\magnification=\magstep1
\centerline{Mathematics 373 -- Applied Mathematics}
\centerline{Introductory Topics in PDE}
\centerline{January 13, 2001}
\bigskip
\noindent
{\it \S 1.  What is a PDE and where do they come from?}
\bigskip
Last semester we studied the beginnings of the theory of
{\it ordinary differential equations}.  This semester, we
turn our attention to the study of {\it partial differential
equations}.  Like an ordinary differential equation,   
a partial differential equation (PDE) is a relation between
derivatives of an unknown function (the ``dependent
variable'').  Only now the
function depends on several independent
variables and other known functions of those variables,
and the derivatives are partial derivatives.
If the independent variables are $(x_1,\ldots,x_n)$,
the relation is assumed to be true on some 
set in ${\bf R}^n$ (possibly for all $(x_1,\ldots,x_n)$).
 
To simplify notation in writing PDE, we will usually
indicate the derivatives with subscripts.  For instance,
${\partial u \over \partial x}$ will be written as $u_x$.
Higher order partial derivatives are indicated the same
way.  For example, if $u = u(x,y)$, then
$u_{xxx}$ means ${\partial^3 u\over \partial x^3}$
and $u_{xy}$ means ${\partial^2 u\over\partial x\partial y}$.
Some first examples of PDE are:
\bigskip
\item{1)} $u_x + 5u_t + u = \sin(x)\cos(t)$, where $u = u(x,t)$,
\medskip
\item{2)} $u_t = ku_{xx}$, where $u = u(x,t)$ again,
\medskip
\item{3)} $u_{xx} + u_{yy} + u_{zz} = \sin(u)$, where $u = u(x,y,z)$,
\medskip
\item{4)} $u_t + uu_x + u_{xxx} = 0$, where $u = u(x,t)$,
\medskip
\item{5)} $u_{tt} = c^2\left(u_{rr} + {1\over r} u_r + u_{\theta\theta}\right)$,
where $u = u(r,\theta,t)$.
\bigskip

As was the case last semester for ODE, we will say a PDE
is {\it linear} if the unknown function and its derivatives
only appear linearly in the equation.  Of the examples above,
1,2,5 are linear PDE (the $c^2/r$ multiplying the $u_r$ term
in equation 5 is OK -- the equation is still linear -- since
the $1/r$ does not depend on the dependent variable $u$).
 But 3 and 4 are not (because of the $\sin(u)$
term in 3 and the $uu_x$ term in 4).  Examples 2 and 5 are
{\it homogeneous linear}, while 1 is an {\it inhomogeneous linear}
equation.  

The {\it order} of a PDE is the highest order derivative that
appears (this is also the same idea as for ODE).  For instance,
in the examples above, 1 is a first order equation, 2, 3, 5 are
second order equations, and 4 is a third order PDE.

Most of the PDE that we will study come from models of physical,
chemical, or biological systems.  A typical situation
that leads to PDE is when we are studying a quantity in 
a system that
depends {\it both on time and on the location in space\/} where
the quantity is measured.  For example, 
we will see shortly that equations such as 2 above
arise in the study of heat conduction along
a long thin wire (when the thermal properties of the wire
are constant along its whole length).  Since 
the temperature will usually depend both
on the location along the wire (measured by the {\it spatial}
coordinate $x$) where the measurement is made, 
{\it and} on the time when the measurement is made,
we naturally have to deal with a function $u(x,t)$ of two 
variables to describe the temperature of the wire
completely.  In a similar fashion, we will see that equation
5 gives a model for the vibrations of a circular membrane
like a drumhead under certain physical assumptions.  Here there 
are two spatial coordinates, $r,\theta$ specifying the location 
of a point on the drumhead
in {\it polar coordinates} -- the natural choice of coordinates
for a circular geometry -- and time $t$ as a third independent 
variable.  The function $u(r,\theta,t)$ in this case will
represent the {\it vertical displacement} of the membrane
at the location $(r,\theta)$ and at time $t$.

Most of our work this semester will be devoted to understanding
different techniques of solving PDE, possibly together with
other side conditions such as specified initial conditions,
or specified values for $u$ at particular values of spatial 
coordinates.  We will begin with the general class of linear
first order PDE in 2 variables.  These already illustrate
some of the key patterns concerning PDE and the differences
between PDE and ODE.  We will not try to provide a complete
treatment of every possible special case.  (For several
reasons to be discussed below, that is very difficult if
not impossible in the most naive sense -- i.e. if we are
looking primarily for analytic formulas for solutions.)
Instead we will give a rapid survey focusing on a few important 
types of these equations.
\bigskip
\noindent
{\it \S 2.  First order linear PDE}
\bigskip
The general first order linear PDE for $u = u(x,t)$ has the 
form:
$$a(x,t)u_x + b(x,t)u_t + c(x,t)u = d(x,t).$$
(Even though we write $t$ here for the second independent
variable, we are not necessarily thinking of $t$ as
representing time.  Both variables could represent 
spatial coordinates in equations of this form.)
We will begin our study of these equations by considering
{\it homogeneous equations} ($d(x,t) \equiv 0$), in which
$c(x,t) \equiv 0$ as well, and in which $a,b$ are constants.
In this case, it will be easy to derive the form of solutions.
Moreover, the features of this special case are representative
of what happens in general.  

For instance, let's consider the PDE 
$$3u_x + 2u_t = 0.\leqno(1)$$
To understand what (1) says about the function
$u(x,t)$, and use that understanding to derive the form of 
solutions, we need to recall some facts from multivariable
calculus.  First, the partial derivatives $u_x,u_t$ are
the components of the {\it gradient vector} 
$$\nabla u(x,t) = (u_x(x,t),u_t(x,t)) = 
\left({\partial u\over\partial x}(x,t),
{\partial u\over\partial t}(x,t)\right).$$
Thinking of the operations of vector algebra, we see that
the equation (1) just says that the vector $(3,2)$ has {\it dot
product} 0 with $\nabla u$:  $0 = (3,2)\cdot (u_x,u_t) = 3u_x + 2u_t$.
Hence, using the geometric interpretation of the dot
product, the vector $(3,2)$ is perpendicular to $\nabla u$
at every point.  The next ingredient we need is the geometric
meaning of the gradient vector of a function.  Recall 
that the gradient $\nabla u$ is perpendicular to the contours or
level curves of the function $u$ at every point.  
Since $(3,2) \perp \nabla u$ and $\nabla u \perp$
the level curves of $u$ (and since we are working in the 2-dimensional
$(u,t)$-plane), we see: $(3,2)$ must be tangent to the
level curves of $u$ at every point.  A moment's thought shows
that this means that {\it the level curves of the function
$u(x,t)$ are the lines with $(3,2)$ as a direction vector}, which
are all the lines of the form $2x - 3t = c$, $c$ constant.
(Note: we chose $(2,-3)$ as a convenient normal vector since
its components contain no fractions.   Any other normal vector
such as $(1,-3/2)$ would work just as well, and the level
curves could also be written as $x - 3t/2 = c$.)
Hence $u(x,t)$ must take constant values along each of the lines
$2x - 3t = c$.  This means that $u$ is a function of $2x - 3t$ only.
In other words, every solution of (1) must have the form 
$u(x,t) = f(2x - 3t)$ for some function $f$ of one variable.

Conversely, now, suppose $u(x,t) = f(2x - 3t)$ for some differentiable
one variable function $f$.  Then by the chain rule:
$$\eqalign{
u_x &= f'(2x - 3t)\cdot {\partial \over \partial x}(2x-3t) = 2 f'(2x-3t)\cr
u_t &= f'(2x - 3t)\cdot {\partial \over \partial t}(2x-3t) = -3 f'(2x-3t)\cr
}$$
As a result, 
$$3u_x + 2u_t = 3\cdot 2f'(2x-3t) + 2\cdot (-3 f'(2x-3t)) = 0$$
so the PDE (1) is satisfied.

In other words, we have shown that the the solutions of (1)
are exactly the functions $u(x,t) = f(2x - 3t)$, where 
$f$ is an {\it arbitrary differentiable function of one variable}.
This is the first big difference between PDE and ODE.  Recall
we saw last semester that the general solution of a first
order ODE has one arbitrary {\it constant}.  The general solution of
a first order PDE contains an {\it arbitrary function of one variable}
instead. This means that PDE typically have a {\it much larger}
collection of solutions than ODE!  In fact, in the 
collection of solutions of (1), we can find a particular
solution agreeing with any given function at points of a
{\it curve that intersects each level curve $2x - 3t = c$ only once}.  

For instance, suppose we wanted to solve (1) together
with the side (initial) condition $u(x,0) = \sin(x)$.  
The line $t = 0$ (the $x$-axis in the $(x,t)$-plane) meets
each level curve of the solutions in just one point.
Hence in the general solution $u(x,t) = f(2x - 3t)$
of (1), if we set $t = 0$ we see that $f(2x) = \sin(x)$,
so $f(x) = \sin(x/2)$.  Therefore the solution of 
(1) and $u(x,0) = \sin(x)$ is the function 
$$u(x,t) = \sin((2x - 3t)/2) = \sin(x - 3t/2).$$ 

For equations $au_x + bu_t = 0$ with $a,b$ constant (and not
both zero, or else the equation is pretty uninteresting!), the 
situation is the same as in the example we just worked out.
The geometric reasoning we applied shows that the 
general solution has the form $u(x,t) = f(bx - at)$
where $f$ is an arbitrary differentiable function of one
variable.  Solutions are constant along the lines
$bx - at = c$, $c$ constant.  

We will also now present
a second, more algebraic, way to derive this fact.
The algebraic method will also be useful for
more general first order linear PDE with constant
coefficients on the $u_x,u_t$ terms.  As
we know from the geometric approach, the lines $bx - at = c$
are important for this problem.  So let's consider
changing coordinates in the $(x,t)$ plane, introducing
$$\eqalign{r &= bx - at\cr
           s &= ax + bt\cr}\leqno(2)$$
as new rectangular coordinates, where $a,b$
are the constant coefficients appearing in the PDE. 
We choose $s$ to have
this form since this makes the lines $s = const$ 
perpendicular to the lines $r = const$, as is the case
for the original $x,t$-coordinate system.  

If $u$ is any function of $r,s$, then we can also think
of $u$ as a function of $x,t$ via (2).  Conversely,
since we assume $(a,b) \ne (0,0)$, the equations (2) can 
also be solved for $x,t$ as functions of $r,s$:
$$\eqalign{
x &= {1\over a^2 + b^2}(br + as)\cr
t &= {1\over a^2 + b^2}(-ar + bs)\cr}\leqno(3)$$
and we can think of any function of $r,s$ as a function
of $x,t$.  

Now we see what becomes of the PDE $au_x + bu_t = 0$
in the new $r,s$ coordinate system.  By the chain rule
and (2),
$$\eqalign{
u_x &= {\partial u\over \partial r}{\partial r\over \partial x} + 
      {\partial u\over \partial s}{\partial s\over \partial x}
    = b u_r + au_s\cr
u_t &= {\partial u\over \partial r}{\partial r\over \partial t} + 
      {\partial u\over \partial s}{\partial s\over \partial t}
    = -a u_r +b u_s\cr}$$
Hence
$$0 = au_x + bu_t = a(bu_r + au_s) + b(-au_r+bu_s) = (a^2+b^2)u_s
\Rightarrow u_s = 0.$$
This says that if we think of $u$ as a function
of $r,s$, then $u$ depends {\it only on $r$, not on $s$}.  Therefore,
$u(r,s) = f(r) = f(bx - at)$ for some function $f$ of one
variable, as we expect.

We ask, what does the form of this solution mean?  First note
that every function $f(bx - at)$ of $bx - at$ alone can also 
be seen as a function
$g(x - {at\over b})$ of $x - {at\over b}$ alone (just rescale
to get $g(x) = f(bx)$).  Even though we
do not always think of $t$ as representing time, a particularly
nice picture emerges if we do think that way now.  At $t = 0$
we have $u(x,0) = g(x)$, so $g(x)$ gives the initial
shape of the solution as function of $t$.  But then for 
other $t$, $u(x,t) = g(x - {at\over b})$, which just gives
a shifted copy of the graph of $g$.  Depending on
the signs of $a,b$, the shifting is either to the left or 
to the right  (left if ${at\over b} < 0$ and right if
${at\over b} > 0$).  The amount of shifting depends
linearly on $t$.  Hence the solutions of equations of 
this form are all ``traveling waves'' with a constant shape
that propagates to the left or right as $t$ increases.
First order linear equations with constant coefficients
are therefore models for what is called {\it transport},
e.g. motion at a constant velocity. 

We next turn to the case where the coefficients of
$u_x$ and $u_t$ are not constant.  In this case, 
the geometric meaning of the PDE 
$$a(x,t)u_x + b(x,t)u_t = 0\leqno(4)$$
is the same:  At each point $(x,t)$, the vector $(a(x,t),b(x,t))$
is perpendicular to the gradient $\nabla u$, hence
tangent to the level curves of $u$.  Since $(a,b)$ is
not a constant vector, though, the level curves are usually 
{\it not straight lines}.  Instead, they form a family of curves
called the {\it characteristic curves} of the PDE.   We can find
the characteristics and the general solution of the PDE
by the following:
\bigskip
\proclaim Method of Characteristics.  To find the general
solution of 
$$a(x,t)u_x + b(x,t)u_t = 0,$$
\item{1)} Solve the following ODE:
$${dx\over dt} = {a(x,t)\over b(x,t)}$$
(If it is more convenient to do so, it is also possible
think of $x$ as the independent variable in this
equation and solve
$${dt\over dx} = {b(x,t)\over a(x,t)}.)$$
\item{2)}  Write the solutions in the implicit form $\phi(x,t) = c$.  
\item{3)}  Then the general solution of (4) will have the form 
$$u(x,t) = f(\phi(x,t))$$
where $f$ is an arbitrary differentiable function of one 
variable.\par

Here is an example illustrating how this process works.  We
look for the general solution of the first order linear 
PDE
$$(x - t)u_x + tu_t = 0\leqno(5)$$
The equation of the characteristic curves is the ODE
$${dx\over dt} = {x-t\over t} = {1\over t}x - 1$$
which is a first order linear equation with $p = {-1\over t}$
and $q = -1$.  The general solution of this ODE is
$$x = c t - t \int {1\over t}\ dt = ct - t\ln|t|$$
Writing this in the required implicit form, we get
$$\phi(x,t) = {x+t\ln|t|\over t} = {x\over t} + \ln|t| = c.$$
This is the equation of the characteristic curves of (5).
So the general solution of (5) looks like
$$u(x,t) = f\left({x\over t} + \ln|t|\right)$$
for an arbitrary differentiable function $f$ of one variable. 
(Note that no solution is defined at $t = 0$.)

As in the constant coefficient case we discussed above, 
particular solutions taking specified values at the points
of ``non-characteristic'' curves (curves intersecting
characteristics only once) can be found from the general
solution by solving for $f$.  For instance, the line 
$t = 2$ intersects the characteristic ${x\over t} + \ln|t| = c$
only at $x = (c - \ln(2))/2$.  Hence
a particular solution of (5) satisfying the condition
$u(x,2) = e^x$ (for instance) can be determined by substituting
$t = 2$ in the general solution.  We get
$$e^x = f(x/2 + \ln(2))$$ 
Hence $f(x) = e^{2x - 2\ln(2)} = e^{2x}/4$, and our desired
particular solution is 
$$u(x,t) = e^{{2x\over t} + 2\ln|t|}/4 = {1\over 4} t^2 e^{2x\over t}.$$

As you can probably see, being able to produce an 
analytic formula for $u(x,t)$ depends
on being able to solve the ODE defining the characteristic curves.
It is very easy to write down examples where that equation has 
no solutions in elementary functions.  In those cases the method
fails in the sense that we cannot use it to get a formula
(unless we define new ``special functions'' to solve the problem). 
{\it However, it is also
very important to understand the underlying geometry of
the method of characteristics from 
a conceptual standpoint.  That is always valid even if the
method cannot be followed to yield a formula for the general
solution of the PDE.  In other words, the method of characteristics
is as much a way to understand what the solutions of first
order linear PDE look like conceptually and graphically
as it is an algorithm to produce analytic solutions.}

We will conclude this section with two additional examples
illustrating how some first order linear equations involving
$u$ terms and inhomogeneous terms can be solved by the methods
we have introduced.  We will not consider the most general
equations of these forms, which frequently have no elementary
solutions.  Instead, we will focus on the linear equations
with constant coefficients.  

Our first example is the general constant coefficient
homogeneous linear PDE:
$$au_x + bu_t + cu = 0\leqno(6)$$
To solve this, we use the change of coordinates from (2) above:
$$\eqalign{r &= bx - at\cr
           s &= ax + bt.\cr}$$
As before, we have 
$$\eqalign{
u_x &= {\partial u\over \partial r}{\partial r\over \partial x} + 
      {\partial u\over \partial s}{\partial s\over \partial x}
    = b u_r + au_s\cr
u_t &= {\partial u\over \partial r}{\partial r\over \partial t} + 
      {\partial u\over \partial s}{\partial s\over \partial t}
    = -a u_r +b u_s\cr}$$
So the equation (6) transforms to
$$a(bu_r + au_s) + b(-au_r + bu_s) + cu = (a^2+b^2)u_s + cu= 0$$
in the new coordinate system.  For PDE like this one involving
a derivative with respect to only one of the independent
variables, we can solve using methods from ODE.  This is 
essentially just a separable ODE for $u$, but the arbitrary
``constant'' will now be a function of the other variable $r$
(this is how the arbitrary function enters the solution in 
this case).  We have 
$$u_s = {-c\over a^2 + b^2}u$$
so 
$$u(r,s) = g(r)e^{-cs\over a^2 + b^2},$$
where $g$ is an arbitrary differentiable function of one variable.
In terms of the original $x,t$ variables, this becomes:
$$u(x,t) = g(bx - at)e^{-c(ax+bt)\over a^2+b^2}\leqno(7)$$
Caution: In this case the lines $bx - at = const$ 
are not level curves for the solution.  Because of the
other exponential factor $u(x,t)$ is not a function of 
$r = bx - at$ alone.  Thus, even though many authors
call the lines $bx - at = const$ characteristic curves
here too, the characteristics are not the level curves
for equations of this form.   

Finally, we consider an example of this type where the 
equation is not homogeneous:
$$u_x + 2u_t + 3u = e^{x - 4t}.$$
As we saw for linear ODE last semester, it is also true
for PDE that the general solution of an inhomogeneous
equation can be written as 
$$u = u_h + u_p$$
where the subscripts here do {\it not} stand for partial
derivatives, but rather for ``homogenenous'' and ``particular'':
$u_h$ is the general solution of the corresponding homogeneous
equation with right side equal to zero, and $u_p$ is any 
particular solution of the equation.  (The proof of this
assertion is the same as for ODE: Since the equation is 
linear, if $u$ and $u_p$ are any two solutions, then
$u - u_p$ solves the corresponding homogeneous equation.)

We find the general solution of the homogeneous equation
as in the solution (7) of the general equation (6).  We
have $a = 1, b = 2, c = 3$, so 
$$u_h(x,t) = g(2x - t)e^{-3(x + 2t)\over 1^2 + 2^2} 
= g(2x-t)e^{-3x-6t\over 5}.$$

To find a particular solution, we can use a variant of the
method of undetermined coefficients from ODE.  Be aware
that this is possible only because of the special form of the 
right hand side of this equation, though.  By the chain
rule, we have
$${\partial\over\partial x} e^{x-4t} = e^{x-4t}\qquad
{\partial\over\partial t} e^{x-t} = -4e^{x-4t}$$
Hence we see that a particular solution of the form
$$u_p = A e^{x-4t}$$
will exist.  We must choose $A$ so that $A - 4A + A = -2A = 1$,
or $A = -1/2$.  Therefore we get a general solution
$$u(x,t) = g(2x-t)e^{-3x-6t\over 5} - {1\over 2}e^{x - 4t}.$$

A particular solution satisfying the initial condition $u(x,0) = 0$
(for instance) can be found from the equation:
$$0=u(x,0) = g(2x)e^{-3x/5} - {1\over 2}e^x$$
Hence
$$g(2x) = {1\over 2}e^x\cdot e^{3x/5} = {1\over 2}e^{8x/5}$$
so 
$g(x) = {1\over 2}e^{4x/5}$.
\bigskip
\noindent
{\it \S 3.  D'Alembert's solution of the wave equation
with one spatial dimension}
\bigskip
In \S 10.2 of Nagle, Saff, and Snider, the wave equation
with one space variable is given in the form
$$u_{tt} = c^2 u_{xx}$$ 
where $c > 0$ is a constant.  We will begin this section
by (very) briefly indicating how this PDE is derived.   
We consider a long, thin, flexible, elastic, wire or 
string (think of a piano or guitar string) made of 
a material that is uniform along the length of the string.
In the problems we will solve later, we will often assume
that the ends of the string are fixed.
If the string is plucked and released, then the tension
each segment of the string exerts on its neighboring segments
will cause the string to move back towards its rest position
at all times.  Hence the string will tend to 
oscillate or vibrate around its rest position.

We will assume that the string 
vibrates only in a single plane.  (If the plucked position of the
string is planar, an actual string will tend to vibrate in just 
one plane,  at least to a good approximation.)
We will use an ordinary rectangular $xy$ coordinate system
in this plane, where the rest position of the string
is on the $x$-axis ($y = 0$).    If we assume that the 
string is stretched taut to begin with, then the shape of the
string will be given by some graph $y = u(x,t)$
at all later times.  In other words the string
is only vibrating ``up and down'' (transversely) and
not ``from side to side'' (longitudinally) in the plane.  
The function $u(x,t)$ can 
also be understood as the {\it vertical
displacement function} -- it  gives the 
difference between the actual location of
the point on the string with coordinate $x$ at time $t$ 
and the rest position of that point.

Because of our assumptions on the flexibility of the
string, if we look at any finite segment, say from 
$x = x_1$ to $x$, then the tension forces
exerted by the rest of the string on the segment
will be represented by vectors ${\bf T}(x_1,t)$ and
${\bf T}(x,t)$ that are {\it tangent} to the string
at $x=x_1$ and $x$ respectively (for all $t$).
Let $T(x,t)$ be the magnitude of the tension force
at $x$ at time $t$.  Because the string is not vibrating
longitudinally, only the vertical component of the tension
is producing motion.  At each point the vertical component
of the tension is $T\sin(\theta)$, where $\theta$ is the 
angle between the horizontal and the tangent to the string.
We have $\sin(\theta) = u_x/\sqrt{1 + u_x^2}$.  But
if we assume that the vertical displacement is {\it small
compared to the length of the string}, then the values
of $u_x$ will also be small, and we can use the {\it approximation}
$$\sqrt{1 + u_x^2} \doteq 1.$$

Then Newton's Law $F = ma$ implies that
$$\int_{x_1}^{x} \rho u_{tt}\ dx = Tu_x |_{x_1}^x$$
where $\rho$ is the mass density of the string (a constant).  (We 
ignore the approximation made above and take this as an exact relation.)
Differentiating both
sides with respect to $x$ gives $\rho u_{tt} = T u_{xx}$
or
$$u_{tt} = c^2 u_{xx}$$
where $c = \sqrt{T\over \rho}$.  This is the wave equation
with one space variable.

Our next goal is to derive some first information about
the solutions of this second order linear PDE.  A physically
meaningful problem to be solved would be one where we
have a string of finite length, $L$ say.  We can take
the string to correspond to the interval $[0,L]$ on
the $x$-axis.  Moreover, we might assume that the ends of the
string are fixed at  0 displacement, and the initial shape
of the string and the initial velocity at each point of
the string are known.  Mathematically, this translates into
an {\it initial-boundary value problem}:
$$\eqalign{u_{tt} &= c^2 u_{xx}\cr
           u(0,t) &= u(L,t) = 0,\ {\rm all}\ t\cr 
           u(x,0) & = f(x)\cr
           u_t(x,0) &= g(x).\cr}\leqno(8)$$

In most cases here,
we would look for solutions defined for $x\in [0,L]$ and not
care whether the definition could be extended beyond
that interval.  We will see how to bring the ``boundary conditions''
$u(0,t) = u(L,t) = 0$ for all $t$ into the picture in a few
days.  For now, let us see how our work with first order
linear ODE can be applied to solve the wave equation
without the assumptions about the range of realistic values
for $x$.  In other words, let us consider solutions of the
wave equation and initial conditions as in (8) but ignoring
the boundary conditions.  The solutions we get will include
the solutions of the PDE plus the boundary conditions, so 
we will get the solutions we want plus many others.  Interestingly,
though, if the string is very long and we are not near one of the
ends ($x$ is far from $0$ and $L$), then we could think of every 
solution as representing an approximation to a possible behavior 
of the string at that $x$.  The same results we will see 
are also derived by a slightly different method in [NSS] on pages 654-5.

The derivation we will see is based on the following observation.
The wave equation is a second order linear homogeneous equation that
can be written in the ``operator form''
$$\left({\partial^2 \over \partial t^2} 
- c^2 {\partial^2 \over \partial x^2}\right)u = 0\leqno(9)$$
Because the coefficients are constant here and because
the mixed partials of a nice ($C^2$) function satisfy
$u_{xt} = u_{tx}$, this can be rewritten in the following form
by factoring the second order wave equation operator into 
a product of two first order partial differential operators.
The equation (9) is equivalent to
$$\left({\partial\over \partial t} + c{\partial \over \partial x}\right)
\left({\partial\over \partial t} - c{\partial \over \partial x}\right) u = 0\leqno(10)$$
To make use of this, let 
$v = \left({\partial\over \partial t} - c{\partial \over \partial x}\right) u$
Then (10) says $v$ satisfies the first order linear PDE with constant
coefficients
$$\left({\partial\over \partial t} + c{\partial \over \partial x}\right)v 
= cv_x + v_t = 0$$
The results of \S 2 say that the general solution of this equation is
$$v(x,t) = \phi(x - ct)$$ where $\phi$ is an arbitrary differentiable
function of one variable.  As we saw there, we can think of $v$
as a traveling wave of constant shape propagating to the 
right ($c > 0$) at a constant speed $c$.  

Next, the wave equation (9) is equivalent to the equation 
$$u_t - cu_x = v(x,t)$$
where $v(x,t) = \phi(x-ct)$ is the function just found.  This is 
an inhomogeneous first order linear PDE.  The general solution
can be found by the method discussed in the example at the end
of \S 2.  The general solution of the associated homogeneous
equation is $u(x,t) = B(x + ct)$ where $B$ is an arbitrary 
differentiable function.  Note that this represents 
a second traveling wave of constant shape, propagating to the left
($c > 0$) as $t$ increases.  To find a particular solution,
we note that if $u(x,t) = A(x-ct)$ is any function of $x - ct$
alone, then by the chain rule
$$u_t - cu_x = -c A'(x-ct) - cA'(x-ct) = -2c A'(x - ct).$$
Hence, given $\phi$ as above, if $A = {-1\over 2c}\int \phi(s)\ ds$
(that is ${-1\over 2c}$ times an antiderivative of $\phi$), then
$u(x,t) = A(x-ct)$ solves $u_t - cu_x = \phi(x-ct)$.   Since
$\phi$ was general, though, so is $A$.   Our
conclusion is that any function of the form
$$A(x - ct) + B(x + ct)\leqno(11)$$
solves the wave equation (9).  Conversely, it is
easy to check that every function of this form ($A$,$B$ any two
arbitrary differentiable functions) satisfies the wave equation.

In other words, {\it the general solution of the wave equation
on ${\bf R}$ looks like a superposition of two traveling 
waves, one propagating to the left at constant speed $c$, one
propagating to the right at constant speed $c$}.  The shapes
of the two waves are completely arbitrary and need not be the
same.  This result was discovered by the French mathematician
and philosopher D'Alembert in the 18th century.   

Continuing from (11) it can be seen that the solution of the
wave equation and the initial conditions $u(x,0) = f(x)$,
$u_t(x,0) = g(x)$ ($f$,$g$ given) can be written as
$$u(x,t) = {1\over 2}(f(x + ct) + f(x - ct)) + {1\over 2c}
\int_{x-ct}^{x+ct} g(s)\ ds.$$
The two traveling waves are given by $A(x - ct)$ and 
$B(x + ct)$ for 
$$A(x) = {1\over 2}\left(f(x) - {1\over 2c}\int_0^x g(s)\ ds\right)$$
and 
$$B(x) = {1\over 2}\left(f(x) + {1\over 2c}\int_0^x g(s)\ ds\right)$$
See page 655 in [NSS] or the class notes for a proof.
\bye 
