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{\bf ASSIGNED EXERCISES OF MATH544: PDE set  6} 
{\it April, 2001}
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\noindent
{\bf  THE HEAT EQUATION}\\
Ronald B Guenther and John W Lee, Partial Differential Equations of
Mathematical Physics and Integral Equations

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\noindent
{\bf 1}.\, Solve the following initial value problem

\begin{eqnarray}
u_{t}=au_{xx},~~ (a>0), t>0,~~~x \in [0,L],\\
u(0,t)=u(L,t)=0, ~~t \ge 0,\\
u(x,0)=f(x),~~ x \in [0,L]
\end{eqnarray}

\noindent
{\bf step 1}.\, Formal Solution: Using the separation of variables we first
let $u(x,t)=T(t)X(x)$. Then find that
$$T^{\prime}=a k T,~~~ X^{\prime \prime }=k X$$
where $k$ is any constant. It must be a negative number $k=-\lambda^2$.
we find that
$$T=e^{-a\lambda_{n}^2 t}, X=\sin(\lambda_{n} x)~~, n=1,2,\cdots$$
where $\lambda_{n}={n\pi \over L}$. Hence
\begin{equation}
u(x,t)=\sum_{n=1}^{\infty}\, \alpha_{n}\, e^{-a \lambda_{n}^2 t}\, 
\sin (\lambda_{n} x)
\end{equation}
Here $\alpha_{n}$ are constants , will be determined by the initial condition
(assuming the  $t$ goes to zero exists which requires the uniform convergence 
of the sine series)
\begin{equation}
f(x)=\sum_{n=1}^{\infty} \alpha_{n} \sin (\lambda_{n} x),\\
\end{equation}
with
\begin{equation}
\alpha_{n}={2 \over L} \int_{0}^{L}\,f(x) \sin(\lambda_{n} x) dx,~~n=1,2, 
\cdots
\end{equation}

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\noindent
{\bf step 2}.\, Is the formal solution a solution of the initial and 
boundary value problem we started with?. For this the given function
must satisfy certain conditions. Term-by term differentiation of  the
series representing $u(x,t)$ in (4) can be justified at each point
$(\tilde{x}, \tilde{t})$ with $\tilde{t} > 0$ and $ 0 \le \tilde{x} \le L$
 by showing that the series in (4) and the series representing 
$u_{t}$, $u_{x}$ and $u_{xx}$ that result from term-by-term differentiation
, converge uniformly in a region $D$ which contains the point 
$(\tilde{x}, \tilde{t})$.
Such a region is 
$$D=\{(x,t)| 0 \le x \le L,~ t \ge { \tilde{t} \over 2} \}.$$ 
We first assume that the function $f(x)$ is bounded in $[0,L]$
$$|f(x)| \le M, ~~~ for ~all~ x \in [0,L]$$ 
Then from (6) we have 
$$|\alpha_{n}| \le 2M$$
and series (4) satisfies
$$|\sum_{n=1}^{\infty}\, \alpha_{n}\,e^{-a\lambda_{n}^2\,t}\, 
\sin (\lambda_{n} x)\, | \le 2M \sum_{n=1}^{\infty} \lambda_{n}^2\, 
e^{-a \lambda_{n}^2\,\tilde{t}/2}.$$
The series at the right hand side converges uniformly  in $D$. Similar
reasoning applies also for the series representing $u_{t}$, $u_{x}$,
and $u_{xx}$. Consequently (4) satisfies the equations (1) and (2).

To guarantee that (4) will also satisfy the initial condition (5), we
need further assumptions on $f(x)$. For the uniform convergence of
the series in (5) we need $\sum_{n=1}^{\infty} |\alpha_{n}| < \infty$.
If $f(0)=f(L)=0$ then
$$\alpha_{n}=-{2 \over \lambda_{n}^2 L} \int_{0}^{L}\, f^{\prime \prime}
\sin (\lambda_{n} x) dx.$$
Hence , if $f^{\prime \prime } $ is continuous on $0 \le x \le$,
then it is bounded say by $M^{\prime \prime}$, and
$$|\alpha_{n}| \le {2M^{\prime \prime } L^2 \over \pi^2\, n^2}$$
This estimate implies that the series (4) for $u(x,t)$ is uniformly convergent
for $t \ge 0$ and $ 0 \le x \le L$. Since $u(x,t)$ is the uniform limit of 
continuous functions , it is continuous. Hence we have the following:

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\noindent
{\bf Theorem}. {\it If $f(x)$ is bounded and integrable, the series
in (4) yields a solution to (1) and (2) for $t > 0$ and $0 \le x \le L$.
If in addition $f$ has continuous second derivative and $f(0)=f(L)=0$, then
$u(x,t)$ defined in (4) is uniformly convergent and continuous for 
$t \ge 0$ and $ 0 \le x \le L$ and also satisfies the initial condition (3).}

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\noindent
{\bf 2}.\, Solve the following initial and boundary value problem:
\begin{eqnarray}
u_{t}-a u_{xx}=F(x,t),~~~ t >0,~ 0 < x < L,\\
u(0,t)=u(L,t)=0,~~ t \ge 0,\\
u(x,0)=f(x),~~ 0 \le x \le L
\end{eqnarray}
Complete the method of solution given in class. The method of solution
 is similar to previous problem.

\noindent
{\bf 3}.\, Solve the problem:
\begin{eqnarray}
u_{t}-a u_{xx}=F(x,t),~~~0<x<L, ~t>0,\\
u_{x}(0,t)=u_{x}(L,t)=0,~~~t \ge 0,\\
u(x,0)=f(x),~~~ 0 \le x \le L.
\end{eqnarray}
Explain also how you can prove the uniqueness of the solution 
of this problem.

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\noindent
{\bf 4}.\, Let $u(x,t)$ be a solution to the problem
\begin{eqnarray}
u_{t}-a u_{xx}=0,~~~ t>0, ~ 0 \le x \le L,\\
u(0,t)=u(L,t)=0,~~~ t \ge 0\\
u(x,0)=f(x), ~~~ 0 \le x \le L
\end{eqnarray}
let $\tilde{u}$ be a solution of the above problem with $f$ replaced by 
$\tilde{f}$. Denote the respective Fourier coefficients by $\alpha_{n}$
nad $\tilde{\alpha_{n}}$. Show $|\alpha_{n}-\tilde{\alpha_{n}}|
\le 2||f^{\prime \prime}-{\tilde{f}}^{\prime \prime}||/\lambda_{n}^2$ and hence
$$||u-\tilde{u}|| \le 2||f^{\prime \prime}-{\tilde{f}}^{\prime \prime}||
\left( \sum_{n=1}^{\infty}\, {1 \over \lambda_{n}^2} \right)$$
where $||h||=max |h(x)| $ for $ 0 \le x \le L$. Deduce continuous dependence
and hence the well-posedness of the problem.


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\noindent
{\bf 5}.\, For $T>0$ let 
$$Q_{T}=\{(x,t)| 0 <x <L, ~ 0<t \le T \}$$
(this is exactly the region $D$ defined in the lecture), be a rectangle 
with its bottom and two vertical sides removed. Let
$${\bar{Q}}_{T}=\{(x,t)| 0 \le x \le L, ~ 0 \le t \le T \}.$$
Then $B_{T}={\bar{Q}}_{T}-Q_{T}$ consists of the three missing sides of the 
original rectangle $Q_{T}$. Suppose that $w(x,t)$ satisfies the differential inequality
$$w_{t}-a(x,t)w_{xx}-b(x,t)w_{x}-c(x,t)w < 0~~ (x,t) \in Q_{T}$$
where $a(x,t) \ge 0$ and $c(x,t) \le 0$  in $Q_{T}$. Then $w(x,t)$ can not 
achieve a positive local maximum in $Q_{T}$

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\noindent
{\bf 6}.\, Let $a,b$ and $c$  be as in the previous problem. Suppose that 
$u(x,t)$ is continuous on ${\bar{Q}}_{T}$ and satisfies 
\begin{eqnarray}
u_{t}-a(x,t)u_{xx}-b(x,t)u_{x}-c(x,t)u \le 0, ~~~(x,t) \in Q_{T},\\
u(x,t) \le 0,~~~ (x,t) \in B_{T}
\end{eqnarray}
Then $u(x,t) \le 0$ on ${\bar{Q}}_{T}$.

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\noindent
{\bf 7}.\, Let $a,c$ and $c$ be as in Prob.4. Suppose that $u(x,t)$ is 
continuous on ${\bar{Q}}_{T}$ and satisfies 
$$u_{t}-a(x,t)u_{xx}-b(x,t)u_{x}-c(x,t)u=0$$
in $Q_{T}$. If $u$ has a positive maximum , it is assumed on the boundary 
$B_{T}$ and if $u$ has a negative minimum , it is assumed on $B_{T}$.

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\noindent
{\bf 8}.\,Let $a,b$ and $c$ be as in Prob. 4. Suppose that $F(x,t)$ is 
defined in $Q_{T}$ and bounded by the constant $N$ there. Suppose that
$u(x,t)$ is continuous on ${\bar{Q}}_{T}$, $M=max_{B_{T}} |u(x,t)|$,
and $u(x,t)$ satisfies
$$u_{t}-a(x,t)u_{xx}-b(x,t)u_{x}-c(x,t)u=F(x,t)$$
in $Q_{T}$. The $|u(x,t)| \le M+TN $ in $Q_{T}$.

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\noindent
{\bf 9}.\, Let $a(x,t) >0$ and $c(x,t) \le 0$ for all $0 <x < L$ and $t> )$.
Show that the initial boundary value problem
\begin{eqnarray}
u_{t}- a(x,t)u_{xx}-b(x,t)u_{x}-c(x,t)u=F(x,t),~~~ 0< x , L,~~ t > 0,\\
u(0,t)=g(t),~~ u(L,t)=h(t),~~ t \le 0,\\
u(x,0)=f(x),~~~ 0 \le x \le L
\end{eqnarray}
has at most one solution which is continuous on $ 0 \le x \le L$ and $t \ge 0$
[Notice that if there is , in fact a continuous solution, then the boundary and initial data must be compatible. That is ,$ g(0)=f(0), ~~h(0)=f(L)$].



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