\input HEAD.TEX
\def\Fr{Fr\'{e}chet}
\footline = {\tenrm \folio \hfil {\sevenrm These solutions rev. 6/10/2001.\ 
\ All rights reserved.}}
\centerline{\notsolarge SOLUTIONS FOR CH. 1 PROBLEMS}
\bigskip

{\bf \S1.1 \#2}: 2a is linear.  2b is not linear. 2c is not linear. 2d
is not linear.  2e is linear.\par %ALL OK
\medskip

{\bf \S1.1 \#3 (a)}: Rewriting the equation as $u_t - u_{xx} = -1$, we
see that it is a linear {\bf in}homogeneous equation.  The operator ${\eusm L}u
= u_t - u_{xx}$ is linear; its order (and therefore the order of the
equation) is 2.\par %ALL OK
\medskip

{\bf (b)}: If ${\eusm L}u = u_t - u_{xx} + xu$, then ${\eusm L}$ is a linear operator. 
The equation $u_t - u_{xx} + xu = 0$ is a linear homogeneous equation of
order 2.
\medskip

{\bf (c)}: The operator ${\eusm N}u = u_t - u_{xxt} + uu_x$ is {\bf non}linear,
because the term $uu_x$ is present.  The equation $u_t - u_{xxt} + uu_x
= 0$ is of order 3.
\medskip

{\bf (d)}: The operator ${\eusm L}u = u_{tt}- u_{xx}$ is linear (it is the
1-dimensional {\bf wave operator}).  The equation $u_{tt}-u_{xx} = -x^2$
is a linear, {\bf in}homogeneous equation of order 2.
\medskip

{\bf (e)}: ${\eusm L}u = iu_t - u_{xx} + \ds{u\over x} = 0$ is a linear
homogeneous equation of order 2.
\medskip

{\bf (f)}: ${\eusm N}(u) = u_x (1 + u_x^2)^{-1/2} + u_u(1+u_y^2)^{-1/2}$ is a
{\bf non}linear operator and the equation ${\eusm N}u = 0$ is of order 1.
\medskip

{\bf (g)}: $u_x + e^yu_y = 0$ is a linear homogeneous equation of order
1.
\medskip

{\bf (h)}: ${\eusm N}(u) = u_t + u_{xxxx} + \sqrt{1+u}$ is a {\bf non}linear
operator and the equation ${\eusm N}u=0$ is of order 4.
\medskip

{\bf \S1.1 \#4}: Suppose that ${\eusm L}$ is a linear operator and that
${\eusm L}u_1 = g$ and ${\eusm L}u_2 = g$.  If $v=u_1-u_2$, then

$$\eqalignno{{\eusm L}(v) = {\eusm L}(u_1-u_2) &= {\eusm L}u_1 - {\eusm
L}u_2 &{\rm (by\ linearity)} \cr &= g - g = 0 \; .  \cr }$$

\noindent Thus ${\eusm L}v=0$. % ALL OK
\medskip

{\bf \S1.1 \#5 (a)}: Let $V = \S(a,0,c): a,c\in\real\}$.  To show that
$V$ is a vector space we have to show that if $u_1=(a_1,0,c_1)$ and
$u_2=(a_2,0,c_2)$ are elements of $V$ and $r$ is a real number, then
$ru\in V$ and $u_1+u_2\in V$.  However, $ru=(ra,0,rc)$, and since the
second co\"{o}rdinate of $ru$ equals zero, $ru\in V$.  Similarly,
$u_1+u_2 = (a_1+a_2,0,c_1+c_2)$ and since the second co\"{o}rdinate of
$u_1+u_2$ is zero, $u_1+u_2\in V$. %ALL OK
\medskip

{\bf (b)}: The set of vectors of the form $(a,1,c)$ where $a,c\in\real$
is {\bf not} a vector space.  To see this, note that $0\cdot(a,1,c) =
(0,0,0)$ {\bf cannot} have 1 as its second co\"ordinate, so that
multiplying an element of this set by a certain scalar ($r=0$) results
in a vector that does not belong to the set.
\medskip

{\bf (c)}: The set $V = \{(a,b,c): a,b,c\in\real,\ ab=0\}$ is {\bf
not} a vector space.  To see this, note that $u=(1,0,0)$ and $v =
(0,1,0)$ are elements of the set but their sum $u+v = (1,1,0)$ is not
(since $1\cdot 1 = 1 \not= 0$).
\medskip

{\bf (d)}: The set $V$ of all linear combinations of the vectors
$(1,1,0)$ and $(2,0,1)$ is, by definition, the set of all vectors of the
form $a(1,1,0) + b(2,0,1)$, where $a$ and $b$ may be any real numbers. 
To prove that this is a vector space, we observe that the sum of two
elements of $V$ has the form
$$ [a_1(1,1,0) + b_1(2,0,1)] + [a_2(1,1,0) + b_2(2,0,1)] =
(a_1+a_2)(1,1,0) + (b_1 + b_2)(2,0,1) $$

\noindent which is again of the form required for elements of $V$, and
similarly for any real number $r$
$$r\cdot[a(1,1,0) + b(2,0,1)] = (ra)(1,1,0) + (rb)(2,0,1)$$

\noindent which again has the form required for membership in $V$.
\medskip

In general, if we are given some finite set $\{u_j: j = 1,\ldots,m\}$ of
vectors in a space $\real^n$, then the set of all linear combinations of
these vectors, \IE, vectors of the form
$$a_1u_1 + a_2u_2 + \cdots + a_mu_m \; , $$

\noindent forms a vector space.  The proof of this fact follows the
same pattern.
\medskip
\goodbreak
{\bf (e)}:  Let $V = \{(a,b,c):c -a = 2b,\ a,b,c\in\real\}$.  Then $V$
is a vector space.  This is most easily seen by rewriting the defining
equation for elements of $V$ in the form
$$V = \{(a,b,c): a + 2b - c = 0,\ a,b,c\in\real\} \; . $$

\noindent Then if $(a,b,c) = (a_1,b_1,c_1) + (a_1,b_2,c_2)$ where the
two vectors on the r.~h.~s.~are known to be in $V$, we will have
$(a,b,c) = (a_1+a_2,b_1+b_2,c_1+c_2)$ and therefore
$$\eqalign{ a + 2b - c &= (a_1+a_2) + 2(b_1+b_2) -(c_1+c_2) \cr
&= (a_1 + 2b_1 - c_1) + (a_2 + 2b_2 - c_2) = 0 + 0 = 0 \;, \cr }$$

\noindent showing that $(a,b,c)\in V$.  Similarly, if we multiply an
element $(a,b,c)$ known to belong to $V$ by a real scalar $r$, obtaining
the vector $(ra,rb,rc)$, then we shall have
$$ (ra) + 2(rb) - (rc) = r\cdot(a + 2b - c) = r\cdot 0 = 0 $$

\noindent showing that if $(a,b,c)\in V$ then $r(a,b,c)\in V$.  Thus we
have shown that $V$ meets the requirements for a vector space. %ALL OK
\medskip

{\bf \S1.1 \#6}: Recall that a set of $n$ vectors belonging to $\real^n$
will be linearly independent if and only if the determinant of the
$n\times n$ matrix formed by using these vectors as rows (if they are
row vectors) or columns (if they are column vectors) is nonzero.  If
(and only if) a set of $n$ vectors belonging to $\real^n$ is linearly
independent, it will form a {\bf basis} of $\real^n$: any vector in
$\real^n$ can be written in one and only one way as a linear combination
of the vectors in that set.
\medskip

Thus to test the linear independence of the given vectors we may form
the determinant
$$\left\vert\matrix{1 &2 &3 \cr -2 &0 &1 \cr 1 &10 &17 \cr }\right\vert
= \left\vert\matrix{1 &2 &3 \cr 0 &4 &7 \cr 0 &8 &14 \cr}\right\vert =
0 \; .$$

\noindent (Recall that adding multiples of one row of a matrix to
another row does not change the value of the determinant of the matrix).
These vectors fail the test for linear independence and are {\bf not}
linearly independent and therefore can {\bf not} span all vectors.
\medskip

{\bf \S1.1 \#7}: A good way to test the linear (in)dependence of a finite
set of highly differentiable functions is the {\bf Wronskian
determinant}, familiar to many people (if not to all) from o.~d.~e.~courses. 
If the functions are $\{f_1,\ldots,f_n\}$ one forms the determinant
$$W(x) = \left\vert\matrix{f_1(x) &f_2(x) &\cdots &f_n(x) \cr
f_1'(x) &f_2'(x) &\cdots &f_n'(x) \cr \vdots &\vdots &\ddots &\vdots \cr
f_1^{(n-1)}(x) &f_2^{(n-1)}(x) &\cdots &f_n^{(n-1)}(x) \cr}\right\vert \;. $$

\noindent If there exist constants $\{\alpha_1,\ldots,\alpha_n\}$ such
that $\ds{\sum_{j=1}^n \alpha_jf_j(x)\equiv 0}$, then the columns of the
Wronskian will be linearly dependent and it will be identically zero.
(The converse is true---in general---only when the $f_j$'s are solutions
of a single linear homogeneous o.~d.~e., but we don't have to worry
about that here.)  Thus if the Wronskian is not identically zero the
functions cannot be linearly dependent.  The Wronskian of the three
functions of this problem is
$$ W(x) = \left\vert\matrix{1+x &1-x & 1+x+x^2 \cr 1 &-1 &1+2x \cr 0 &0
&2 \cr}\right\vert \equiv -4 \not= 0 \; . $$

\noindent Thus these functions must be linearly independent.
\medskip
\goodbreak
{\bf \S1.1 \#9}: Just to proceed na\"{\i}vely at first (because anybody
can see that $\sin^2x = 1 - \cos^2x$ implies that $\{1,\cos^2x\}$ is a
basis for this space of functions) let's set up the Wronskian again.  We
get
$$ W(x) = \left\vert\matrix{1 &\sin^2x & \cos^2x \cr 0 &2\sin x\cos x
&-2\sin x \cos x \cr 0 &2(\cos^2x - \sin^2x) &-2(\cos^2x - \sin^2x)
\cr}\right\vert \equiv 0 \; .  $$

\noindent This says that linear dependence of the generating set
$\{1,\sin^2x,\cos^2x\}$ is a possibility, and we would want to proceed
further, obtaining what the Pythagorean identity told us.
\medskip

{\bf \S1.1 \#10}: This is a linear homogeneous o.~d.~e., and it is a part
of {\bf everybody's} o.~d.~e.~course to show that if $u_1$ and $u_2$ are
solutions of such an o.~d.~e.~and $\alpha_1$, $\alpha_2$ are constants,
then $\alpha_1u_1+\alpha_2u_2$ is also a solution.  That makes the set
of solutions of such an o.~d.~e.~a vector space of functions.  Now one can
{\bf find} a basis for the solution space of such an o.~d.~e.~by finding a
linearly independent set of solutions of the equation, such that the
number of elements of that set equals the order of the
equation.\footnote{\SS{1}}{\sevenrm This is part of the uniqueness
theorem for o.~d.~e.'s, so we take it for granted in this course.} A
standard way to do that with constant-coefficient equations is to
substitute $e^{rt}$ into the equation and see what values of $r$ will
make it be a solution.  For this particular equation, this approach
leads to the equation
$$ e^{rt}(r^3 - 3r^2+4) = 0$$

\noindent which can hold identically if and only if $r$ is one of the
roots of $(r^3 - 3r^2+4) = (r+1)(r-2)^2=0$.  Thus $e^{-t}$ and $e^{2t}$
are solutions of this equation, and because of the repeated root,
$te^{2t}$ is also a solution.  The Wronskian of these three functions is
$$\left\vert\matrix{e^{-t} &e^{2t} &te^{2t} \cr -e^{-t} &2e^{2t}
&e^{2t}+2te^{2t} \cr e^{-t} &4e^{2t} &3e^{2t}+4e^{2t} \cr }\right\vert =
6e^{3t} \not\equiv 0 \; , $$

\noindent and so these three functions form a basis of the solution
space.
\medskip

{\bf \S1.1 \#11}: If $f(x)$ and $g(y)$ are differentiable functions and
$u(x,y) = f(x)g(y)$, then $u_x = f'(x)g(y)$, $u_y = f(x)g'(y)$, and
$u_{xy} = f'(x)g'(y)$.  It follows that
$$u\cdot u_{xy} = f(x)g(y)\cdot f'(x)g'(y) = [f'(x)g(y)]\cdot[f(x)g'(y)]
= u_xu_y \; . $$

{\bf \S1.1 \#12}: If $u_n(x,y) = \sin nx \sinh ny$ then for every $n>0$
$$u_{xx} + u_{yy} = (-n^2 \sin nx)(\sinh ny) + (\sin nx)(n^2\sinh y) = (-n^2
+ n^2)\sin nx \sinh ny = 0 \; . $$
%\S1.1 ENDS HERE

{\bf \S1.2 \#1}: This equation says that $u$ is constant along any curve
satisfying $\ds{{dt}\over{d\tau}}=2$, $\ds{{dx}\over{d\tau}}=3$, or
equivalently any line $x=\ds{3\over 2}\,t + C$.  Thus, ``solving for
$C$'' as in the examples of \S1.2, we find that $\ds{u(x,t) = f\left(x -
{3\over 2}\,t\right)}$ satisfies the equation for any choice of the
(differentiable) function $f$.  The initial condition $u(x,0)\equiv \sin
x$ says $f(x) = \sin x$, and the solution satisfying it is therefore
$$u(x,t) = \sin\left(x - {3\over 2}\,t\right) \; . $$

{\bf \S1.2 \#2}: Let $v=u_y$; then if $3u_y + u_{xy} = 0$ we have $v_x +
3v= 0$.  This is a linear homogeneous 1st order o.~d.~e.~in the variable
$x$, and its solution is $v(x,y) = C(y)e^{-3x}$, where the value of the
``constant of integration'' normally called $C$ may depend on the value
of $y$.  We now know $u_y(x,y) = C(y)e^{-3x}$---another o.~d.~e., of even
simpler form, with $y$ as the independent variable---so if $G(y)$ is a
differentiable function of $y$ for which $G'(y)= C(y)$ and $F(x)$ is any
function of $x$ (the ``constant of integration'' for this o.~d.~e.) then
$$u(x,y) = G(y)e^{-3x} + F(x) $$

\noindent is a solution of the original equation.  It is easily verified
that any function having this form is a solution of the original
equation.
\medskip

{\bf \S1.2 \#3}:  To solve $(1+x^2)u_x+u_y=0$, we solve $\ds{
{{dy}\over{dx}} = {1\over{1+x^2}}}$, whose solutions are of the form
$y(x) = \arctan(x) + C$.  Then $u(x,y(x)) = f(C)$, or $u(x,y) =
f(y-\arctan(x))$, where $f(\cdot)$ is an arbitrary differentiable
function.
\medskip

{\bf \S1.2 \#3}: The same method that we applied to \#1 above says that
$u$ is constant along any curve satisfying $\ds{{dx}\over{dt}}=1 + x^2$,
$\ds{{dy}\over{dt}}=1$, or equivalently the graph of any function $y(x)$
satisfying $\ds{ {{dy}\over{dx}}={1\over{1+x^2}} }$.  These are the
functions $y = \arctan x + C$.  Solving for the constant shows that
$u(x,y) = f(y- \arctan x)$ will be a solution for any (differentiable)
function $f$.  The characteristic curves of this equation are the
vertical translates of the graph of the function $\arctan x$.
\medskip

{\bf \S1.2 \#4}: If $f(\cdot)$ is a differentiable function and $u(x,y)
= f(e^{-x}y)$, then the chain rule gives $u_x = -e^{-x}y\cdot
f'(e^{-x}y)$, $u_y = e^{-x} \cdot f'(e^{-x}y)$, and thus $u_x + yu_y =
[-e^{-x}y\cdot f'(e^{-x}y)] + y\cdot[e^{-x} \cdot f'(e^{-x}y)] = 0$. 
\medskip

{\bf \S1.2 \#5}: Solutions of $\sqrt{1-x^2}u_x + u_y=0$ are constant on
curves satisfying $\ds{ {{dx}\over{dt}}=\sqrt{1-x^2}}$, $\ds{
{{dy}\over{dt}} = 1}$, or $\ds{ {{dy}\over{dx}}={1\over\sqrt{1-x^2}} }$.
The solutions of that o.~d.~e.~are $y = \arcsin x + C$, and thus functions
of the form $u(x,y)=g(y-\arcsin x)$ are solutions of the p.~d.~e.  For
$u(0,y) = g(y) = y$ the solution is therefore simply $u(x,y) =y-\arcsin
x$.
\medskip

{\bf \S1.2 \#6 (a)}: The same method that we've applied twice says that
$u$ is constant along any curve satisfying $\ds{{dx}\over{dt}}=y$,
$\ds{{dy}\over{dt}}=x$; the graphs of functions $y(x)$ satisfying $\ds{
{{dy}\over{dx}}={x\over y} }$ will have this property.  Formal solution
of this o.~d.~e.~leads to $y^2 = x^2 + C$.  Depending on whether $C$ is
positive, negative or zero, solving for $y$ will lead to the family of
hyperbolas $y = \pm\sqrt{x^2+C}$ opening ``up and down,'' the family of
hyperbolas $x = \pm\sqrt{y^2-C}$ opening ``left and right,'' and the two
$45\INT$ lines $y = \pm x$, respectively.  If we want $u(0,y) =
e^{-y^2}$, then ``solving for $C$'' gives $u(x,y) = e^{x^2-y^2}$ as a
solution satisfying that condition, and in the region $|y| > |x|$ this
is the only possible solution, since the characteristic curve through
any point of that region (one of our first family of hyperbolas) also
passes through the $y$-axis.  {\bf (b)} However, in the region
$|y|\le|x|$ we could consider the function that is identically 1.  This
satisfies the p.~d.~e.~and takes the same values that $e^{x^2-y^2}$ does
on the bounding $45\INT$ lines $y = \pm x$.  Setting
$$u(x,y) = \cases{e^{x^2-y^2} &\ \ \ for $|y| > |x|$ \cr 1 &\ \ \ for
$|y|\le|x|$ \cr }$$

\noindent would then give us a function---not differentiable along the
lines $y = \pm x$, but still continuous---satisfying the p.~d.~e.~and the
condition on $u(0,y)$.  Do we want to admit this function, with its
``crinkle'' along the $45\INT$ lines, as a solution?  We seem to admit
things similar to it in the D'Alembert solution of the wave equation
(see ahead to \S2.1 and its ``plucked string'' Example 2, p.~35 ff.).  If
we do, we have lost uniqueness.
\medskip

{\bf \S1.2 \#7}: If we apply to this equation (even though it has a
``$u$ term'') the ``change of co\"{o}rdinates method,'' exactly as given
on p.~7 of Strauss's book, we shall find that $au_x + bu_y =
(a^2+b^2)u_{x'}$, exactly as he does there.  The equation thus
transforms into $(a^2+b^2)u_{x'} + cu = 0$ in the new (``primed'')
variables.  This is a {\bf linear homogenenous o.~d.~e.~of first order},
having the form $Au' + Cu = 0$, or $u' + \ds{C\over A}\,u = 0$.  We know
how to solve those: the solution has the form $\ds{u = {\rm
(const.)}\cdot\exp\left(-{C\over A}\,x'\right)}$, though in this case
the ``constant of integration'' may be a function of the ``missing
co\"ordinate'' or ``parameter'' $y'$.  We thus have
$$u(x',y') = g(y')\cdot\exp\left(-{c\over {a^2+b^2}}\,x'\right) \; . $$

\noindent Transforming the r.~h.~s.~back into $(x,y)$-co\"ordinates gives
$$u(x,y) = g(bx-ay)\cdot\exp\left(-{c\over {a^2+b^2}}\,(ax+by)\right)
\eqno{(*)}$$

\noindent which may not look like the solution in the back of the book. 
However, under the assumption that $a\not=0$ the algebraic manipulation
$${{cx} \over a} - {{c(ax+by)}\over{a^2+b^2}} =
{{bc}\over{a^2+b^2}}\,(bx-ay) $$

\noindent tells us that 
$$\exp\left(-{c\over {a^2+b^2}}\,(ax+by)\right) = \exp\left({{-cx}\over
a}\right) \cdot\exp\left( {{bc}\over{a^2+b^2}}\,(bx-ay)\right) $$

\noindent and thus
$$\eqalign{ u(x,y) &= g(bx-ay)\cdot\exp\left(-{c\over
{a^2+b^2}}\,(ax+by)\right) \cr &= \left\{ g(bx-ay)\cdot \exp\left(
{{bc}\over{a^2+b^2}}\,(bx-ay)\right)\right\}\cdot\exp\left({{-cx}\over
a}\right) \cr }$$

\noindent where the last line has the form
$\ds{f(bx-ay)\cdot\exp\left({{-cx}\over a}\right)}$, as in the back of
the book.\footnote{\SS{2}}{\sevenrm The solution in the back of the
book, at least in my printing, gives nonsense in the case ${\scriptstyle
a=0,\> b \not= 0}$, but the solution ${\scriptstyle (*)}$ given
here---without the manipulations needed to make it conform to the
back-of-the-book solution, of course---checks out in all cases.}
\medskip

{\bf \S1.2 \#8}: The new twist in this equation $u_x+u_y+u = e^{x+2y}$
is the presence of the 0-th order term on the l.~h.~s.  One can handle
it using the ``integrating factor'' approach familiar from o.~d.~e.: an
equation of the form $w'(t) + a(t)w(t) = f(t)$ is handled by multiplying
both sides by $\exp(A(t))$, where $A(t)$ is any antiderivative of
$a(t)$.  The l.~h.~s.~of the equation then becomes the derivative of a
product: $$\eqalign{e^{A(t)}w'(t) + e^{A(t)}a(t)w(t) &= e^{A(t)}f(t)\cr
{d\over{dt}}\left[ e^{A(t)}w(t)\right] &= e^{A(t)}f(t)\cr e^{A(t)}w(t)
&= \int_0^t e^{A(\xi)}f(\xi)\,d\xi + C\cr w(t) &= e^{-A(t)}\int_0^t
e^{A(\xi)}f(\xi)\,d\xi + Ce^{-A(t)}\cr}$$

\noindent gives the general solution of the equation.  So in the
p.~d.~e.~at hand we can multiply by $e^x$:
$$\eqalign{ u_x+u+u_y &= e^{x+2y} \cr e^xu_x+e^xu+e^xu_y &= e^{2x+2y}
\cr {\partial\over{\partial x}}\,[e^xu]+{\partial\over{\partial
y}}\,[e^xu] &= e^{2x+2y} \cr {{\partial v}\over{\partial x}}+{{\partial
v}\over{\partial y}} &= e^{2x+2y} \hbox{\ \ where $v(x,y) = e^xu(x,y)$.}
\cr }$$

\noindent The l.~h.~s.~of this equation we know how to handle:
introducing new co\"{o}rdinates $\xi=x+y$, $\eta = x-y$ so that
$x = (\xi+\eta)/2$, $y = (\xi-\eta)/2$ and 
$\ds{ {\partial\over{\partial x}} = {\partial\over\partial\xi}}
+{\partial\over\partial\eta}$,
$\ds{ {\partial\over{\partial y}} = {\partial\over\partial\xi}}
-{\partial\over\partial\eta}$, we turn the equation into
$$\eqalign{ v_\xi + v_\eta + v_\xi - v_\eta &= e^{2\xi} \cr 
2v_\xi &= e^{2\xi} \; . \cr }$$

\noindent Since $\ds{\partial\over\partial\eta}$ is absent from this
equation, integrating it is an o.~d.~e.~problem with the solution
$v(\xi,\eta) = \ds{{e^{2\xi}}\over 4} + g(\eta)$ (the ``constant of
integration'' can---in general will---depend on $\eta$).  Transforming
back to $(x,y)$ co\"{o}rdinates gives the general solution
$$\eqalign{ e^xu(x,y) = v(x,y) &= {{e^{2x+2y}}\over 4} + g(x-y) \cr
u(x,y) &= {{e^{x+2y}}\over 4} + e^{-x}g(x-y) \; . \cr }$$

\noindent The condition $u(x,0)\equiv 0$ gives
$$\eqalign{ 0 = u(x,0) &= {{e^{x}}\over 4} + e^{-x}g(x) \cr 
{{-e^{2x}}\over 4} &= g(x) \cr }$$

\noindent so we have identified the function $g(\cdot)$, and the
solution function we seek is therefore
$$ u(x,y) = {{e^{x+2y}}\over 4} - e^{-x}\cdot{{e^{2(x-y)}}\over 4} =
{{e^{x+2y}}\over 4} - {{e^{x-2y}}\over 4} = {{e^x}\over 2} \sinh 2y\; . 
$$

Other approaches to the solution are possible.  For example,
if one does \#7 first, then one knows that the general solution of the
associated homogeneous equation $u_x + u_y + u = 0$ is $u(x,y) =
e^{-x}\cdot f(x-y)$.  Just as in the o.~d.~e.~situation, one can then seek
a {\bf particular solution} of \#8 that has the form
``constant-times-r.~h.~s.''  Plugging $u = Ce^{x+2y}$ into the equation
and attempting to solve for $C$, one gets success for $C = 1/4$.  Thus
the general solution of \#8 has the form
$$u(x,y) = {1\over 4}\,e^{x+2y} + e^{-x}\cdot f(x-y)$$

\noindent and applying the condition $u(x,0)\equiv 0$ to determine the
function ``$f(x)$'' leads to the same result as before.  Or one might
try the change-of-co\"{o}rdinates approach on the equation $u_x+u_y+u =
e^{x+2y}$ directly, just to see what happens.  Under the change of
co\"{o}rdinates $\xi=x+y$, $\eta = x-y$ (etc.) the equation transforms
into $\ds{u_\xi + {1\over 2}u = {1\over 2}\,e^{(3\xi - \eta)/2}}$, an
o.~d.~e.~with $\xi$ as its independent variable and $\eta$ as a parameter. 
Using an integrating factor $e^{\xi/2}$ in the manner described
above makes the equation into $[e^{\xi/2}u]_\xi = \ds{1\over
2}\,e^{(4\xi - \eta)/2}$, which integrates to $e^{\xi/2}u = \ds{1\over
4}\,e^{(4\xi - \eta)/2} + \widetilde{g}(\eta)$ or $u = \ds{1\over
4}\,e^{(3\xi - \eta)/2} + e^{-\xi/2}\widetilde{g}(\eta)$, which back in
$xy$-co\"{o}rdinates is $u = \ds{1\over 4}\,e^{x + 2y} +
e^{-(x+y)/2}g(x-y)$.  One then finishes the problem in the same way as
described above. 
\medskip

{\bf \S1.2 \#11}:  We shall use the co\"ordinate method to solve the
equation
$$ u_x + 2u_y + (2x-y)u = 2x^2+3xy - 2y^2 \; . $$

\noindent Following the method described on p.~7 of the text, we set $x'
= x+2y$ and $y' = 2x-y$; the inverse equations are easily seen to be
$$x = {{x'+2y'}\over 5} \hbox{\ \ \ and\ \ \ } y = {{2x'-y'}\over 5}\; . 
$$

\noindent To help keep the books in order, write $U(x',y')$ for the
function defined by substituting the values of $x$ and $y$ given by the
inverse equations into the function $u(x,y)$; thus $u(x,y) = U(x',y')$
when $(x,y)$ and $(x',y')$ are related by those equations.  Applying the
chain rule as on p.~7 gives
$$u_x = U_{x'}+2U_{y'} \hbox{\ \ \ and\ \ \ } u_y = 2U_{x'} - U_{y'}\; . 
$$

\noindent In terms of the primed co\"ordinates, the r.~h.~s.~transforms
into
$$2x^2 + 3xy - 2y^2 = 2\cdot\left({{x'+2y'}\over 5}\right)^2 +
3\left({{x'+2y'}\over 5}\right)\left({{2x'-y'}\over 5}\right) - 2\left(
{{2x'-y'}\over 5} \right)^2 = x'y'\; .  $$

\noindent Thus, when the equation is rewritten in terms of (functions
of) the primed co\"ordinates, it reads
$$5 U_{x'} + y'U = x'y' \; . \eqno{(*)}$$

As we might have expected, this is an o.~d.~e.~with independent variable
$x'$ (regarding $y'$ as a parameter).  Its ``integrating factor'' is
$\ds{ \exp\left({{y'}\over 5}\,x'\right)}$, and multiplied by it the
equation becomes
$$ {\partial\over{\partial x'}}\,[e^{y'x'/5}U] = e^{y'x'/5}U_{x'} +
{{y'}\over 5}\,e^{y'x'/5}U = {1\over 5}\,e^{y'x'/5}y'x'\;.\eqno{(**)}$$

\noindent The r.~h.~s.~of ${(**)}$ has the form $\int e^{at} at\,dt$, so
its antiderivatives are easily obtained by partial integration and have
the corresponding form $te^{at} - \ds{{e^{at}}\over a} + {\rm
(const.)}$.  Thus we may integrate both sides of $(**)$ with respect to
$x'$, obtaining
$$\eqalign{ e^{x'y'/5}U(x',y') &= x'\exp\left({{x'y'}\over 5}\right) -
{5\over {y'}}\,\exp\left({{x'y'}\over 5}\right) + h(y') \cr
U(x',y') &= \left(x' - {5\over
{y'}}\right) + h(y')\cdot \exp\left(-{{x'y'}\over 5}\right) \cr
u(x,y) = U(x',y') &=\left(x + 2y + {5\over{y-2x}}\right) +
h(2x-y)\exp\left(- {{2x^2+3xy-2y^2}\over 5}\right) \; . \cr }$$

\noindent Here $h(y')$ is the constant of integration---which may depend
on the ``absent variable'' or ``parameter'' $y'$, and be an arbitrary
(differentiable) function of its argument.
\medskip
%\S1.2 ENDS HERE

{\bf \S1.3 \#2}: The approach to this problem that I wrote up initially
went as follows.  Set up rectangular space axes with the $u$-axis
pointing to the right, the $x$-axis pointing downward, and the chain of
length $\ell$ attached at the origin.  Consider the part of the hanging
chain between $x$ and $x+\Delta x$ (with $\Delta x > 0$).  The
horizontal component of its mass-times-acceleration is given by
$$ ma_{\rm horiz} = \int_x^{x+\Delta x} \rho\,{{\partial^2
u(\xi,t)}\over{\partial t^2}}\,d\xi \; .  $$

\noindent The force it feels at the $x$ end (from the attachment at the
origin) is the weight of the chain below that point, $g\rho(\ell-x)$,
directed upward; at the $x+\Delta x$ end it feels the weight
$g\rho(\ell-(x+\Delta x))$, directed downward.  The horizontal component
of the force it feels (assuming $\Delta x$ small) is then
$$ F_{\rm horiz} = g\rho\left[(\ell-(x+\Delta x))\cdot {{u_x(x+\Delta
x,t)}\over\sqrt{1 + u_x(x+\Delta x,t)}} -(\ell - x)\cdot
{{u_x(x,t)}\over\sqrt{1 + u_x(x,t)}}\right] \; .$$

\noindent For small oscillations we may ignore the (second-order)
difference between $\sqrt{1 + u_x}$ and 1, and ``linearize'' the
(horizontal component of the) equation $F=ma$ to
$$ \int_x^{x+\Delta x} \rho\,{{\partial^2 u(\xi,t)}\over{\partial
t^2}}\,d\xi = g\rho\left[(\ell-(x+\Delta x))\cdot u_x(x+\Delta x,t)
-(\ell - x)\cdot u_x(x,t)\right] \; .  $$

\noindent Dividing this by $\Delta x \to 0^+$ and remembering the
fundamental theorem of the integral calculus on the l.~h.~s.~and the
definition of the derivative on the r.~h.~s., we get
$$\eqalign{ \rho\,{{\partial^2 u(x,t)}\over{\partial t^2}} &=
g\rho\,{\partial\over\partial x}\,\left[ (\ell - x)\cdot u_x(x,t)\right]
\cr {{\partial^2 u(x,t)}\over{\partial t^2}} &=
g\,{\partial\over\partial x}\,\left[ (\ell - x)\cdot u_x(x,t)\right] \;
.  \cr }$$

One of the solutions subimtted when this problem was assigned in a
previous offering of the course went much faster: look at the derivation
of the vibrating-string equation, Example 2 on pp.~11--12.  The outcome
of the physics is the relation
$$ (Tu_x)_x = \rho u_{tt} \; . $$

\noindent The only difference between the vibrating-string situation and
the present situation is that the tension varies with $x$: $T = g\rho
(\ell-x)$ because the tension in a given small section of the chain is
proportional (with proportionality constant $g\rho$) to the amount of the
string below $x$.  Voila:
$$ [g\rho (\ell-x)u_x]_x = \rho u_{tt} \; ! $$

\noindent Out of the hat comes the rabbit.  A little too fast for a
worked-out solution, but quite perceptive. 
\medskip

{\bf \S1.3 \#5}: Let us give a different derivation of the ``simple
transport'' equation, Strauss's Example 1, pp.~10--11, and then try to
carry that derivation over to the case at
hand.\footnote{\SS{3}}{\sevenrm The fact that your humble instructor did
not think to use a ``moving-frames'' approach to this problem on 2/1/2001 is
a sign of encroaching senility.} Let $u(x,t)$ give the density of dye at
$x$-co\"{o}rdinate $x$ and time $t$.  If the liquid is moving to the
right in the pipe at constant velocity $V$ and we install a new
co\"{o}rdinate $\xi$ whose origin moves to the right at the same
velocity, then relative to the co\"{o}rdinates $(\xi,t)$ the function
$u$ will be constant in time.  Clearly (this is just the usual relation
for translation of the origin of co\"{o}rdinates) $x = \xi+Vt$.  That
says $\ds{ {\partial\over{\partial t}}\,u(\xi+Vt,t) = 0}$.  Now (to keep
from being confused) let us denote the partial derivatives of the
function $u(\cdot,\cdot)$ with respect to its first and second arguments
by $u_1$ and $u_2$ respectively.  Then the chain rule gives

$$ {\partial\over{\partial t}}\,u(\xi+Vt,t) = u_1(\xi+Vt,t)\cdot V +
u_2(\xi+Vt,t) = 0 $$

\noindent as an expression of that constancy in time.  We can now
replace $\xi+Vt$ by $x$ everywhere it occurs in the relation on the
set-off line above---this relation holds identically in the first and
second arguments in $u$---and it then becomes legitimate to use the
subscripts $x$ and $t$ for the partial derivatives again.  So the
equation we just derived becomes
$$V\cdot u_x(x,t) + u_t(x,t) = 0 \; , $$

\noindent which is the transport equation.  To turn this into the
``diffusion equation with drift,'' we need only observe that relative to
our moving-frame space co\"{o}rdinate $\xi$, all the change of
concentration in time that we can see\footnote{\SS{4}}{\sevenrm In that
co\"{o}rdinate frame.} is due to diffusion (there is no transport
relative to this frame, and we are tacitly assuming that the rate of
diffusion is not influenced by transport).  Now, when we compute the
rate of change of concentration in the $(\xi,t)$-co\"{o}rdinates, we get
$$ u_1(\xi+Vt,t)\cdot V + u_2(\xi+Vt,t) = {\partial\over{\partial
t}}\,u(\xi+Vt,t) = {{\partial^2u(\xi+Vt,t)}\over{\partial\xi^2}} =
u_{22}(\xi+Vt,t)\; .  $$

\noindent Again replacing $\xi+Vt$ by $x$ and using subscript $x$ and
$t$ to indicate partial differentiation, we get from the two ends of
that set-off relation
$$\eqalign{ u_t(x,t) + V\cdot u_x(x,t) &= u_{xx}(x,t) \cr u_t(x,t) &=
u_{xx}(x,t) - V\cdot u_x(x,t) \cr }$$

\noindent which is our ``diffusion equation with transport.''
\medskip
\goodbreak

{\bf \S1.3 \#5}: To consider the case in which the liquid is being
propelled to the right at constant velocity $V$, think instead of the
pipe---marked with the same $x$-co\"{o}rdinate---being moved to the left
at the same speed.  Then for the mass of dye in the interval $[x_0,x_1]$
one sees
$$M(t) = \int_{x_0-Vt}^{x_1-Vt} u(x,t)\,dx \hbox{\ \ instead\ of\ \ }
M(t) = \int_{x_0}^{x_1} u(x,t)\,dx \; . $$

\noindent Thus with the substitution $\xi=x+Vt$, $d\xi=dx$ we have
$$\eqalign{ {{dM(t)}\over{dt}} &= {d\over{dt}}\int_{x_0-Vt}^{x_1-Vt}
u(x,t)\,dx = {d\over{dt}}\int_{x_0}^{x_1} u(\xi+Vt,t)\,d\xi \cr &=
\int_{x_0}^{x_1} [u_x(\xi+Vt,t)\cdot V + u_t(\xi+Vt,t)]\,d\xi \; .  \cr }$$

\noindent This equals $k\cdot[u_x(x_1+Vt,t) - u_x(x_0+Vt,t)]$ as in the
derivation of diffusion without ``drift,'' so division by
$(x_1+Vt)-(x_0+Vt)=x_1-x_0$ and passage to the limit as $x_1\to x_0$
results in
$$\eqalign{ u_x(x_0+Vt,t)\cdot V + u_t(x_0+Vt,t) &=
k\,u_{xx}(x_0+Vt,t) \cr
u_t(x_0+Vt,t) &=
k\,u_{xx}(x_0+Vt,t) - u_x(x_0+Vt,t)\cdot V \cr
u_t(x,t) &=
k\,u_{xx}(x,t) - u_x(x,t)\cdot V \hbox{\ when\ $x_0+Vt$\ is\ everywhere\
replaced\ by\ $x$,} \cr
}$$

\noindent by the same reasoning as before.  If the sign convention seems
wrong, consider the case where $k=0$ so no diffusion takes place and
everything is transport.  The equation becomes $u_t + Vu_x = 0$ and by
the usual argument the solutions are constant on lines $x = C + Vt$, so
$u(x,t) = u(C,0) = u(x-Vt,0)$: to see the concentration at $x_0$ when
the time is $t$ you have to ``look back'' to see what it was at
$x=x_0-Vt$ when the transport began.  Thus one simply ``adds the
diffusion term to the transport term.''
\medskip

{\bf \S1.3 \#6}:  In a separate set of notes\footnote{\SS{5}}{\sevenrm Or
on pp.~150--151 of Strauss's book \dots} you can find the details of
rewriting the two-dimensional Laplace operator in terms of polar
co\"ordinates: 
$$\nabla^2 u = u_{xx} + u_{yy} = u_{rr} + {1\over r}\,u_r +
{1\over{r^2}}\,u_{\theta\theta} \; .  $$

\noindent Since cylindrical co\"ordinates in space are simply ``polar in
the $xy$ directions and rectangular in the $z$ direction,'' it is clear
that the 3-dimensional Lapalce operator in cylindrical co\"ordinates has
the form
$$\nabla^2 u = u_{xx} + u_{yy} + u_{zz}= u_{rr} + {1\over r}\,u_r +
{1\over{r^2}}\,u_{\theta\theta} + u_{zz}\; .  $$

The 3-dimensional heat equation $u_t = k\nabla^2 u$ then must take the
form
$$u_t = k\left[u_{rr} + {1\over r}\,u_r +
{1\over{r^2}}\,u_{\theta\theta} + u_{zz} \right] $$

\noindent in cylindrical co\"ordinates.  The assumption that
$u(r,\theta,z,t)$ does not depend on $\theta$ or $z$ implies that all
derivatives with respect to those variables are zero, and the heat
equation becomes
$$u_t = k\left[u_{rr} + {1\over r}\,u_r \right]\; .  $$

{\bf \S1.3 \#7}: Similarly, in the solution of {\bf \S2.2 \#6} (which
may be posted elsewhere) one can find a derivation of the relation
$$\nabla^2 f(r) = f_{rr} + {{n-1}\over r}\,f_r $$

\noindent for functions of the distance to the origin in $\real^n$, \IE,
the spherical co\"ordinate that is usually called ``$\rho$'' but that
Strauss insists on calling ``$r$.'' For $n=3$ the heat equation applied
to such functions becomes
$$u_t = k\,\nabla^2 u = k\left[u_{rr} + {2\over r}\,u_r \right]\; .  $$

{\bf \S1.3 \#9}: If $\VEC{F} = r^2\VEC{x}$, where $r^2 = x^2+y^2+z^2$
and $\VEC{x} = x\,\VEC{i} + y\,\VEC{j}+z\,\VEC{k}$, then
$$ \nabla\dotprod\VEC{F} = \PARSH{r^2x}{x} + \PARSH{r^2y}{y}+
\PARSH{r^2z}{z} = r^2 + 2x^2 + r^2+2y^2 +r^2+2z^2 = 5r^2\; . $$

\noindent If $D$ is the ball of radius $a$ and center $\VEC{0}$, then
$$\VEC{F}\dotprod\VEC{n} = a^2\VEC{x}\dotprod{{\VEC{x}}\over a} =
a\,(\VEC{x}\dotprod\VEC{x}) = a^3 \hbox{\ \ on\ \ } \partial D\; .  $$

\noindent (Here $\VEC{n}$ is the unit outward normal, which is
$\ds{{\VEC{x}}\over a}$ on $D$.)  Thus
$$ \int\!\int_{\partial D} \VEC{F}\dotprod\VEC{n}\,dS =
\int\!\int_{\partial D} a^3\,dS = a^3\cdot({\rm surface\ area\ of\
}\partial D) = 4\pi a^5 \; . $$

\noindent On the other hand,
$$ \int\!\int\!\int_D \nabla\dotprod\VEC{F}\,dV = \int\!\int\!\int_D
5r^2\,dV = \int_0^{2\pi}\int_0^\pi\int_0^a 5r^2\cdot
r^2\sin\phi\,dr\,d\phi\,d\theta = \int_0^a (4\pi)\,5r^4\,dr =
4\pi a^5 \; , $$

\noindent exactly what the divergence theorem would lead us to
believe.\footnote{\SS{6}}{\sevenrm This is a volume integral in
spherical co\"ordinates, with element of volume ${\scriptstyle dV =
r^2\sin\phi\,dr\,d\phi\,d\theta}$, where $\scriptstyle\phi$ is
co-latitude and $\scriptstyle\theta$ is longitude.  Note that
Strauss---and physicists---interchange these labeling conventions so
that for them $\scriptstyle\theta$ is co-latitude and $\scriptstyle\phi$
is longitude.  As a result of this inconsistency, one can spend many
happy hours trying to figure out what various formulas mean.}
\medskip

{\bf \S1.3 \#10}:  The definition of ``integrating over all of
$\real^3$'' is
$$ \int\int\int_{\real^n} \nabla\dotprod \VEC{f}\,dV = \lim_{R\to\infty}
\underbrace{\int\int\int}_{x^2+y^2+z^2\le R^2} \nabla\dotprod \VEC{f}\,dV
\; . $$

\noindent The divergence theorem gives us
$$ \underbrace{\int\int\int}_{x^2+y^2+z^2\le R^2} \nabla\dotprod
\VEC{f}\,dV = \underbrace{\int\int}_{x^2+y^2+z^2 = R^2}
\VEC{f}\dotprod\VEC{n}\,dS \; , $$

\noindent an integral over the surface of the sphere of radius $R$
centered at the origin.  For $\Vert\VEC{x}\Vert = R$ we can use the
given estimate $\ds{\Vert \VEC{f}(\VEC{x})\Vert \le
{1\over{\Vert\VEC{x}\Vert^3+1}} }$ and the Schwarz inequality to deduce
that the same estimate applies to the component $\VEC{f}\dotprod\VEC{n}$
of $\VEC{f}$ along the outward normal to that sphere:
$$|\VEC{f}\dotprod\VEC{n}| \le \Vert\VEC{f}\Vert\,\Vert\VEC{n}\Vert =
\Vert\VEC{f}\Vert \le {1\over {\Vert\VEC{x}\Vert^3+1}} =
{1\over{R^3+1}}$$

\noindent when $\VEC{x}$ is on the surface of that sphere.  Its surfact
integral must therefore be smaller than that estimate times the area of
the sphere:
$$ \left\vert\underbrace{\int\int}_{x^2+y^2+z^2 = R^2}
\VEC{f}\dotprod\VEC{n}\,dS\right\vert \le {1\over{R^3+1}}\cdot 4\pi R^2
\; .  $$

\noindent Thus
$$\eqalign{\left\vert \int\int\int_{\real^n} \nabla\dotprod
\VEC{f}\,dV\right\vert &= \left\vert\lim_{R\to\infty}
\underbrace{\int\int\int}_{x^2+y^2+z^2\le R^2} \nabla\dotprod \VEC{f}\,dV
\right\vert \cr &= \lim_{R\to\infty}
\left\vert\underbrace{\int\int}_{x^2+y^2+z^2 = R^2}
\VEC{f}\dotprod\VEC{n}\,dS\right\vert \le \lim_{R\to\infty} {{4\pi
R^2}\over{R^3+1}} = 0 \; .  \cr}$$
%\S1.3 ENDS HERE

{\bf \S1.4 \#1}: It is easy to do this problem by assuming a solution
that is a polynomial of degree 2 in $x$ and $t$, then plugging it into the
p.~d.~e.~and the initial condition to find the values of the coefficients. 
However, I thought I would use the problem as a simple example of the use of
the Laplace transform in solving the heat equation (for people who know
what the Laplace transform is, or might be willing to look back in their
old o.~d.~e.~book to learn it---it's easy).  The basic idea is that since
heat-equation problems are typically initial-value problems for $0\le t
< \infty$, applying the Laplace transform {\bf in the time variable}
while leaving the space variable(s) free looks like a viable technique. 
If we follow the usual ``$F(s) = \int_0^\infty e^{-st}f(t)\,dt$''
convention, we would write ``$U(x,s)$'' for the Laplace transform of
$u(x,t)$ in the time variable---$\ds{U(x,s) = \int_0^\infty
e^{-st}u(x,t)\,dt}$---and then have
$$\eqalign{ \PARSH{u(x,t)}{t} &= {{\partial^2u(x,t)}\over{\partial x^2}}
\cr s\cdot U(x,s) - u(x,0) = \int_0^\infty e^{-st}\,\PARSH{u(x,t)}{t}\,dt
&= \int_0^\infty e^{-st}\,{{\partial^2u(x,t)}\over{\partial x^2}}\,dt =
{{\partial^2}\over{\partial x^2}}\,U(x,s) \cr}$$

\noindent where the first equal-sign on the second line is due to the
famous Laplace-transform integration by parts
$$s\cdot F(s) - f(0) = \int_0^\infty e^{-st}\,{{df(t)}\over{dt}}\,dt $$

\noindent that ``turns differentiation into multiplication-by-$s$ and
the initial condition,'' while the last equal-sign on the second line is
due to Leibniz's rule, which we hope can be applied to integrals on
infinite intervals.  Since the problem gives us $u(x,0)=x^2$, our
Laplace-transformed equation now reads (l.~h.~and r.~h.~sides have been
interchanged)
$$ {{\partial^2U(x,s)}\over{\partial x^2}} -sU(x,s) = -x^2 \; . 
\eqno{(*)}$$

\noindent Regarded as an equation in $x$ only, this looks like
$$ w''- sw = -x^2$$

\noindent so we need to find a particular solution and the general
solution of the associated homogeneous equation---the ``arbitrary
constants'' will actually depend on $s$, though.  To get a particular
solution we try the usual trick: try $Cx^2+Dx+E$ and pick the constants
correctly.  Plugging in $U(x,s) = Cx^2+Dx+E$ {\bf and treating $s$ as
constant} leads to 
$$2C - s[Cx^2+Dx+E] = -x^2 $$

\noindent which is satisfied identically in $x$ if and only if $C =
\ds{1\over s}$, $D\equiv 0$ and $E=\ds{2\over{s^2}}$.  This gives us
$$\eqalignno{U(x,s) &= {1\over s}\,x^2 + {2\over {s^2}} \cr 
\hbox{whence\ \ \ }u(x,t) &= x^2 + 2t &\hbox{(use\ a\ table,
\ or\ {\tt MATLAB})} \cr }$$

\noindent and it is easy to verify that this does indeed satisfy the
heat equation.  It is our great good fortune (!) that it also satisfies
the initial condition $u(x,t) = x^2$, so we don't have to investigate
the inverse Laplace transforms of the solutions of the homogeneous
o.~d.~e.~associated with $(*)$.
\medskip

{\bf \S1.4 \#2}: This is basically just a ``look-up'' question on the
physical laws involved.  Thus in {\bf (a)} or the first half of {\bf
(c)}, the assumption that there is no flow of heat out at the end of the
rod implies that the temperature gradient---which in this 1-dimensional
situation is just $u_x$---is zero there.  In 2- or 3-dimensional space,
one can only require that the {\bf component of the temperature gradient
normal to (or across) the boundary} equal zero, \IE, $\ds\PARSH{u}{n}=0$, since (for
example) a sphere hot at the equator and cold at the poles, even though
surrounded by an insulating medium, would experience heat flow along its
surface.  Fick's law leads to similar conclusions for {\bf (b)} or the
second half of {\bf (c)}.
\medskip

{\bf \S1.4 \#3}: If we integrate both sides of the heat equation over
the body and assume that Leibniz's rule is valid in many variables (it
is), then we get
$$\eqalign{ {\partial\over{\partial t}}\int\int\int_D u(x,y,z,t)\,dV &=
\int\int\int_D u_t(x,y,z,t)\,dV = k \int\int\int_D \nabla\bullet[\nabla
u(x,y,z,t)]\,dV \cr &= k \int\int_{\partial D} [\nabla u(x,y,z,t)]\bullet
\VEC{n}\,dS = 0 \cr }$$

\noindent where the transition to the second line is the divergence
theorem and the zero is due to the fact that the object is insulated. 
It follows that the integral $\ds{\int\int\int_D u(x,y,z,t)\,dV }$ is
constant in time, and thus for all $t\ge0$ it has the value
$\ds{\int\int\int_D f(x,y,z)\,dV }$ it had at $t=0$.  Now to see that
the temperature of the body is constant in space in steady state,
observe that the steady-state version of the heat equation is
$\nabla\bullet[\nabla u] = 0$.  If $u$ satisfies this equation, then we
have (since $\nabla u\bullet \VEC{n} \equiv 0$ on the boundary $\partial
D$)
$$\eqalign{ \nabla\bullet[u\,\nabla u] &= \nabla u \bullet \nabla u +
\nabla\bullet[\nabla u] = \nabla u \bullet \nabla u \cr 0 =
\int\int_{\partial D} [u\,\nabla u]\bullet\VEC{n}\,dS &= \int\int\int_D
\nabla\bullet[u\,\nabla u]\,dV= \int\int\int_D [\nabla u \bullet \nabla
u]\,dV \; .  \cr }$$

\noindent But $\nabla u \bullet \nabla u = \Vert\nabla u\Vert^2 \ge 0$
everywhere in $D$, and if the integral of a nonnegative continuous
function over $D$ is zero then the function must be zero everywhere in
$D$.  Thus $u$ has zero gradient in $D$ and must therefore be a constant
function.  But then at the steady state we have 
$$\eqalign{ u\cdot {\rm vol}(D) &= \int\int\int_D u\,dV = \int\int\int_D
f(x,y,z)\,dV \cr u &\equiv {1\over{\rm vol}(D)} \int\int\int_D
f(x,y,z)\,dV \; , \cr }$$

\noindent the (volume) average of the initial temperature.
\medskip

{\bf \S1.4 \#5}: In both of the rods, the steady-state form of the heat
equation is $0 = u_t = ({\rm const.})\cdot u_{xx}$, so the steady-state
equation (involving only the space variable) is $u''(x) = 0$: its
solutions are straight lines.  One usually thinks of straight lines in
the form $u(x) =ax+b$, but for the second rod it will be convenient to
think in the form $u(x) = a(L_1+L_2-x)+b$, which is equivalent but
easier to work with for the boundary conditions.  We thus have two
steady-state solutions of the 1-dimensional heat equation
$$ u_1(x) = a_1x + b_1 \hbox{\ \ \ and\ \ \ } u_2(x) = a_2(L_1+L_2-x) +
b_2 \; . $$

\noindent At the left end $x=0$ of the assembly we have $u(x)=u_1(x)=0$,
and thus $b_1=0$.  At the right end $x=L_1+L_2$ of the assembly we have
$u(x)=u_2(x)=T$, and thus $T = a_2\cdot 0 + b_2$ or $b_2=T$.  If we view
the junction from the left, we see that the flux is $c\rho
k_1\cdot(u_1)_x(L_1) = c\rho k_1a_1$ (we already know $b_1=0$, and
anyway the derivative of a constant is zero).  If we view the junction
from the right, we see that the flux is $-c\rho k_2\cdot (u_2)_x(L_1) =
c\rho k_2a_2$.  Equality of the heat transfer viewed from both sides
gives the equation
$$k_1a_1 = - k_2a_2 \;. $$

\noindent It goes without saying that the temperatures viewed from both
sides of the junction must be the same: that gives the equation
$$a_1L_1 = T + a_2L_2 \; . $$

\noindent Using the constants given in the problem, we have the two
equations
$$\eqalign{ 2\cdot a_1 &= -1\cdot a_2 \cr a_1\cdot 3 &= 10 + a_2\cdot 2
\cr }$$

\noindent and solving these for $a_1 = \ds{{10}\over 7}$, $a_2 =
\ds{{-20}\over 7}$ is routine.  Thus $u(x) = \ds{{10}\over 7}\,x$ for
$0\le x \le 3$ and $u(x) = 10-\ds{{20}\over 7}\,(5-x) = \ds{{10}\over
7}\,(2x-3)$ for $3 \le x \le 5$, agreeing with the back of the book. 
\medskip

{\bf \S1.4 \#6 (a)}: If we take the curl of both sides of equation (4)
on p.~22, the fact that the space variables are independent of $t$
implies that we get (using the $\nabla$ notation throughout)
$$\eqalign{ \nabla \times \left( \PARSH{\VEC{v}}{t} \right)+
{{c_0^2}\over{\rho_0}}\nabla\times(\nabla\rho) &= \VEC{0} \cr
\PARSH{\nabla\times\VEC{v}}{t} + \VEC{0}
&= \VEC{0} \cr }$$

\noindent because the curl of a gradient, like
$\nabla\times(\nabla\rho)$, is always zero.\footnote{\SS{7}}{\sevenrm
This fact simply amounts to the statement that it doesn't matter in what
order one takes cross-partials.  Write out ${\scriptstyle
\nabla\times(\nabla\rho)}$ in component form, if you've never done that,
and you'll see this.} So the time derivative of the vector field $\nabla
\times \VEC{v}$ is zero, which says that the curl of $\VEC{v}$ is
identically a constant (vector).  Therefore if the curl equals $\VEC{0}$
at time $t=0$, then the curl remains zero thereafter. 
\medskip

{\bf (b)}: The time derivatives of equations (4) and (5) are 

$$\eqalignno{ {{\partial^2\VEC{v}}\over{\partial t^2}} +
{{c_0^2}\over{\rho_0}}\,\nabla\PARSH{\rho}{t} &= \VEC{0} \hbox{\ \ \ \ \
and} &{(4_t)}\cr {{\partial^2\rho}\over{\partial t^2}} + \rho_0
\nabla\dotprod \PARSH{\VEC{v}}{t} &= 0 &{(5_t)} \cr }$$

\noindent respectively.  If we plug $(4)$ into $(5_t)$ we get
$$\eqalign{ 0 = {{\partial^2\rho}\over{\partial t^2}} + \rho_0
\nabla\dotprod \PARSH{\VEC{v}}{t} &= {{\partial^2\rho}\over{\partial
t^2}} + \rho_0
\nabla\dotprod\left(-\,{{c_0^2}\over{\rho_0}}\right)\nabla\rho \cr
{{\partial^2\rho}\over{\partial t^2}} &= c_0^2\,\nabla^2\rho \cr }$$

\noindent which is the wave equation (with velocity $c_0$) for $\rho$. 
Similarly, if we plug $(5)$ into $(4_t)$ we get
$$\eqalign{ \VEC{0} &= {{\partial^2\VEC{v}}\over{\partial t^2}} +
{{c_0^2}\over{\rho_0}}\nabla\left(-\rho_0
\nabla\dotprod\VEC{v}\right) \cr 
{{\partial^2\VEC{v}}\over{\partial t^2}} &= c_0^2
\nabla\left(\nabla\dotprod\VEC{v}\right) \; . \cr }$$

\noindent It is a simple if tedious exercise for the
reader\footnote{\SS{8}}{\sevenrm If (s)he has never done it before \dots
} to compute out the identity $\nabla(\nabla\dotprod\VEC{v}) =
\nabla\times(\nabla\times\VEC{v}) + (\nabla\dotprod\nabla)\VEC{v}$. 
Under the assumption that the curl of $\VEC{v}$ is $\VEC{0}$, this
identity says that $\nabla(\nabla\dotprod\VEC{v})$ takes the Laplacian
component-by-component, so this last equation is the wave equation on
all components of $\VEC{v}$. 
\medskip
%\S1.4 ENDS HERE

{\bf \S1.5 \#1}: The solutions of $u'' + u = 0$ all have the form $A\cos
t + B\sin t$.  The boundary condition $u(0) = 0$ forces $A= 0$, and the
other boundary condition then becomes $B\sin L = 0$.  This forces $B=0$
(and uniqueness of the solution, which can then only be the
identically-zero function) only if $\sin L\not=0$, which fails if and
only if $L=k\pi$ for some integer $k$.  But if that happens, then any
multiple of $\sin x$ will satisfy the o.~d.~e.~and the boundary
conditions, so uniqueness fails---under, and only under, that condition
on $L$.
\medskip

{\bf \S1.5 \#2}: This problem is less pleasant than its predecessor but
is still elementary.  {\bf (a)} It is easy to see that uniqueness will
fail for the inhomogeneous equation $u'' + u' = f$ if and only it fails
for the homogeneous equation $u''+u' = f$ (and the boundary conditions). 
The general solution of the associated homogeneous o.~d.~e.~is $u(x) = A +
Be^{-x}$.  The boundary condition $u(0) = u'(0)$ is
$$\eqalign{ A + Be^{-x}\bigr\vert_{x=0} &= -Be^{-x}\bigr\vert_{x=0} \cr
A+B &= -B \cr A + 2B &= 0 \; . \cr }$$

\noindent The boundary condition $u'(0) = \ds{{u'(\ell)+u(\ell)}\over
2}$ is
$$\eqalign{ -B &= {{-Be^{-\ell} + A + Be^{-\ell}}\over 2} = {A\over 2}
\cr A + 2B &= 0 \; , \cr }$$

\noindent the same condition (and the two of them taken together imply
the remaining condition).  Thus there is really only one boundary
condition; the solutions of the homogeneous equation that satisfy the
boundary condition all look like $B(e^{-x}-2)$, any choice of $B$ works,
and there is no uniqueness.
\medskip

{\bf (b)} The question of existence is slightly trickier, but not much. 
Suppose a solution $u(x)$ of the o.~d.~e.~exists.  Integrating both sides
of the o.~d.~e.~gives
$$ u'(x) + u(x) = \int_0^x f(\xi)\,d\xi + u(0) + u'(0) \eqno{(*)}$$

\noindent which must hold for all values of $x$.  With $x=\ell$ we get
$$ u'(\ell) + u(\ell) = \int_0^\ell f(\xi)\,d\xi + u(0) +u'(0)\; .  $$

\noindent Using $\ds{ u'(0) = u(0) = {{u(\ell) + u'(\ell)}\over 2}}$ in
combination with this, we see that the condition
$$0 = \int_0^\ell f(\xi)\,d\xi $$

\noindent must be satisfied by any solution of the o.~d.~e.~with these
boundary conditions: it is a {\bf necessary} condition for solvability
of the equation with the boundary value conditions on the solution.  Is
it also {\bf sufficient}? Suppose it holds.  We know (as a consequence
of the solvability of the {\bf initial}-value problem for o.~d.~e.) that
the o.~d.~e.~has a solution satisfying $u'(0)=u(0)=0$.  Denote that
solution by $u(x)$.  Assuming that integral-zero condition, at $x=\ell$
the relation $(*)$ (which is the integrated version of the o.~d.~e.) reads
$$ u'(\ell) + u(\ell) = \int_0^\ell f(\xi)\,d\xi = u(0) + u'(0) = 0 \; . 
$$

\noindent Thus $\ds{{u(\ell)+u'(\ell)}\over 2} = u(0) = u'(0)$ holds
automatically under the assumption $\ds{\int_0^\ell f(\xi)\,d\xi =0}$. 
\medskip
\goodbreak

{\bf \S1.5 \#4}: {\bf (a)} If you add a constant to $u$ you don't change
$\Delta u$ (also known as $\nabla^2u$ and as $\nabla\bullet\nabla u$)
and you don't change $\ds\PARSH{u}{n}$ (which is $\nabla u \bullet
\VEC{n}$), so the boundary conditions are still satisfied and the
``$f$'' on the r.~h.~s.~is undisturbed.\footnote{\SS{9}}{\sevenrm Note
that you can't add two solutions of this equation to get another
solution: the equation is not linear-homogeneous.}  Thus the
solution is only ``unique up to an additive constant.''  This is hardly
surprising when you think of the steady-state heat equation for an
insulated body: the notion of ``zero temperature'' is only relative. 
For example, the Kelvin and Celsius scales have degrees of the same
size, but ``absolute zero'' is not the same as $0\INT$C.
\medskip

{\bf (b)} The divergence theorem gives
$$ \int\int\int_D f(x,y,z)\,dV = \int\int\int_D \nabla\bullet\nabla
u(x,y,z)\,dV = \int\int_{\partial D} \nabla u\bullet\VEC{n}\,dS = 0$$

\noindent (the first equality is the p.~d.~e., the last is the boundary
condition on the normal derivative), so if the Neumann problem has a
solution the r.~h.~s.~function $f(x,y,z)$ must have a zero integral over
$D$.
\medskip

{\bf (c)} $\nabla^2u = f(x,y,z)$ says that there are sources and sinks
of dye (for diffusion) or heat (for heat flow) but that $u(x,y,z)$ is a
steady-state solution of the heat (or diffusion) equation.  The
r.~h.~s.~function $f(x,y,z)$ is the volume density of sources and sinks. 
If there is no transfer of dye (or heat) across the boundary, which is
what $\ds\PARSH{u}{n} = 0$ says, then the sources must be balanced by
the sinks,\footnote{\SS{10}}{\sevenrm The most general version of this
statement, in completely unrecognizable form, can be found in
${\scriptstyle Perturbation\ of\ harmonic\ struc}$- ${\scriptstyle
tures\ and\ an\ index}$-${\scriptstyle zero\ theorem}$,
Ann.~Inst.~Fourier (Grenoble) 20,1 (1970), pp.~317--359; Thm.~4.1.4,
p.~353 ff.} which is what $\ds{\int\int\int_D f\,dV = 0}$ says. 
\medskip
%\S1.5 ENDS HERE

{\bf \S1.6 \#1(a)}:  Combining the cross-partial terms, we can write the
equation as
$$ u_{xx} - 4 u_{xy} + u_{yy} + 2u_y + 4u = 0 \; . $$

\noindent The polynomial $x^2 - 4xy + y^2 = x^2 - 4xy + 4y^2 - 3y^2 = (x
- 2y)^2 - 3y^2 = [x - (2 - \sqrt{3})y][x - (2 + \sqrt{3})y]$, so the
2nd-order part of the operator factors as $[\partial_x - (2 -
\sqrt{3})\partial_y][\partial_x - (2 + \sqrt{3})\partial_y]$ and the
equation is hyperbolic.  The ``discriminant'' test ${\cal D} =
2^2-1\cdot 1 = 3$ yields the same result, as does analysis of the
eigenvalues of the matrix $\qmatrix{1 &-2 \cr -2 & 1}$.
\medskip

{\bf \S1.6 \#1(b)}: The homogeneous 2nd-degree polynomial whose
coefficients are those of the highest-order terms of the differential
operator\footnote{\SS{11}}{\sevenrm This polynomial (or a closely related
polynomial) is sometimes called the ${\scriptstyle symbol}$ of the
operator.} is $9x^2 + 6xy + y^2 = (3x+y)^2$ and the operator is
parabolic. 
\medskip

{\bf \S1.6 \#3}: Before we start: in an earlier version of the course,
people seemed to be having trouble with the logic of this problem.  What
does the following statement of the problem {\it mean?}
\medskip
\item{} Among all the equations of the form (1), show that the only ones
that are unchanged under all rotations {\it (rotation invariant)\/} have
the form $a(u_{xx} + u_{yy}) + bu = 0$.
\medskip

\noindent What I get out of this is that you have to start with a
(constant-coefficient) equation of the form
$$a_{11}u_{xx} + 2a_{12}u_{xy} + a_{22}u_{yy} + a_1u_x + a_2u_y + a_0u = 0$$

\noindent and show that the assumption that the equation is invariant
under rotation forces $a_{11} = a_{22}$, $a_{12}=0$, etc.  The converse
proposition, that such an equation {\it is\/} invariant under rotation,
is much easier: see Strauss, p.~150; but the direct proposition, that
rotational invariance implies a certain form for the differential
operator, is what the problem asks you to prove.  Certainly my intention
in assigning it was to get people either (a) to work through the kind of
ugly computation that is occasionally necessary, or (b) to see that
linear algebra is relevant in many places in analysis.
\medskip

Some people argued, usually more intuitively than rigorously, that since
the 2nd-degree polynomials of the form
$$a_{11}x^2 + 2a_{12}xy + a_{22}y^2 + a_1x + a_2y + a_0u = 0$$

\noindent with $a_{11}=a_{22}$, \dots, were the only ones invariant
under rotation, the same must be true of the operators.  Yes, this is
true, but nonetheless it must be checked: if $\xi = \alpha x + \beta y$, $\eta
= \gamma x + \delta y$ with $\alpha\delta - \beta\gamma \not= 0$, how do
polynomials in $(x,y)$ transform into polynomials in $(\xi,\eta)$? In
the same way as operators, or does the transpose or inverse of the
matrix $\qmatrix {\alpha &\beta\cr\gamma&\delta\cr}$ come into play?
Nobody checked.  It can be dangerous to jump to conclusions, even
correct ones.  For those who know about tensors: contravariant and
covariant ones transform differently (upper indices vs.~lower indices!).
\medskip

Rotation through an angle $\vartheta$ is given by
$$\eqalign{ \xi &= (\cos\vartheta) x + (\sin\vartheta) y \cr
\eta &= (-\sin\vartheta) x + (\cos\vartheta) y \cr }$$

\noindent so the differential operators transform as
$$\eqalign{ {\partial\over{\partial x}} &= \cos\vartheta
\,{\partial\over{\partial\xi}} - \sin\vartheta\,{\partial\over{\partial
\eta}} \cr {\partial\over{\partial y}} &= \sin\vartheta\,
{\partial\over{\partial\xi}} + \cos\vartheta\,
{\partial\over{\partial\eta}} \hbox{\ \ \ (note the transposed matrix
here!)} \cr
{{\partial^2}\over{\partial x^2}} &= \cos^2\vartheta
\,{{\partial^2}\over{\partial\xi^2}} -
2\sin\vartheta\,\cos\vartheta\,{{\partial^2}\over{\partial\xi\partial\eta}}
+\sin^2\vartheta\,{{\partial^2}\over{\partial\eta^2}}\cr
{{\partial^2}\over{\partial x\partial y}} &=
\cos\vartheta\,\sin\vartheta \,\left[{{\partial^2}\over{\partial \xi^2}} 
- {{\partial^2}\over{\partial \eta^2}}\right] + (\cos^2\vartheta -
\sin^2\vartheta)\, {{\partial^2}\over\partial\xi{\partial\eta}} \cr
{{\partial^2}\over{\partial y^2}} &= \sin^2\vartheta
\,{{\partial^2}\over{\partial\xi^2}} +
2\sin\vartheta\,\cos\vartheta\,{{\partial^2}\over\partial\xi{\partial\eta}}
+\cos^2\vartheta\,{{\partial^2}\over{\partial\eta^2}}\; . \cr
}$$

\noindent The orders of the operators are left invariant under any
linear constant-coefficient change of coordinates, so we can consider
the 2nd-, 1st- and 0-th-order parts of the operator separately.  $a_{11}
u_{xx} + 2a_{12}u_{xy} + a_{22}u_{yy}$ transforms under rotation to
$$\eqalign{ &a_{11}\,[\cos^2\vartheta \,u_{\xi\xi} -
2\sin\vartheta\,\cos\vartheta\,u_{\xi\eta} +
\sin^2\vartheta\,u_{\eta\eta}] \cr +
&2a_{12}\,\{\cos\vartheta\,\sin\vartheta \,\left[u_{\xi\xi} -
u_{\eta\eta} \right] + (\cos^2\vartheta - \sin^2\vartheta)\,u_{\xi\eta}\}
\cr + &a_{22}\,[\sin^2\vartheta \,u_{\xi\xi} +
2\sin\vartheta\,\cos\vartheta\,u_{\xi\eta}
+\cos^2\vartheta\,u_{\eta\eta}] \cr &= [a_{11}\cos^2\vartheta +
a_{12}\,2\sin\vartheta\cos\vartheta + a_{22}\sin^2\vartheta]\,u_{\xi\xi}
\cr &+ [- a_{11}\,2\sin\vartheta\cos\vartheta +
2a_{12}(\cos^2\vartheta-\sin^2 \vartheta) +
a_{22}\,2\sin\vartheta\cos\vartheta]\,u_{\xi\eta} \cr &+
[a_{11}\sin^2\vartheta - a_{12}\,2\sin\vartheta\cos\vartheta +
a_{22}\cos^2\vartheta]\,u_{\eta\eta} \; .  \cr }$$

\noindent The easiest place to apply rotational invariance is the
cross-partial term: for this to remain invariant, we would have to have
$$ 2a_{12} = [- a_{11}\,2\sin\vartheta\cos\vartheta +
2a_{12}(\cos^2\vartheta-\sin^2 \vartheta) +
a_{22}\,2\sin\vartheta\cos\vartheta] = (a_{22} - a_{11})\sin2\vartheta +
2a_{12}\cos 2\vartheta $$

\noindent for all choices of $\vartheta$.  Taking $\vartheta = \pm\pi/4$
fixes the l.~h.~s., changes the sign of the $(a_{22}-a_{11})$-term on
the r.~h.~s.~and kills the $a_{12}$-term on the r.~h.~s., so
$a_{11}=a_{22}$ must hold and so must $a_{12}=0$.  Thus the 2nd-order
part of the operator must have the form $a(u_{xx} + u_{yy})$, where $a =
a_{11}=a_{22}$ (and $a_{12}=0$).  The 1st-order part is much easier to
dispose of, since we would have to require that the coefficients match in
$$a_1u_x + a_2u_y = a_1\,[\cos\vartheta \,u_{\xi}
-\sin\vartheta\,u_\eta + a_2\,[ \sin\vartheta\,u_\xi +
\cos\vartheta\,u_\eta] = [a_1\cos\vartheta +a_2\sin\vartheta]\,u_\xi +
[-a_1\sin\vartheta + a_2\cos\vartheta]\,u_\eta $$

\noindent which would give $a_1 = -a_1$ and $a_2 = -a_2$ for
$\vartheta=\pi$.  The 0-th-order term is automatically rotationally
invariant.
\medskip

$\{$I wrote this out in co\"{o}rdinate form so that the reader could see
that elementary methods handle this problem.  However, that is not the ``right way'' to do the
problem: the right way uses the relations (7) and (8) of Strauss's \S1.6
and works in any number of dimensions.  Recall that a linear change of
variables with matrix $B$ transforms the equation whose 2nd-order part
has matrix-of-coefficients $A$ to one whose 2nd-order part has
matrix-of-coefficients $BAB^T$.  A {\bf rotation} is a linear change of
variables whose matrix $Q$ is an {\bf orthogonal} matrix, \IE, one whose
inverse is its transpose, and which also has the property that its
determinant is 1---these are also called {\bf proper orthogonal}
matrices.  Thus invariance under rotations is equivalent to $A = QAQ^T =
QAQ^{-1}$ for every proper orthogonal matrix, or equivalently: $AQ=QA$
for every proper orthogonal matrix.  But $A$ is a symmetric matrix, so
there exists a proper orthogonal matrix $Q$ for which $Q^TAQ = \Lambda$
where $\Lambda$ is a diagonal matrix.  However, since $Q$ commutes with
$A$, this says $\Lambda = Q^TAQ = Q^TQA = A$ and thus $A$ is already a
diagonal matrix---which commutes with all proper orthogonal matrices. 
In particular $A$ commutes with all permutation matrices that
interchange two basis elements and change the sign of one of them, and
that is possible\footnote{\SS{12}}{\sevenrm The reader should verify this
assertion, at least in the ${\scriptstyle 3\times 3}$ case.} if and only
if all the diagonal entries of $A$ are equal.  Thus $A = \lambda I$ for some
$\lambda \not= 0$ (can't be zero if $A\not=0$).  The argument just given
applies to all dimensions, not just to the 2-dimensional case.  Taking
care of the 1st-order terms involves the same argument in any dimension
that it does in dimension 2; the reader might want to check this.$\}$
\medskip

{\bf \S1.6 \#5}: Before one starts on this one might want to know one of
Leibniz's other rules: for the second derivative of a product
$$(f(x)\cdot g(x))'' = f''(x)g(x) + 2f'(x)g'(x) + f(x)g''(x) $$

\noindent and more generally (although we won't need this fact right
away)
$$ (f(x)\cdot g(x))^{(n)} = \sum_{k=0}^n {n\choose k} f^{(n-k)}(x)\,
g^{(k)}(x) \; .  $$

\noindent For $u = e^{\alpha x+\beta y} v$ we thus have
$$\eqalign{ u_x &= e^{\alpha x+\beta y} v_x + \alpha e^{\alpha x \beta
y} v \cr u_y &= e^{\alpha x+\beta y} v_y + \beta e^{\alpha x+\beta y} v
\cr u_{xx} &= e^{\alpha x+\beta y} v_{xx} + 2\alpha e^{\alpha x+\beta y}
v_x + \alpha^2 e^{\alpha x+\beta y} v \cr u_{yy} &= e^{\alpha x+\beta y}
v_{yy} + 2\beta e^{\alpha x+\beta y} v_y + \beta^2 e^{\alpha x+\beta y}
v \; .  \cr }$$

\noindent Plugging all this into the equation, followed by toiling
patiently with little hope of reward, results in
$$\eqalign{ &u_{xx} + 3u_{yy} - 2u_x + 24u_y + 5u = \cr &e^{\alpha x
+\beta y}\,\{v_{xx} + 2\alpha v_x + \alpha^2 v + 3[v_{yy} + 2\beta v_y +
\beta^2 v] - 2[v_x + \alpha v] + 24[v_y + \beta v] + 5v\} = \cr
&e^{\alpha x + \beta y}\,\{v_{xx} + 3v_{yy} + [2\alpha - 2]\,v_x +
[6\beta + 24]\,v_y + [\alpha^2 + 3\beta^2 -2\alpha + 24\beta + 5]v\} \;
.  \cr }$$

\noindent To drop out the coefficients of $v_x$ and $v_y$ respectively,
we want\footnote{\SS{13}}{\sevenrm This does not agree with the back of
the book, of course, but the back of the book is wrong.  This fact can
also be found on Strauss's errata sheet, obtainable through a link from
the course web page.  People in the course would be well advised to
print out a copy to keep with the book; the book is great, aber der
Herrgott hat selber gefehlt.} to choose $\alpha=1$ and $\beta = -4$. 
With these choices the equation for $v$ becomes $v_{xx}+3v_{yy} - 44v =
0$.  To make the coefficient of the ``$v_{yy}$'' term equal 1 we would
want to replace $y$ by $\ds{y \over \sqrt{3}}$.\par

{\bf \S1.6 \#6}: {\bf (a)} The 2nd-order part of the operator $u_{xy} +
3u_y$ is hyperbolic: one way to see this is that its matrix of
coefficients $\qmatrix{0 &1/2 \cr 1/2 &0 \cr}$ has one eigenvalue of
each sign, and another is that Strauss transforms the wave operator
$u_{xx} - u_{yy}$ (which we know is hyperbolic) into the operator
$2u_{\xi\eta}$ on p.~33.
\medskip

{\bf (b)} If we let $v = u_y$ the equation for $v$ becomes $v_x + 3v =
0$.  This is an o.~d.~e.~(in which we can simply treat $y$ as a
constant) with integrating factor $e^{3x}$: it is equivalent to
$[e^{3x}v]_x = 0$, or $e^{3x}v(x,y) = g(y)$, or $v(x,y) = e^{-3x}g(y)$
where $g(y)$ is an arbitrary function.  However, we are not done yet
since this relation simply says $u_y(x,y) = e^{-3x} g(y)$.  But we can
integrate both sides with respect to $y$ (dummy variable of integration
$\eta$) to get
$$ u(x,y) = f(x) + e^{-3x} \int_0^y g(\eta)\,d\eta \hbox{\ \ \ (note that
the ``constant of integration'' depends on $x$})$$

\noindent as the general solution.  The initial conditions $u(x,0) =
e^{-3x}$ and $u_y(x,0) = 0$ force $f(x) = e^{-3x}$ and $e^{-3x}g(0)=0$
respectively, so (since $e^{-3x}\not=0$) the only restriction on $g(y)$
is that $g(0)=0$.  Uniqueness of solution fails spectacularly for this
problem! it looks as though we have given an ``initial position'' and
``initial velocity'' and yet our choice of $g(y)$ is almost
unrestricted.  The reason is that $y=0$ is a {\bf characteristic line} of this
operator: trying to determine solutions by prescribing values on a
characteristic line is asking for difficulties with uniqueness (and
possibly also with existence).  Consider what happened in 
{\bf \S1.2 \#6}.
\medskip

By the way, when this problem was assigned in an earlier version of this
course {\it many\/} people missed the ``constant of integration
depending on $x$'' called ``$f(x)$'' in the solution just given, so
their general solutions contained only one arbitrary function.  Some
were able to recover from this oversight by finding two solutions to the
given quasi-initial-value-problem anyway; the others might have have
been able to continue if there had been more ``degrees of freedom''
available.  In a problem that was hyperbolic and was second-order in
$y$, wouldn't it have seemed natural to have two ``arbitrary functions''
in the general solution, by analogy with what we know happens with the
wave equation? Loosely speaking, just as one expects one arbitrary
constant in a first-order o.~d.~e.~and two constants in a second-order
o.~d.~e., one would expect one arbitrary function in a first-order
p.~d.~e.~and two functions in a second-order p.~d.~e.\par

\vfill \supereject \bye

