\def \O {\Omega}
\def \¦ {\partial}
\def \D {\Delta}
\def \o {\omega}
\def \e {\epsilon}
\def \z {\zeta}
\def \d {\delta}
\def \l {\lambda}
\def \G {\Gamma}
\def \t {\tau}
\def \C {{\bf C}}
\def \a {\alpha}
\def \b {\beta}
\def \Re {{\bf R}}
\def \s {\sigma}
\def \S {{\bf S}}
\def \T {{\bf T}}
\def \x {\bf x}

\font\biggrm=cmbx12 scaled \magstep 2
\font\bigrm=cmbx12 scaled \magstephalf
\font\bigrmi=cmmi10 scaled \magstep 2
\font\bfit=cmbxti10 scaled \magstep 1
\font\bfrm=cmbx12
\font\bfss=cmbxti10

\centerline{\biggrm Corrections for First and Second Printings}
\medskip
\centerline{\bigrm   Partial
Differential Equations: Methods and Applications}
\smallskip
\centerline{\bigrm by Robert McOwen}
\smallskip
\centerline{November 1998}

\bigskip\noindent\bf
Introduction\rm

\itemitem{p.\ 7.} Notice that $C_B(\Omega)\subset C(\Omega)$

\itemitem{} For example, $C(\Omega,{\bf R}^N)$ denotes 

\bigskip\noindent\bf
1. First-Order Equations\rm

\item{  } {\it There are some unfortunate inconsistencies in my
discussion of noncharacteristic space curves and the solvability of the Cauchy
problem that occurs on pages 15-16, 18, and 32-3. Corrections are given below.}

\smallskip
\itemitem{p.\ 15} {\it Replace the first paragraph on the page by the following two
paragraphs:}

\hsize 5.5 in\leftskip=1in
In fact, it is also
true that an  integral surface $S$ of the
vector field $V=\langle a,b,c\rangle$ is \it always \rm
a union of characteristic
curves: this can be proved using the uniqueness
theorem for solutions of ordinary differential equations
(see Exercise 1).  Since the solutions of
the characteristic equations are unique, we find that the
integral surface $S$ is unique. This
means that we have proved the following.

\proclaim
Theorem. If $\Gamma$ is noncharacteristic, then the vector field
$V=\langle a,b,c\rangle$ admits a unique integral surface $S$ containing
$\Gamma$.\par

 
To find the solution $u(x,y)$ of (1), it remains only to
replace the variables $s$ and $t$ by expressions
involving $x$ and $y$. Theoretically, this can be achieved 
exactly when the integral surface $S$ is the graph of a function.
We shall discuss this in more 
detail in the next two subsections.

\leftskip=0in
\hsize 6.5in
\smallskip
\itemitem{p.\ 16} \dots satisfied when $\G$ is noncharacteristic {\it and}
$\gamma$ has no stationary points.

\smallskip
\itemitem{} {\bf Example 1.} \dots and we see that (9) holds: $2\not=0$.

\smallskip
\itemitem{p.\ 18} {\it Replace:} Since this geometric condition \dots near $\G$.
{\it By:}\par
\hsize 5.5 in\leftskip=1in 
Therefore, we can solve the Cauchy
problem provided $\Gamma$ is noncharacteristic and $\gamma=(f(s),g(s))$ has no
stationary points.

\leftskip=0in
\hsize 6.5in
\smallskip
\itemitem{p.\ 23.} Exercise 1. \dots through $P$. (Assume the vector field
$V$ is $C^1$.)

\smallskip
\itemitem{} Exercise 5. \dots determine the values of $x$, $y$, and $z$
for which it exists:

\smallskip
\itemitem{p.\ 26.} {\it Displayed formula:} $u(x,y)=$

\smallskip
\itemitem{} \dots for all values of $(x,y)$.

\smallskip
\itemitem{p.\ 28.} Exercise 4. Find a continuous weak solution.

\smallskip
\itemitem{p.\ 29.} Exercise 8.
$$\rho(x,0)=\left\{\eqalign{
{1\over 2}\rho_{max}\quad&\hbox{for}\quad x<0\cr
\rho_{max}\quad&\hbox{for}\quad x>0.}\right.$$

\smallskip
\itemitem{p.\ 32.} \dots (29) and (30) have been fixed, the integral surface $S$
always exists \dots

\smallskip
\itemitem{p.\ 33.} {\it Replace the Theorem by:}

\hsize 5.5in\leftskip=1in
\proclaim
Theorem.  If $\Gamma$ is noncharacteristic for (24)
and functions $\phi, \psi$ exist satisfying (29) and (30),
then there is an integral surface $S$ containing $\Gamma$
(which is unique for the choice of $\phi$ and $\psi$.)\par

\noindent
In order to obtain $u(x,y)$ from $S$, we need to have the Jacobian condition
(8), just as in the quasilinear case.

\hsize 5.5in\leftskip=0in

\smallskip
\itemitem{p.\ 37.} \hangindent=1in
$$x^2+y^2=(1\pm ct^2).\leqno(41)$$

\smallskip
\itemitem{} We may also study solutions of (34) \dots complete integral
for (34) 

\smallskip
\itemitem{p.\ 38-9.} {\bf Exercises 2 \& 6.} Clairaut 

\bigskip\noindent\bf
2. Principles for Higher-Order Equations\rm

\smallskip
\itemitem{p.\ 41.} Differentiating these functions with respect \dots

\smallskip
\itemitem{} \dots use the general solution discussed in Section 1.1.e.

\smallskip
\itemitem{p.\ 45.} \dots and hence $u_{yyxx}(x,0)=0=u_{yxyx}(x,0)=\cdots$

\smallskip
\itemitem{} \dots use to determine $u_{yyy}(x,0)=1+2x-x^2$.

\smallskip\hsize 5.5in
\itemitem{p.\ 46.}\hangindent=1in
$$\partial_y^j\partial_x^k u=(i\partial_x)^j\partial_x^k u
\qquad\hbox{for all}\ j,k\geq 0.$$
$$u(x,y)\sim\cdots {{(i)^jg^{(j+k)}(0)}\over{j!\,k!}}\,y^jz^k
\leqno(4)$$
$$(a+b)^m=\sum_{j+k=m}{{m!}\over{j!\,k!}}\,b^ja^k$$
$$\sum_{m=0}^\infty\sum_{j+k=m}{{m!}\over{j!\,k!}}\,b^ja^k=\sum_{m=0}^\infty(a+b)^m$$


\smallskip\hsize 5.5in\leftskip=0in
\itemitem{p.\ 52.} {\it Every instance of a $\nu$ should be $\eta$, and}
$$B=2a\mu_x\eta_x+{b}(\mu_x\eta_y+\mu_y\eta_x)
+2c\mu_y\eta_y. $$

\smallskip
\itemitem{p.\ 54.}
{\bf Example 8.} Consider $u_{xx}-u_{yy}=0$ as in Example 1 \dots

\smallskip
\itemitem{p.\ 55.} \dots all functions depend continuously on their variables.

\smallskip
\itemitem{p.\ 56.} \par
\hsize 5.5in\leftskip=1in
\dots \it strictly hyperbolic \rm if the $N\times N$ matrix
$A$ has $N$ distinct real eigenvalues \dots (19) is \it hyperbolic \rm if $A$
has all real eigenvalues \dots confirms that $A$ has real eigenvalues and a
basis of eigenvectors; \dots

\hsize 5.5in\leftskip=0in\smallskip
\itemitem{p.\ 58.} Nevertheless, a solution can be shown to exist (see Section
10.1).

\smallskip
\itemitem{p.\ 59.} Exercise 3.
$$u(x,y)=\left\{\eqalign{&0\qquad\hbox{if}\quad x\leq y\cr
(x&-y)^2 \quad\hbox{if}\quad x>y.}\right.$$

\smallskip
\itemitem{ } Exercise 10. \dots i.e.\ $W=\sqrt{4ac-b^2}$, \dots 
new coordinates $\mu,\eta$ transforming \dots

\smallskip
\itemitem{p.\ 62.}\hangindent=1in
$$\int_{\Re^2}\tilde u\, Lv\,d\mu d\eta=
\int_{\mu> 0}\mu^2\, v_{\mu\eta}\,d\mu d\eta=
-\int_{\mu>0}2\mu\, v_\eta\, d\mu d\eta=0$$
$$u_{\mu\eta}=d(\mu,\eta,u,u_\mu,u_\eta),\leqno(39)$$


\smallskip
\itemitem{p.\ 63.}\hangindent=1in
$$\int_{\Omega^{\pm}} u^\pm\, L'v\,d\mu d\eta=
\int_{\Omega^{\pm}} L(u^{\pm}) v\,d\mu d\eta +
\int_{\partial \Omega^\pm} B^\pm (u^\pm ,v)\, ds,
\leqno(41)$$
where $ds$ is arclength and $B^\pm (u,v)$ is an expression \dots

\smallskip
\itemitem{p.\ 65.}\par
\hsize 5.5in\leftskip=1in\noindent
\dots Notice
that we can easily take any number of distributional
derivatives of $\delta(x)$; in fact, $D^\a\delta (x)$
is the object satisfying $\int
D^\a\delta(x)\,v(x)\,dx$ $=(-1)^{|\a|}\int\delta(x)D^\a
v(x)\,dx=(-1)^{|\a|}D^\a v(0)$.

\leftskip=0in\hangindent=1in
$$\int_{\Re^n} \delta_\mu (x)
v(x)\,dx= \int_{\Re^n} \delta (x-\mu) \, v(x)\,dx=
\int_{\Re^n} \delta (y) \, v(y+\mu)\,dy=v(\mu).$$
\dots Therefore, a
distribution should simply be thought of as a linear
mapping $C_0^\infty(\Re^n)\to\Re$.

\hsize 5.5in\leftskip=0in\smallskip
\itemitem{p.\ 66.} {\it After (47):} \dots if $f,g\in L^1_{\ell oc}(\Re^n)$, \dots

\smallskip
\itemitem{p.\ 67.}\hangindent=1in
$$D^\a(F\star G)=(D^\a F)\star G=F\star D^\a G,
\leqno(52)$$
so derivatives of a convolution can be taken on either
factor. In particular, if $f\in C^k(\Re^n)$ and $g\in L^1_{\ell
oc}(\Re^n)$, one of which has compact support, then the lemma above
implies $f\star g\in C^k(\Re^n)$.

\smallskip
\itemitem{p.\ 69.} Assuming that $f\in L^1(\Re^n)$ has compact support, \dots

\smallskip
\itemitem{p.\ 70.}

\hsize 5.5in\leftskip=1in\noindent\hangindent=.2in
1.\ (a) For complex-valued functions $u$ and $v$ on $\O$,
let $\langle u,v\rangle=\int_\O u\overline v dx$, where
$\overline v$ denotes the complex conjugate of $v$. If the coefficients
$a_\a(x)$ in (32a) are complex-valued functions, define the adjoint
$L^*$ so that $\langle Lu,v\rangle=\langle u,L^*v\rangle$ for all $u\in
C^m(\O)$, $v\in C_0^m(\O)$.

\noindent \hsize 5.5in\leftskip=1in\hangindent=.2in
(b) For (complex) vector-valued functions $\vec
u(x)=(u_1(x),\dots,u_N(x))$ on $\O$, let $\langle \vec
u,\vec v\rangle=\int_\O (u_1\overline v_1+\cdots u_N\overline v_N) dx$.
If the coefficients $a_\a(x)$ in (32a) are $N\times N$ matrix-valued
functions, then (32a) defines a \it system \rm of $m$th-order operators.
Define the adjoint $L^*$ so that $\langle L\vec u,\vec v\rangle=\langle
\vec u,L^*\vec v\rangle$ for all $\vec u\in C^m(\O,\Re^N)$, $\vec v\in
C_0^m(\O,\Re^N)$.

\smallskip\leftskip=0in
\itemitem{p.\ 70.} Exercise 2. Let $Lu=u_{\mu\xi}$\dots (b)\dots
with respect to $d\mu$ and then\dots

\smallskip
\itemitem{p.\ 71.} Exercise 5.
(b) For $\O=\Re^n$, find an infinite sequence
$\{\xi_k\}_{k=1}^\infty$ \dots

\smallskip
\itemitem{} Exercise 10.
(b)
$$F(x)=\left\{\eqalign{a^{-1}\sinh&\, ax\qquad\hbox{if}\quad x>0 \cr
&0 \qquad\quad \hbox{if}\quad x<0.}\right.$$

\smallskip
\itemitem{p.\ 72.} Exercise 13.
$$\eqalign{F_1(\xi,\eta)&=\,H(\xi)\,H(\eta),
\qquad
F_2(\xi,\eta)=-\,H(\xi)\,H(-\eta), \cr
F_3(\xi,\eta)&=-\,H(-\xi)\,H(\eta),
\qquad
F_4(\xi,\eta)=\,H(-\xi)\,H(-\eta).}$$


\bigskip\noindent\bf
3. The Wave Equation\rm

\smallskip
\itemitem{p.\ 76.} In fact, if $F(\mu)$ and\dots rectangle ABCD in the
$\mu\eta$-plane\dots Using $u(\mu,\eta)=F(\mu)+G(\eta)$\dots

\smallskip
\itemitem{p.\ 82.} \dots a particular solution for a nonhomogeneous equation.

\smallskip
\itemitem{} Exercise 3.
Consider the initial/boundary \dots

\smallskip
\itemitem{p.\ 89.} \dots (see Exercise 7) with $c_n=(n-2)(n-4)\cdots 1$:\par
\hangindent=1in
$$\eqalign{u(x,t)=&
{{1}\over {c_n\o_n}} {{\¦ }\over {\¦ t}}
\left(\left({{1}\over {t}} {{\¦ }\over {\¦ t}}\right)
^{{n-3}\over {2}}\, t^{n-2}\,
\int_{|\xi|=1}g(x+ct\xi)\,dS_\xi\right)\cr
&+{{1}\over {c_n\o_n}}\left({{1}\over {t}}
{{\¦ }\over {\¦ t}}\right)^{{n-3}\over {2}}\,t^{n-2}
\int_{|\xi|=1}h(x+ct\xi)\,dS_\xi.}\leqno(42)$$

\smallskip
\itemitem{p.\ 90.} Exercise 2. \dots (a) by using (37), and \dots

\smallskip
\itemitem{} Exercise 5.
$$\left\{\eqalign{&v_{tt}=c^2\Delta v -m^2 v
\quad\hbox{for}\; x\in\Re^2\;\hbox{and}\;t>0 \cr
&v(x,0)=g(x),
\qquad v_t(x,0)=h(x).}\right.$$

\smallskip
\itemitem{p.\ 93.}
Notice that $\partial\Omega_\tau=C_\tau\cup(\overline{B}_0\times\{0\})\cup
(\overline{B}_\tau\times\{\tau\})$\dots
 Moreover, the exterior unit
normal $\nu$ on $\partial\Omega_\tau$ is given on \dots

\smallskip
\itemitem{p.\ 97.}\hangindent=1in
$$u_{\xi\eta}-{{\a+\b}\over 4}u_\xi+{{\a-\b}\over 4}u_\eta-{\gamma\over
4}u=0.$$
$$w_{\xi\eta}+{\l\over 4}w=0\quad\hbox{where}\ \l={{\a^2-\b^2-4\gamma}\over 4}
\leqno(62)$$

\bigskip\noindent\bf
4. The Laplace Equation\rm

\smallskip
\itemitem{p.\ 100.} \dots div$\vec E$ has a natural interpretation as \dots

\smallskip
\itemitem{p.\ 102.}\par
\hsize 5.5in\leftskip=1in\noindent
The equation $X''+n^2X=0$ has solutions $X_0=c_0t+d_0$ and
$X_n(t)=c_ne^{nt}+d_ne^{-nt}$ for $n=1,\dots$. This means that
$u_0(r,\theta)=-c_0\log r+d_0 $ and
$u_n(r,\theta)=(a_n\cos{n\theta}+b_n\sin{n\theta})
(c_nr^{-n}+d_nr^{n})$ for $n=1,\dots$

\smallskip\leftskip=0in
\itemitem{p.\ 109.}
Taking $c_2$ so that
$\bar c_2\o_n=-1$, we find \dots

\smallskip
\itemitem{p.\ 110.}\hangindent=1in
$$u_\xi(x)=\int_{{\bf R}^3}{{\delta_\xi(y)}\over{|x-y|}}\,dy=
-4\pi\int_{{\bf R}^3}K(x,y)\delta_\xi(y)\,dy.\leqno(26)$$ Now let us replace
$\delta_\xi(y)$ by
 $\rho(y)$, \dots
$$u(x)=\int_{\O}{{\rho(y)}\over{|x-y|}}\,dy=
-4\pi\int_{\O}K(x,y)\rho(y)\,dy\leqno(27)$$
\dots compactly supported distribution $-4\pi\rho$; by the results of Section
2.3, $u$ is a distribution solution of $\D u=-4\pi\rho$. \dots \it expect \rm
that $u\in C^2(\O)$ with $\D u=-4\pi\rho$ in $\O$, \dots
$$\int_{\partial\O}{{\rho(y)}\over{|x-y|}}\,dS_y
\quad\hbox{and}\quad
\int_{\partial\O}{{\partial}\over{\partial\nu_y}}(|x-y|^{-1})\mu(y)\,dS_y
\leqno(28)$$
\dots Poisson equations $\D u=-4\pi\rho\d_{\partial\O}$ and 
$\Delta u=4\pi\mu{\partial\over{\partial\nu}}
\delta_{\partial\O}$

\smallskip
\itemitem{p.\ 116.}
\bf Proof. \rm Taking $u\equiv 1$ in (33) shows \dots

\smallskip
\itemitem{p.\ 117.} \dots and $\overline{B_a(0)}\subset\O$, then (44)
implies \dots

\smallskip
\itemitem{p.\ 118.}\hangindent=1in
$$\max_{\xi\in\Gamma}|D^\a u(\xi)|\leq \ \dots \hskip 2in\leqno(49{\rm b})$$

\smallskip
\itemitem{p.\ 119.} If we let $M=\max\{|u(x)|:|x|=a\}$ \dots

\smallskip
\itemitem{p.\ 120.}
Exercise 3. (c)\dots
(In Section 8.2 we shall show \dots

\smallskip
\itemitem{p.\ 129.} \dots will be found to be \it convergence in square
mean: \rm

\smallskip
\itemitem{p.\ 131.}\hangindent=1in
$$\hskip 1in \dots\quad
B_n={{b_n}\over{\sqrt{\l_n}}}={1\over{\sqrt{\l_n}}}\int_\O h(x)\phi_n(x)\,dx.
\leqno(76)$$

\bigskip\noindent\bf
5. The Heat Equation\rm

\smallskip
\itemitem{p.\ 134.} \dots discuss the existence and
uniqueness of solutions to these \dots

\smallskip
\itemitem{p.\ 137.} Exercise 3. Suppose $f(x,t)=\sum_{n=1}^\infty
a_n(t)\phi_n(x)$ \dots

\smallskip
\itemitem{p.\ 138.} Exercise 5. \dots initial temperature distribution
$u(x,y,0)=g(x,y)$, \dots heat diffusion $u(x,y,t)$.

\smallskip
\itemitem{p.\ 144.}\par
\hsize 5.5in\leftskip=1in\noindent
It only remains to show that $u$ is continuous.
For $s$ and $t$ satisfying $0\leq s<t$, let us define
$w(x,t,s)=\int_{\Re^n}K(x,y,t-s)\,f(y,s)\,dy$. We
can apply Theorem 1 with $g(y)=f(y,s)$
to conclude that $w(x,t,s)$
is continuous in $x\in \Re^n$ and $0\leq s\leq t$. Therefore 
$$u(x,t)=\int_0^t w(x,t,s)\,ds$$

\smallskip\leftskip=0in
\itemitem{p.\ 145.} Exercise 4.
$$u(x,t)=\sum_{k=0}^{\infty}
{{1}\over{(2k)!}}x^{2k}{{d^k}\over{dt^k}}e^{-1/t^2}$$

\smallskip
\itemitem{p.\ 146.} Exercise 10. \dots of the nonhomogeneous heat equation
\dots

\itemitem{} Exercise 11. \dots where $g$ is continuous and bounded.

\smallskip
\itemitem{p.\ 147.} We could start with $u\in C^{2;1}(U)$ \dots

\itemitem{}
Now choose $\psi\in C^\infty(\Re^{n+1})$ with
$\psi\equiv 0$ on $B_\e (0)$ and $\psi\equiv 1$ on
$\Re^{n+1}\backslash B_{2\e}(0)$. \dots
But for $(x,t)\in B_\e (\xi)$ and
$(y,s)\in\hbox{supp}\,v\subset B_{4\e}(\xi)\cap (\Re^{n+1}\backslash
B_{3\e}(\xi))$,
we have \dots

\bigskip\noindent\bf
6. Linear Functional Analysis\rm

\smallskip
\itemitem{p.\ 153.} (ii) $\langle x,y \rangle =\overline{\langle y,x\rangle}$
\ (where \dots

\smallskip
\itemitem{p.\ 165.} {\bf Remarks.} \dots the regularity of weak solutions in
Chapter 8 using \dots

\smallskip
\itemitem{p.\ 166.}\hangindent=1in
$$\int_\Omega(-\nu\nabla\vec u:\nabla \vec v+\vec f\cdot\vec
v)\,dx=-\int_\Omega p\,\hbox{div}\vec v\,dx, \leqno(17)$$

\smallskip
\itemitem{p.\ 168.}\hangindent=1in
$$ B_L(u,v)=\int_\Omega \left( \sum_{i,j=1}^n
a_{ij}(x){{\¦ u}\over {\¦ x_i}}
{{\¦ v}\over {\¦ x_j}}\, - \, c(x)uv\right)\, dx
\leqno(23)$$
(and a similar summation sign is intended in the displayed equation after
(25)).

\smallskip
\itemitem{p.\ 178.}
Since $\|\nabla u\|_p\leq C\|u\|_{1,p}$, we obtain\dots

\smallskip
\itemitem{p.\ 179.}\hangindent=1in
$$\eqalign{\int_{\bar B_{t\sigma}} |\nabla u(\bar z)|\,d\bar z
&\leq \left(\int_{\bar B_{t\sigma}} d\bar z\right)^{1/p'}
\left(\int_{\bar B_{t\sigma}} |\nabla u(\bar z)|^p\,d\bar
z\right)^{1/p}\cr&={\bar \o_n}^{1/p'}(t\sigma)^{n/p'}\left(\int_{\bar
B_{t\sigma}} |\nabla u(\bar z)|^p\,d\bar z\right)^{1/p},}$$
and dominating the last term by $\|\nabla u\|_p$, we obtain
$$\left|{1\over{\bar \o
\sigma^n}}\int_{B_{\sigma}}u(z)\,dz-u(x)\right|\leq
2\sigma^{1-{n\over p}}\, {\bar \o_n}^{-{1\over p}}\,\left(\int_0^1 t^{-{n\over
p}}\,dt\right)\, \|\nabla u\|_p.$$

\smallskip
\itemitem{p.\ 180.} \dots
where $C=4\,{\bar \o_n}^{-{1\over p}}\,\int_0^1 t^{-n/p}\, dt<\infty$ since
$n/p<1$.

\smallskip
\itemitem{p.\ 182.}\hangindent=1in
$$\|u\|_{3/2}\leq {1\over 3}\sum_{i=1}^3\|u_{x_i}\|_1
\leq {1\over {\sqrt{3}}}\,\|\nabla u\|_1,$$
\dots moreover, we
have explicitly found the constant $C(3,1)=1/\sqrt{3}$.
$$\|\,|u|^r\,\|_{3/2}\leq{r\over
\sqrt{3}}\|\,|u|^{r-1}\nabla u\,\|_1.$$
$$\|\,|u|^r\,\|_{3/2}\leq{r\over
\sqrt{3}}\|\,|u|^{r-1}\,\|_{p'}\,\|\nabla u\|_p
\leqno(54)$$
$$\left(\int|u|^{{3p}\over {3-p}}\,dx\right)^{2/3}\leq
{{2p}\over {\sqrt{3}(3-p)}}\left(\int|u|^{{3p}\over
{3-p}}\,dx\right)^{{p-1}\over p}\,\|\nabla u\|_p.$$
$$\|u\|_q\leq {{2p}\over {\sqrt{3}(3-p)}}\|\nabla u\|_p
\qquad \hbox{where} \quad q={{3p}\over {3-p}},$$
which is the Sobolev inequality with $C(3,p)={{2p}\over
{\sqrt{3}(3-p)}}$.

\smallskip
\itemitem{p.\ 183.} \dots with the value
$C(n,p)={{p(n-1)}\over {\sqrt{n}(n-p)}}$.

\eject\noindent\bf
7. Differential Calculus Methods\rm

\vglue 2pt
\itemitem{p.\ 204.} \dots is a compact embedding by Corollary A \dots

\vglue 2pt
\itemitem{p.\ 207.}  Notice that the
coercive functional of Example 1 satisfies this condition \dots

\vglue 2pt
\itemitem{p.\ 208.} Exercise 8. (b)
$(1+u_y^2)u_{xx}-2u_xu_yu_{xy}+(1+u_x^2)u_{yy}=0$

\vglue 2pt
\itemitem{p.\ 209.} Exercise 10. Show that the coercive functional of Example 1
satisfies \dots.

\vglue 2pt
\itemitem{p.\ 216.} \dots this is called the \it maximin characterization
\rm of the eigenvalues.
\vglue 2pt
\itemitem{} \qquad The monotonicity with respect to the domain does \it not \rm apply
to Neumann eigenvalues; see Exercise 6 for a counterexample. However, the maximin 
 \dots

\vglue 2pt
\itemitem{p.\ 217.} Exercise 6. Let $\O$ be the square $\{(x,y): -1<x,y<1\}$
and $\hat \O$ be the slitted circle $\{(r,\theta): 0\leq r<1,\ -\pi<\theta<\pi\}$.
Although $\hat\O\subset\O$, show that the Neumann eigenvalues do \it not \rm satisfy
$\mu_n(\hat\O)\geq\mu_n(\O)$ for all $n\geq 1$.

\vglue 2pt
\itemitem{} Exercise 8. $\Delta^2 u-\lambda u=0$

\vglue 2pt
\itemitem{p.\ 220.} \dots the \it perturbation theory \rm of
PDEs; i.e., if $F(0)=0$ and the \it linear \rm equation \dots

\vglue 2pt
\itemitem{p.\ 223.}\hangindent=1in
$$F'(u)v=\hbox{div}\left({{\nabla v}\over{(1+|\nabla
u|^2)^{1/2}}}-{{(\nabla u\cdot\nabla v)\,\nabla u}\over{(1+|\nabla
u|^2)^{3/2}}}
\right). \leqno(54)$$

\vglue 2pt
\itemitem{p.\ 224.} Exercise 4.
(b) Use ${\cal T}_z x-{\cal T}_z y=\int_0^1
{{d }\over{dt}}{\cal T}_z(y+t(x-y))\,dt = \int_0^1
{\cal T}_z'(y+t(x-y))\,(x-y)\,dt$

\bigskip\noindent\bf
8. Linear Elliptic Theory\rm

\vglue 2pt
\itemitem{p.\ 226.}\hangindent=1in
$$\eqalign{
\langle u,&v\rangle_2
=\int_{\T^n}\sum_{|\gamma|=2}D^{\gamma}uD^{\gamma}vdx+
\langle u,v\rangle_1\cr
&=(2\pi)^n \sum_\a
\left(\a_1^4+\a_1^2\a_2^2+\a_2^4+\cdots
+\a_n^4+|\a|^2+1\right)u_\a v_{-\a}\cr
&\|u\|_2^2=(2\pi)^n \sum_\a
(\a_1^4+\a_1^2\a_2^2+\a_2^4+\cdots
+\a_n^4+|\a|^2+1)|u_\a|^2,}\leqno(6{\rm a})$$

\vglue 2pt
\itemitem{p.\ 232.} Exercise 4. (a) Determine the eigenvalues $\l$ for
which\dots

\vglue 2pt
\itemitem{p.\ 236.}\hangindent=1in
$$ \l^{-1}\int_\O\sum_{j,k}^n a_{jk}
{{\partial u}\over{\partial x_j}}{{\partial u}\over{\partial x_k}}\,dx
\leqno(37)$$

\vglue 2pt
\itemitem{p.\ 240.} Exercise 1.
$\|\nabla u\|_{0;\O_1}^2\leq
C\,\left(\|\D u\|_{0;\O}^2+\|u\|_{0;\O}^2\right)$

\vglue 2pt
\itemitem{} Exercise 5. \dots together imply $\partial u/\partial x_j\in
H^k(\O)$.

\vglue 2pt
\itemitem{}
Exercise 8. $$\|u\|_{2;\O}^2\leq
\,C\,\left(\,\|\D
u\|_{0;\O}^2+\|u\|_{1;\O}^2\,\right)\quad\hbox{valid for all}\ u\in
H^{2}(\O).$$


\vglue 2pt
\itemitem{p.\ 241.}
\dots apply the theorem to the inequality $L_0 u\geq -cu\geq 0$ which
\dots

\vglue 2pt
\itemitem{p.\ 242.}\hangindent=1in
$$\max_{x\in\overline\O}u(x)\leq\max_{x\in\¦ \O}u^+(x)
\qquad\hbox{where}\; u^+(x)=\max(u(x),0).
\leqno(49)$$

\vglue 2pt
\itemitem{p.\ 243.} {\it Delete:} notice that the maximum principle (applied to
$-w$) shows $w(x)\geq N\geq 0$.
\itemitem{} {\it Now reads:}
 \dots how to construct $w(x)$.
Now let $v=u-w$. The calculation \dots

\bigskip\noindent\bf
9. Two Additional Methods\rm

\smallskip
\itemitem{p.\ 261.}\hangindent=1in
$$\{\vec u,\vec w,\vec v\}=
\int_\O(\vec u\cdot\nabla)\vec w\cdot\vec v\,dx
\qquad\hbox{for}\ \vec u,\vec w,\vec v\in 
\tilde H_0^{1,2}(\O,\Re^n).\leqno(7{\rm a})$$

\smallskip
\itemitem{p.\ 269.} \dots which is a consequence of $\int_{{\bf R}^n}K(x,y,t)\,dy=1$,
guarantees \dots

\smallskip
\itemitem{p.\ 281.} Exercise 6. 
(c) \dots $\beta_0=\lim_{t\to\infty}(\log\|S(t)\|)/t$.

\bigskip\noindent\bf
11. Linear and Nonlinear Diffusion\rm

\smallskip
\itemitem{p.\ 310.} A calculation shows $Lv-v_t=
\exp(-Ar^2)[4A^2(x-\xi)^2-2aA-2bA(x-\xi)+c+2A(t-\t)]-c\exp(-A\e^2).$
\dots achieve its maximum on $\partial B_\delta$, since $w(x_0,t_0)=\eta v(x_0,t_0)=0$.

\smallskip
\itemitem{p.\ 312.} 4. {\it Proof of (a).} \dots such that $u(x_0,t)<M$ for $t<t_1$,
\dots

\itemitem{} {\bf Corollary.} \dots \sl are nonnegative \dots\rm

\smallskip
\itemitem{p.\ 319.} {\it The proof of Theorem 4 should be replaced with the
following.}

\hsize 5.5in\noindent\hangindent=1in
\hskip 1in \bf Proof. \rm  The {\it Gagliardo-Nirenberg inequality} generalizes (35)
of Chapter 6: $$\eqalign{\|u\|_{\ell,q}&\leq
C\|u\|_{m,p}^\theta\|u\|_r^{1-\theta},\quad \hbox{for all}\ u\in H^{2,2}(\O),\
\hbox{provided}\cr & \ell<m,\ \ell/m\leq\theta\leq 1,\ \hbox{and}\ 
\ell-{n\over q}<\theta(m-{n\over p})-(1-\theta){n\over r};}
\leqno(18{\rm a})$$ 
for a proof of (18a) see [Friedman] (where the desired estimate is even proved
when $\ell-{n\over q}\leq\theta(m-{n\over p})-(1-\theta){n\over r}$, provided
$m-\ell-n/p\not\in{\bf N}$). Let $Y=H^{1,q}(\O)$, take $m=2=p=r$, and
use the elliptic estimate $\|u\|_{2,2}\leq
C\|Au\|_2$ for $u\in D(A)$, to obtain 
$$\|u\|_Y\leq
C\|Au\|_{2}^\theta\|u\|_2^{1-\theta},\quad
\hbox{for all}\ u\in D(A). \leqno(18{\rm b})$$

\hangindent=1in\hskip 1in 
Using Young's inequality, (18b) implies the existence of $K$ for which
$$\eqalign{\|u\|_Y\leq &
\e\,\|Au\|_2+K\,\e^{-\theta/(1-\theta)}\,\|u\|_2\quad\hbox{for all}\ \e>0\ \hbox{and}\ u\in
D(A),\cr
&\hbox{provided}\ 1/2\leq\theta\leq 1\ \hbox{and}\ 1-{n\over q}<2\theta-{n\over 2}.}
\leqno(18{\rm c})$$

\hangindent=1in\hskip 1in
 Another (more general) way of defining $A^{-\a}$ for $\a>0$ is inspired by the
Laplace transform:
$$A^{-\a}={1\over{\Gamma(\a)}}\int_0^\infty e^{-sA}s^{\a-1}\,ds.
\leqno(19)$$
Using (19), we may show that $A^{-\a}:Y\to Y$ is bounded when $\theta<\a\leq 1$; hence
$X_\a\subset Y$ as claimed in (17a). In fact,
$$\|A^{-\a}u\|_Y\leq
{1\over{\Gamma(\a)}}\int_0^\delta\|e^{-sA}s^{\a-1}u\|_Y\,ds+
\int_\delta^\infty\|e^{-sA}s^{\a-1}u\|_Y\,ds.
$$
Use (18c) with $\e=s^{1-\theta}$ and (44) of Section 9.2 to estimate
$\|e^{-sA}s^{\a-1}u\|_Y$ by $C\,s^{\a-\theta-1}\|u\|_2$, which is
integrable near
$s=0$ since $\a-\theta>0$; so the integral over $(0,\d)$ is bounded by
$C\|u\|_2$,
which in turn is bounded by $ C\|u\|_Y$ since $Y\subset L^2(\O)$. Since $e^{-sA}$
is $\b$-contractive with $\b<0$, the integral
over $(\d,\infty)$ is  bounded by $C\|u\|_Y$. Thus
$\|A^{-\a}u\|_Y\leq C\|u\|_Y$.

\hangindent=1in\hskip 1in
The proof of (17b) is similar; in fact, if $2\a>1$ and $m=0$, then
(17a) may be used to obtain (17b).
$\spadesuit$


\hsize 6.5in\smallskip
\itemitem{p.\ 324.} For convenience, let us assume $\int_\O\,dx=1$,
and use the Poincar\'e inequality to conclude 
$$V(t)=\dots\qquad\qquad\qquad\dots \geq(k-\e)\|u\|_{1,2}^2-C_\e.$$

\smallskip
\itemitem{p.\ 325.} \par
\hsize 5.5in\hangindent=1in\noindent\hskip 1in
Taking $\e=k/2$, we obtain
$$\|u\|_{1,2}^2\leq {2\over k}(V(t)+C).
\leqno(29)$$
 \dots the desired bound (24) with $K(t)\equiv 2k^{-1}(V(0)+C)$, \dots

\smallskip
\itemitem{p.\ 326.} {\it Between (35) and (36):} Integrating, we obtain

\bigskip\noindent\bf
13. Nonlinear Elliptic Equations\rm

\smallskip
\itemitem{p.\ 364.} Moreover, $G_w(\l_*,0,0)\colon w\to Lw+QH_u(\l_*,0)w=Lw$,
since $H(\l,u)=F(\l,u)-Lu$ implies $H_u(\l_*,0)=F_u(\l_*,0)-L=0$.

\smallskip
\itemitem{p.\ 369.} {\it After (29), delete} If we take \dots {\it and substitute :}
\par \hangindent=1in\noindent\hskip 1in
Without loss of generality, we
may assume
$g\equiv 0$ in (13); this is achieved by the substitution $v=u-g$, where $g$ has been
extended to $\overline{\O}$. Then $\Delta v=f(x,v+g)+\D g\equiv\tilde f(x,v)$, where
$\tilde f$ is again uniformly bounded.

\hangindent=1in\hskip 1in
 With $g\equiv 0$ in (13), we may take $u_{\pm}\in
C^{\infty}(\overline\O)$ to be solutions of
$$\left\{\eqalign{\D u&=\mp M \qquad\hbox{in}\ \O\cr
u&=0 \qquad\hbox{on}\ \¦ \O.}\right.$$ 
 By the maximum principle, $u_+\geq 0\geq u_-$, so
that Theorem 1 implies that we have a solution
$u\in C^{2,\a}(\overline\O)$ of (13) with $g\equiv 0$. This yields the following
result (for general $g$): 

\smallskip
\hangindent=1in\noindent\hskip 1in
 \bf Theorem 2. \sl If $f(x,u)$ satisfies (16) and
(29), and $g(x)$ satisfies (18),
then there is a solution $u\in C^{2,\a}(\overline\O)$ of (13). \rm

\smallskip
\itemitem{p.\ 370.} {\it Before (32):} so that (13) with $g\equiv 0$ may be written

\smallskip
\itemitem{} {\it After (34):} 
\par \hangindent=1in\noindent\hskip 1in
A solution of (13) with $g\equiv 0$ and $f$
satisfying (30) must satisfy the same equation with $f$ replaced by $f_0$. But we
may then apply Theorem 2. Since $f(x,u)$ satisfies (30) if and only if $\tilde
f(x,u)\equiv f(x,u+g)+\D g$ does, we may extend this solvability to $g\not\equiv 0$:

\smallskip
\hangindent=1in\noindent\hskip 1in\bf Theorem 3. \sl Suppose $f(x,u)$ satisfies (16)
and (30), and $g$ satisfies (18). Then (13) admits a unique solution $u\in
C^{2,\a}(\overline\O)$. \rm

%%%%%%%%%%%%%%%%%%%%%%%

\bigskip
\noindent\bf
Hints and Solutions for Selected Exercises\rm

\smallskip\noindent\hskip.1in
Section 1.1

\itemitem{4.} (a) $u(x,y)={1\over 2}y^2+x^2e^{-2y}$ exists for all $x,y$.

\itemitem{5.} (a) $u(x,y,z)=h(xe^{-z},ye^{-z})e^z$ exists for all $x,y,z$.

\itemitem{6.} (a) $u(x,y)=(3y+1)^{1/3}$.

\itemitem{7.} (a) $u=f((x+u)/(y+u))$, (b) $u=yf(x^2y+y^3+u^2y)$.

\smallskip\noindent\hskip.1in
Section 1.2

\itemitem{3.} The shock curve is given by $x=\xi(y)=1-(1+u_0y)^{1/2}$. The solution is
$u(x,y)=0$ for $x<\xi(y)$, and $u(x,y)=u_0(x-1)/(1+u_0y)$ for $x>\xi(y)$.

\itemitem{7.} (b) Assuming a car has length $\ell$, then $\rho(x,0)=1/\ell$
for
$|x-x_0|\leq\ell/2$, $\rho(x,0)=0$ for $|x-x_0|>\ell/2$ describes a single
car with midpoint at $x_0$. (One could also use a point-mass model for
traffic flow, in which case $\rho(x,0)=\delta_{x_0}(x)$, using the delta
function of Section 2.3.)

\itemitem{8.} The shock curve is $x = \xi(t) = -{1\over 2}\,ct$. (See above
correction for p.\ 29.)

\smallskip\noindent\hskip.1in
Section 1.3

\itemitem{2.} $u(x,y)=\pm y+{1\over 2}(1-x^2)$.

\itemitem{3.} It is possible to solve the equation if $0<a\leq {1\over 4}$. If
$0<a<{1\over 4}$, then there are two solutions:
$u(x,y)=(\sqrt{a}x\pm{\sqrt{1-4a}\over 2}y)^2$. But, for $a={1\over 4}$, there
is a unique solution $u(x,y)={1\over 4}x^2$.

\smallskip\noindent\hskip.1in
Section 2.1

\itemitem{5.}
The solution itself is a polynomal (which converges everywhere):
$$u(x,t)=\sum_{j,k=0}^{2j+k=n}{{(2j+k)!}\over{j!k!}}a_{2j+k}t^jx^k.$$

\smallskip\noindent\hskip.1in
Section 2.2

\itemitem{2.} {(a) $u(x,y) = f(y-x-\cos(x)) + g(y+x-\cos(x))$.}

\itemitem{6.} $u_1(x,t)=x+5t$, $u_2(x,t)=-4t$.

\itemitem{7.} (a) decouples, but (b) does not.

\itemitem{8.} Let $\mu=x-{t\over{\sqrt{LC}}}$, 
$\eta=x+{t\over{\sqrt{LC}}}$, and $k=R/L$. Then
$I(x,t)={1\over 2}[I_0(\mu)+I_0(\eta)+\sqrt{C\over L}V_0(\mu)-
\sqrt{C\over L}V_0(\eta)]e^{-kt}$, and
$V(x,t)={1\over 2}[\sqrt{L\over C}I_0(\mu)-\sqrt{L\over C}I_0(\eta)
+ V_0(\mu)+V_0(\eta)]e^{-kt}$.

\smallskip\noindent\hskip.1in
Section 2.3

\itemitem{1.} {(a) $L^*v = \sum_{|\a|\leq m}
(-1)^{|\alpha|} \,D^\alpha(\overline{a}_{\alpha} v)$.}

\itemitem{}{(b) $L^*\vec v = \sum_{|\a|\leq m}
(-1)^{|\alpha|} \,D^\alpha(\overline{a}_{\alpha}^T\vec v)$, where
$\overline{a}_{\alpha}^T$ denotes the conjugate transpose matrix for
$a_\a$.}

\itemitem{3.}{(a) \dots a weak solution if
$\int_\Omega f(\mu)\,v_\eta(\mu,\eta)\,d\mu\,d\eta = 0$ for all \dots}
\quad{(c) No.}

\itemitem{5.} {(b) \dots extra assumptions on the coefficients $a_k$.}

\itemitem{6.} {(a) \dots Use this fact to conclude $|f\ast
g(y) - f\ast g(x)| \leq \epsilon\,\int|g(z)|\,dz$, whenever \dots}
\itemitem{}{ (b)
Consider the distribution $G=\delta'$ on ${\bf R}^1$ and the function
$f(x)=|x|$. }

\itemitem{7.}{(b) $\langle a\partial_\nu \delta_\Gamma,v\rangle = -\int_\Gamma
\ \nabla (a(z) v(z)) \cdot \nu(z)\,dz$.}

\itemitem{8.} {Use the mean value theorem.}

\itemitem{10.} {(a) Use $e^{-ax}$ as an integrating factor.}

\itemitem{11.} {(a) Apply the Lemma of this section to show that $u
\in C({\bf R})$.}
{ (b)  Write $u(x)={1\over 2}\int_{-\infty}^x(x-y)f(y)\,dy
+{1\over 2}\int_x^{\infty}(y-x)f(y)\,dy$, and show
$h^{-1}[u(x+h)-u(x)]\to{1\over 2}\int_{-\infty}^xf(y)\,dy-{1\over
2}\int_x^\infty f(y)\,dy.$ Finally, use the hypotheses on $f$ to verify
$u'\in C(\Re)$.}
{ (c) Use (b) to compute $u''$.}

\itemitem{12.} {(a) Use $u=u_p+u_h$ to find
$u(x)=u_0-{1\over 2}\int_0^\infty |y|f(y)\,dy
+(u_0'+{1\over 2}\int_0^\infty f(y)\,dy)x+{1\over 2}\int_0^\infty
|x-y|f(y)\,dy$.}

\itemitem{13.} {(b) Use $v_{tt}-c^2 v_{xx}=-4c^2 v_{\mu\eta}$.
One fundamental
solution is
$F_1(x,t)=(2c)^{-1}H(ct+x)H(ct-x)$, having support in
$-ct<x<ct$ and $t>0$.} (See above correction
for p.\ 72.)

\smallskip\noindent\hskip.1in
Section 3.1

\itemitem{1.} (a) $u(x,t)=x^3+3c^2xt^2+c^{-1}\sin(x)\sin(ct)$.

\itemitem{2.} The Fourier series solution is  $$u(x,t) =
{2\over\pi}\sum_{n=1}^\infty {1-\cos(n\pi) \over {n^2}}\,\sin(nx)\,
\sin(nt).$$   

\itemitem{3.} {(a) The Fourier series solution is
$$u(x,t) = {\pi\over 2} + {2\over \pi}\sum_{k=1}^\infty {1 \over
k^2}\,[\cos(k\pi)-1]\,\cos(kx)\,\sin(kt).$$
(b) Use the parallelogram rule for $u_x$ in $R_1$ and $R_2$. The solution
is \it not \rm unique; it may be $C^0$, but it is \it not \rm $C^1$.}

\itemitem{4.} {Show that the solution given by d'Alembert's
formula satisfies the boundary condition $u(0,t)=0$.}

\itemitem{7.} {The solution is given by (9) with $L=\pi$. }
{(a)
$a_n(t)=c_n\sin\l_nt+d_n\cos\l_nt$ where $\l_n=(c^2n^2+m^2)^{1/2}$. }
{(b) If
$c^2<m^2$, then $a_1(t)$ is either trivial, or unbounded.}

\smallskip\noindent\hskip.1in
Section 3.2

\itemitem{1.} {(a) $|\a|^2=c^{-2}$.}
{ (b) $\nabla g(x)=\a F'(\a\cdot x)=-\a
h(x)$.}
{ (c) $u(x,t) = 1 + x_1 - x_2 \pm c\sqrt{2}\,t$.}

\itemitem{2.} \dots and we can solve it by (39).

\itemitem{5.} {Let $u(x_1,x_2,x_3,t) = \cos({m\over c}x_3)\,v(x_1,x_2,t)$, and
show $u$ satisfies a three-dimensional wave equation with initial
conditions.  Compute $u$ from (37), and then use this to compute $v$.
(See above correction for p.\ 90.)}

\itemitem{6.}
{ (b) No. This is a
consequence of Huygen's principle, and may be proved by using (39).}

\smallskip\noindent\hskip.1in
Section 3.3


\itemitem{1.} {Use energy as a measure of the solution, and show a small
change in the Cauchy data (on a compact set) results in only a small change
at a later time.}

\itemitem{2.} {Differentiate ${\cal E}_\O(t)$ with respect to $t$, and then use
the boundary condition to show $\int_\O\sum u_{x_i}u_{x_it}\,dx=-\int_\O
u_t\D u\,dx$.}

\itemitem{4.} {(a) The energy integral is ${\cal E}(t) = \int (|u_t|^2 +
c^2|\nabla u|^2 + q(x)\,u^2)\,dx$, and may be used to define global and
local energy.  (b) Duplicate the proof of Theorem 2 for the energy defined
above.}

\smallskip\noindent\hskip.1in
Section 3.4

\itemitem{5.} $\pm\sqrt{k^2+\l}g'(x)=h(x)$.

\smallskip\noindent\hskip.1in
Section 4.1

\itemitem{2.} {You should get a series of the form $a_oy+\sum_{k=1}^\infty
a_k\,\cos(kx)\,\sinh(ky)$.}

\itemitem{3.} {Let $w=u_1-u_2$ and use 
$\int_\O|\nabla w|^2\,dx=\int_{\partial\O}w{{\partial w}\over{\partial\nu}}\,dS$.}

\itemitem{4.} {(b) Consider
$u(r,\theta)=r(a\cos\theta+b\sin\theta)$.}

\itemitem{5.} Use (9) with either Dirichlet or Neumann boundary condition to
conclude $\int(qu^2+|\nabla u|^2)\,dx=0$.


\itemitem{8.} { Let $v(x) = \exp(-\alpha|x-x_1|^2) - \exp(-\alpha\epsilon^2)$
for $x\in B_\e(x_1)$ where $\a>0$. For $\a$ sufficiently large, show that
$\D v<0$ in $B_{\e/2}(x_0)$. For $\eta>0$ sufficiently small, show $u+\eta
v\leq u(x_0)$ on $B_\e(x_1)\cap B_{\e/2}(x_0)$. Conclude that $\¦ (u+\eta
v)/\¦ \nu\geq 0$ at $x_0$. Compute $\¦ v/\¦ \nu$ at $x_0$.}

\smallskip\noindent\hskip.1in
Section 4.2

\itemitem{7.} {Define $\tilde u(x_1,\cdots,x_n)$ by $\tilde
u(x_1,\dots,x_n) = -u(x_1,\dots,-x_n)$ if
$x_n \leq 0$ and $\tilde u =
u$ if $x_n \geq 0$.  Show \dots.}

\itemitem{12.} Use $n^k=(1+\cdots 1)^k=\sum_{|\a|=k}(|\a|!/\a!)$
to show $|\a|!\leq n^{|\a|}\a!$ for each $\a=(\a_1,\dots,\a_n)$.

\smallskip\noindent\hskip.1in
Section 5.2

\itemitem{1.} {(a) Direct calculations. (b) and (c) Use the substitution
$z=(x-y)/2\sqrt{t}$. (d) Write $u(x,t)-g(x)=\int K(x,y,t)(g(y)-g(x))\,dy$
and imitate the proof of Theorem 3 in Section 4.2.}

\smallskip\noindent\hskip.1in
Section 7.2

\itemitem{6.} For $\hat \O$ you will encounter the half-integral
Bessel function $J_{1/2}(z)=({2\over{\pi z}})^{1/2}\sin z$.

\smallskip\noindent\hskip.1in
Section 7.3

\itemitem{7.} {Show $\langle F'(u)v,\phi\rangle=
-\int_\O {{\nabla v\cdot\nabla\phi}\over{(1+|\nabla u|^2)^{1/2}}}\,dx
+\int_\O{{(\nabla u\cdot\nabla v)\,(\nabla u\cdot\nabla \phi)}\over{(1+|\nabla
u|^2)^{3/2}}}\,dx.$ }

\smallskip\noindent\hskip.1in
Section 9.2

\itemitem{1.} (b) $\dot{\cal E}(t)=\dot{y}\,\ddot{y}+ky\dot{y}=-c\dot{y}^2. $
\itemitem{} (c) With $\mu=\sqrt{4k-c^2}/ 2$,
$${\bf x}(t)=\pmatrix{y\cr{\dot{y}}\cr}=
e^{-ct/2}\pmatrix{
{c\over{2\mu}}\sin\mu t+\cos \mu t&{1\over\mu}\sin\mu t\cr
-{k\over\mu}\sin\mu t & -{c\over{2\mu}}\sin\mu t+\cos\mu t\cr
}{\bf x_0}$$
\itemitem{} (d) 
$${\bf x}(t)=e^{-ct/2}
\pmatrix{1+{c\over 2}t& t\cr -kt & 1-{c\over 2}t\cr
}{\bf x_0}$$

\smallskip\noindent\hskip.1in
Section 11.3

\itemitem{2.} \dots at least by $t=2/\sqrt{\e}$.

\bigskip\noindent\bf
References\rm

\smallskip
\itemitem{p.\ 412.}
[Kevorkian]\  J.\ Kevorkian, \it Partial Differential Equations:
Analytical Solution Techniques\rm , Wads\-worth \& Brooks/Cole,
1990.

\bigskip\noindent\bf
Index\rm

\smallskip
\itemitem{p.\ 418.}
\noindent
Pohozaev's identity, 379

\vfill\eject




\bye