Math 309B/C Winter 2012, Homework 6
Due February 22
- Show that any function of the form $c_{1}e^{\mu x}+c_{2}
e^{-\mu x}$ may be expressed as $d_{1}\cosh \mu t+ d_{2}\sinh
\mu t$ for appropriate constants $d_{1}$ and $d_{2}$.
- In this problem, you will solve the heat equation with
initial temperature $f$ and insulated ends: \[\alpha^{2}
u_{xx}=u_{t} \] \[ u_{x}(0,t)=u_{x}(L,t)=0\] \[ u(x,0)=f(x).\]
- Use separation of variables to show that if
$u(x,t)=X(x)T(t)$ solves $\alpha^{2}u_{xx}=u_{t}$, then
\[X''=-\lambda X, \;\; T'=-\alpha^{2} \lambda T\] for some
constant $\lambda$.
- Note that $X'(0)=u_{x}(0,t)=0$ and $X'(L)=u_{x}(L,t)=0$.
Show that the only values of $\lambda$ that allow a nonzero
solution to $X''=-\lambda X$ with this initial data are
$\lambda = \frac{n^{2}\pi^{2}}{L^{2}}$ for $n\geq 0$.
Hint: Recall how we did this with the original
homogeneous equation. Test the cases $\lambda=0$,
$\lambda<0$, and $\lambda>0$. Note that for
$\lambda=0$, however, there is a nonzero solution! Your
solutions should be \[X(x)=\cos \frac{n\pi x}{L}, \;\;
n=0,1,2,...\] Note that if $n=0$, then $X(x)=1$.
- For each such $\lambda$, find out what $T$ is (i.e. solve
$T'=-\alpha^{2}\lambda T=-\alpha^{2}
\frac{n^{2}\pi^{2}}{L^{2}}$ for each $n\geq 0$. (Don't
forget $n=0$!)
- Conclude that
$e^{-\frac{\alpha^{2}n^{2}\pi^{2}t}{L^{2}}}\cos \frac{n\pi
x}{L}$ is a solution for each $n\geq 0$, and verify this by
plugging it into $-\alpha^{2}u_{xx}=u_{t}$.
- Hence, any linear combination of solutions is also a
solution, i.e.
\[\frac{c_{0}}{2}+\sum_{n=1}^{\infty}c_{n}e^{-\frac{\alpha^{2}n^{2}\pi^{2}t}{L^{2}}}\cos
\frac{n\pi
x}{L}\]
is also a solution. (You don't need to prove
anything in this part, I'm just stating a consequence of
what you've shown so far). Note that the $c_{0}$ term
corresponds to $\lambda=0$.
- Now we can for what $c_{0}$ and $c_{n}$ need to be so
that $u(x,0)=f(x)$. If we plug in $t=0$, note that
\[f(x)=u(x,0)=\frac{c_{0}}{2}+\sum_{n=1}^{\infty}c_{n}
\cos\frac{n\pi x}{L}.\] Extend $f$ to $[-L,0]$ by letting
$f(x)=f(-x)$ for $x<0$, so $f$ is even about the origin.
Note that the Fourier coefficients of the sine functions,
$b_{n}$, are all zero for this even function. Hencem
\[f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}a_{n}\cos \frac{n
\pi x}{L}.\] Hence, we let $c_{n}=a_{n}$. Thus, our final
solution is
\[u(x,t)=\frac{c_{0}}{2}+\sum_{n=1}^{\infty}c_{n}e^{-\frac{\alpha^{2}n^{2}\pi^{2}t}{L^{2}}}\cos
\frac{n\pi
x}{L}\]
where
\[c_{0}=a_{0}=\frac{\int_{-L}^{L}f(x)}{L}=\frac{2}{L}\int_{0}^{L}f(x)\]
and
\[c_{n}=b_{n}=\frac{1}{L}\int_{-L}^{L}f(x)\cos
\frac{n\pi x}{L}=\frac{2}{L}\int_{0}^{L}f(x)\cos \frac{n\pi
x}{L}.\] Note that in the third equalities of each line, we
used the fact that the extension of $f$ and hence $f(x)\cos
\frac{ n\pi x}{L}$ are even to simplify the expression.
Now you can use these formulas in solving some of the
problems below.
- Determine whether separation of variables can be used to
replace the given partial differential equations by a pair
of ordinary differential equations (that is, see if you can
replace $u(x,t)=X(x)T(t)$ and find some differential
equations for $X$ and $T$). Note: some of them you can't.
- $xu_{xx}+u_{t}=0$ Answer: Let $u=X(x)T(t)$,
so this equation implies \[xX''T+XT'=0\] \[xX''T=-XT'\]
\[x\frac{X''}{X}=-\frac{T'}{T}=C\] where $C$ is some
constant, hence we have two equations
\[X''=C\frac{X}{x}, \;\;\; T'=-CT.\]
- $u_{xx}+u_{xt}+u_{t}=0$
- $u_{xx} + (x+y)u_{yy}=0$
- $tu_{xx}+xu_{t}=0$ Answer: Letting $u=XT$,
this gives \[tX''T + xXT'=0\] \[ tX'' T=-xXT'\]
\[\frac{X''}{xX}=\frac{-T'}{tT}=C\] so we have the
equations \[X''=CxX, \;\; T'=-CtT.\]
- $u_{xx}+u_{yy}+xu=0$.
In the problems below, you solve some heat conduction
problems. Check out this link for some examples of how to
recognize what type of conduction problem you're dealing with
and how to solve them (please reference your notes for the
general formulas for the solutions).
- Solve the following heat conduction problem:
\[\left\{\begin{array}{cc} 9u_{xx}=u_{t}, & x\in [0,1],
\;\; t>0 \\ u(0,t)=u(1,t)=0, \\ u(x,0)=\sin 2\pi x -\sin
5\pi x.\end{array}\right.\]
- Solve the following heat conduction problem:
\[\left\{\begin{array}{cc} u_{xx}=u_{t}, & x\in [0,2],
\;\; t>0 \\ u(0,t)=u(2,t)=0 \\
u(x,0)=x(1-x).\end{array}\right.\]
- Solve the following heat conduction problem:
\[\left\{\begin{array}{cc} 16u_{xx}=u_{t} & x\in [0,4],
\;\; t>0 \\ u(0,t)=1,\;\; u(4,t)=17 \\ u(x,0)=1+x^2.
\end{array}\right.\] What is the steady state temperature?
(That is, what is $u(x,t)$ converging to as
$t\rightarrow\infty$?)
- Solve the following heat conduction problem:
\[\left\{\begin{array}{cc} u_{xx}=u_{t} & x\in [0,2],
\;\; t>0 \\ u(0,t)=0, \;\; u(2,t)=2 \\
u(x,0)=x\left(1-\frac{x-1}{|x-1|}\right).\end{array}\right.\]
Answer: Note that $u(x,0)=2x$ if $x\in [0,1)$ and
$0$ if $x\in (1,2]$. Let $L(x)=2x$. Let's first solve the
homogeneous problem with initial data $v(x,0)=u(x,0)-L(x)$.
Then \[u(x,0)-L(x)= \left\{\begin{array}{cc} 0 & x\in
[0,1) \\ -2x & x\in (1,2]\end{array}\right.\] So
\[c_{n}=\frac{2}{2}\int_{1}^{2}(-2x)\sin \frac{n\pi x}{2} =
\frac{4}{n\pi}x\cos \frac{n\pi
x}{2}|_{1}^{2}-\int_{1}^{2}\frac{4}{n\pi }\cos \frac{n\pi
x}{2} = \frac{4}{n\pi x}(\cos n\pi- \cos
\frac{n\pi}{2})-\frac{8}{n^{2}\pi^{2}}\sin\frac{n\pi
x}{2}|_{1}^{2} =\frac{4}{n\pi x}(\cos n\pi- \cos
\frac{n\pi}{2})+ \frac{8}{n^{2}\pi^{2}}\sin\frac{n\pi }{2}\]
and so \[v(x,t)=\sum_{n=1}^{\infty}\left(\frac{4}{n\pi
x}(\cos n\pi- \cos \frac{n\pi}{2})+
\frac{8}{n^{2}\pi^{2}}\sin\frac{n\pi
}{2}\right)e^{-n^{2}\pi^{2}t/4}\sin \frac{n\pi x}{2}\] and
our final solution is
\[u(x,t)=L(x)+v(x,t)=2x+\sum_{n=1}^{\infty}\left(\frac{4}{n\pi
x}(\cos
n\pi- \cos \frac{n\pi}{2})+
\frac{8}{n^{2}\pi^{2}}\sin\frac{n\pi
}{2}\right)e^{-n^{2}\pi^{2}t/4}\sin \frac{n\pi x}{2} .\]
Note also that the steady state temperature is
$\lim_{t\rightarrow\infty} u(x,t)=2x$.
- Solve the following heat conduction problem:
\[\left\{\begin{array}{cc} 4u_{xx}=u_{t} & x\in [0,2],
\;\; t>0 \\ u_{x}(0,t)=u_{x}(2,t)=0 \\
u(x,0)=2-x.\end{array}\right.\]
- Solve the following heat conduction problem:
\[\left\{\begin{array}{cc} u_{xx}=u_{t} & x\in[0,\pi],
\;\; t>0 \\ u_{x}(0,t)=u_{x}(\pi,t)=0 \\
u(x,0)=x^{2}.\end{array}\right.\] What is the steady state
temperature?
- Solve the following heat conduction problem:
\[\left\{\begin{array}{cc} u_{xx}=u_{t} & x\in[0,\pi],
\;\; t>0 \\ u_{x}(0,t)=u_{x}(\pi,t)=0 \\ u(x,0)=1+\cos
2x.\end{array}\right.\] What is the steady state
temperature? Answer: Note that this is a heat
insulation problem, so the solution should be of the form
\[u(x,t)=\frac{c_{0}}{2} +\sum_{n=1}^{\infty} e^{-n^{2}t}
\cos n x\] where \[c_{0}=\frac{2}{\pi}\int_{0}^{\pi}(1+\cos
2x)=2\] and \[c_{n}=\frac{2}{\pi}\int_{0}^{\pi}(1+\cos
2x)\cos nx = 0+\frac{2}{\pi}\int_{0}^{\pi}\cos 2x\cos
nx=\left\{\begin{array}{cc} 0 & n\neq 2 &
\frac{2}{\pi} \frac{\pi}{2}=1 & n=2\end{array}\right.\]
and so the final solution is \[u(x,t)=1 + e^{-4t}\cos 2x.\]
The steady state temperature is
$\lim_{t\rightarrow\infty}u(x,t)=1$.
- Note that if a rod has insulated ends, that means no heat
is passing through the ends of the rods. So if no heat is
escaping, then the total heat inside should be constant.
Verify this by showing that, if $u(x,t)$ is a solution to
\[\left\{\begin{array}{cc} \alpha^{2}u_{xx}=u_{t} & x\in
[0,L], \;\; t>0 \\ u_{x}(0,t)=u_{x}(L,t)=0, \\
u(x,0)=f(x),\end{array}\right. ,\] then $\int_{0}^{L}u(x,t)$
does not change in time. What is it always equal to? (
Hint: There are multiple ways of doing this. Here you
are allowed to pull integrals inside of sums, so
$\int\sum\cdots = \sum\int\cdots$, and you may differentiate
under the integral). Answer: The easiest way to
show that this is constant is by showing that
$\frac{d}{dt}\int_{0}^{L}u(x,t)dx=0$:
\[\frac{d}{dt}\int_{0}^{L}u(x,t)dx=\int_{0}^{L}u_{t}(x,t)=\int_{0}^{L}\alpha^{2}u_{xx}(x,t)dx
=
\alpha^{2}u_{x}(x,t)|_{0}^{L}=\alpha^{2}(u_{x}(L,t)-u_{x}(0,t))=0.\]
Here,
we used both the wave equation and the boundary
conditions on $u_{x}$. Hence, the total heat is constant. To
compute the total heat, we need only compute it at $t=0$
(since it will remain the same for all $t$), so this is
\[\int_{0}^{L}u(x,0)=\int_{0}^{L}f(x)dx.\]
- Recall that the two dimensional heat equation is
$\alpha^{2}(u_{xx}+u_{yy})=u_{t}$. Let
$u(x,y,t)=X(x)Y(y)T(t)$, and use the differential equation
to find differential equations for $X,Y,$ and $T$
respectively.