and }\ \partial_1 U(\cdot,t) + \A_2 (\gamma U(\cdot,t))=0\ \hbox{ in }\
B'\ ,\end{array}
\right\} \end{equation}
and
$$\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\}$$
where $\Gamma$ is a prescribed subset of $\partial G$, then a solution $U$
of \eqn{473} satisfies
\begin{equation}\label{eq474}
\left.\begin{array}{l}
\displaystyle \partial_t U- \sum_{i,j=1}^n \partial_j (a_{ij} \partial_iU)
+ \sum_{j=0}^n a_j \partial_j U=F\ \hbox{ in }\ L^2(G\times [0,T])\ ,\\
\noalign{\vskip6pt}
U(s,t) = 0\ ,\qquad t>0\ ,\ \hbox{ a.e. }\ s\in \Gamma\ ,\ \hbox{ and}\\
\noalign{\vskip6pt}
\displaystyle {\partial U(s,t) \over \partial\nu_A} = 0\ ,\qquad
t>0\ ,\ \hbox{ a.e. }\ s\in\partial G\sim\Gamma\ ,
\end{array} \right\}
\end{equation}
where
$${\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
\begin{equation}\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\}
\end{equation}
If we choose $V= \{v\in H^1(G) : \gamma v=$ constant$\}$, then $U$ satisfies
\begin{equation}\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\}
\end{equation}
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}