Examples from Section 18.2

Math 421:03


Date: 6 November 2002

Saddledome

Example 1   Solve the boundary-value problem

$\displaystyle \nabla^2v$ $\displaystyle = 0;$ $\displaystyle (0 < x < L, 0 < y < k)$    
$\displaystyle v(x,0)$ $\displaystyle = v(x,K) = x(L-x);$ $\displaystyle (0 \leq y \leq K)$    
$\displaystyle v(0,y)$ $\displaystyle = v(L,y) = 0.$ $\displaystyle (0 \leq x \leq L);$    

Proof. [Solution]We will find the solution as $ v = v_1 + v_3$, where each $ v_i$ is harmonic (just a fancy word for $ \nabla^2v_i = 0$). We will force $ v_1$ to be zero on the entire boundary except the ``North'' edge, where $ u(x,K) = x(L-x) = f(x)$ for all $ 0 \leq x \leq L$. Likewise, $ v_3$ will be zero all around the boundary except the ``South'' edge, where $ u(x,0) = f(x)$ as well.

The solution for $ v_1$ is then

$\displaystyle v_1(x,y) = \sum_{n=1}^\infty
a^1_n \sin\left(\frac{\pi n x}{L}\right)
\sinh\left(\frac{\pi n y}{L}\right),
$

where

\begin{displaymath}
\begin{split}
a^1_n &= \frac{2}{L \sinh(\pi n K/L)}
\int_0...
...\frac{4L^2(1 - (-1)^n)}{\pi^3 n^3 \sinh(\pi n K/L)}
\end{split}\end{displaymath}

As for $ v_3$,

$\displaystyle v_3(x,y) = \sum_{n=1}^\infty
a^3_n \sin\left(\frac{\pi n x}{L}\right)
\sinh\left(\frac{\pi n}{L}(K-y)\right),
$

where

\begin{displaymath}
\begin{split}
a^3_n &= \frac{2}{L \sinh(\pi n K/L)}
\int_0...
...in\left(\frac{\pi n x}{L}\right) dx \\
&= a^1_n.
\end{split}\end{displaymath}

So we have

$\displaystyle v = \frac{4L^2}{\pi^3}\sum_{n=1}^\infty
\frac{1 - (-1)^n}{n^3 \si...
...left(\frac{\pi n y}{L}\right)
+ \sinh\left(\frac{\pi n}{L}(K-y)\right)\right].
$

See Figure 1 for the graph of $ v$. $ \qedsymbol$

Figure 1: $ v$ from Example 1 ($ L=K=\pi $)
\begin{figure}\begin{center}
\epsfig{file=v.eps,angle=-90,width=\textwidth}
\end{center}
.
\end{figure}

Application to the Heat Equation

Let's see how we can use this to solve the heat equation with inhomogenous boundary conditions. We will solve the boundary value problem

$\displaystyle u_t$ $\displaystyle = k\nabla^2u;$ $\displaystyle (0 < x < L, 0 < y < K, t > 0)$    
$\displaystyle u(x,0,t) = 0;$ $\displaystyle \quad u(x,K,t) = f(x);$ $\displaystyle (0 \leq x \leq L, t > 0)$    
$\displaystyle u(0,y,t)$ $\displaystyle = u(L,y,t) = 0;$ $\displaystyle (0 \leq y \leq K, t > 0)$    
$\displaystyle u(x,y,0)$ $\displaystyle = 0.$ $\displaystyle (0 < x < L, 0 < y < K)$    

So our rectangular region has zeros all around it except at the top. We know that we can solve the PDE part by writing $ u(x,y,t) = U(x,y,t) +
v(x,y)$, where $ U_t = k\nabla^2U$, $ \nabla^2v = 0$. The boundary conditions can be satisfied if

$\displaystyle U(x,0) = U(0,y) = U(x,K) = U(L,y) = 0,$ (1)

and

$\displaystyle v(x,0) =0;$ $\displaystyle \qquad v(x,K) = x(L-x);$ $\displaystyle (0 \leq y \leq K)$    
$\displaystyle v(0,y)$ $\displaystyle = v(L,y) = 0.$ $\displaystyle (0 \leq x \leq L);$    

That is, all the nonzero boundary stuff is given to $ v$. We can satisfy the initial conditions if we require

$\displaystyle U(x,y,0) = - v(x,y).$ (2)

Now $ v$ is easy; it's just $ v^1$ from Example 1.

$\displaystyle v(x,y) = \sum_{n=1}^\infty \frac{b_n}{\sinh(\pi n K/L)} \sin\left(\frac{\pi n x}{L}\right) \sinh\left(\frac{\pi n y}{L}\right),$ (3)

where the $ b_n$ are just the Fourier sine series of $ f$:

$\displaystyle b_n = \frac{2}{L} \int_0^L f(\xi) \sin\left(\frac{\pi p
\xi}{L}\right) d\xi.
$

As for $ U$, we know that the general solution has to be

$\displaystyle U(x,y,t) =
\sum_{p,q=1}^\infty
c_{p,q}
\sin\left(\frac{\pi p x}...
...ht)
\exp\left(-\left(\frac{n^2}{L^2}+\frac{m^2}{K^2}\right)
\pi^2 k t\right),
$

where to fit the inital conditions (2),

$\displaystyle c_{p,q} = - \frac{4}{LK}\int_0^L \int_0^K
v(x,y)
\sin\left(\frac{\pi p x}{L}\right)
\sin\left(\frac{\pi q y}{K}\right)
 dy dx.
$

Because of the series expression for $ v$, we can simplify:

\begin{displaymath}\begin{split}c_{p,q} &= - \frac{4}{LK}\int_0^L \int_0^K \sum_...
...{L}\right) \sin\left(\frac{\pi q y}{K}\right)  dy. \end{split}\end{displaymath}    

The $ x$ integral should be familar; it is nonzero if and only if $ n =
p$, in which case it is $ L/2$. So

$\displaystyle c_{p,q} = -\left(\frac{b_p}{\sinh(\pi n K/L)}\right)
\left(\frac{...
...K
\sinh\left(\frac{\pi p y}{L}\right)
\sin\left(\frac{\pi q y}{K}\right)
 dy.
$

So we are interested in the Fourier sine coefficients on $ [0,K]$ of the function $ y \mapsto \sinh\left(\frac{\pi p y}{L}\right)$. They are

\begin{displaymath}\begin{split}d_{p,q} &= \frac{2}{K}\int_0^K \sinh\left(\frac{...
...ght) (-1)^q - \left(\frac{Lq}{Kp}\right)^2 d_{p,q}. \end{split}\end{displaymath}    

Thus

$\displaystyle d_{p,q} = \frac{\left(\frac{2}{K}\right)
\left(\frac{L^2q}{Kp^2\p...
...right)^2}
= \frac{2L^2 q(-1)^{q+1} \sinh(p \pi K/L)}
{\pi(K^2p^2 + L^2 q^2)}.
$

Putting this back into the general solution, we have

\begin{displaymath}\begin{split}U(x,y,t) &= - \sum_{p,q=1}^\infty \frac{b_p}{\si...
...2}{L^2}+\frac{m^2}{K^2}\right) \pi^2 k t\right).  \end{split}\end{displaymath}    

Putting it all together,

\begin{displaymath}\begin{split}u(x,y,t) &= \sum_{p,q=1}^\infty \frac{2L^2 q(-1)...
...n x}{L}\right) \sinh\left(\frac{\pi n y}{L}\right). \end{split}\end{displaymath}    

Example 2   Solve the heat equation with inhomogenous boundary conditions as those given in Example 1.

Proof. [Solution]We will have to $ U = U_1 + U_3$, and compute each $ U_i$ separately as in the previous development. If we let

$\displaystyle b_n = \frac{2}{L}\int_0^L x (L-x) \sin\left(\frac{\pi n x}{L}\right) dx
= \frac{4L^2(1-(-1)^p)}{\pi^3 n^3}
$

we have

\begin{displaymath}\begin{split}U_1(x,y,t) &= \sum_{p,q=1}^\infty \frac{2L^2 q(-...
...c{n^2}{L^2}+\frac{m^2}{K^2}\right) \pi^2 k t\right) \end{split}\end{displaymath}    

As for $ U_3$, the general solution is now of the form

$\displaystyle U_3(x,y,t) = \sum_{p,q =1}^\infty
c^3_{p,q} \sin\left(\frac{\pi p...
...ght)
\exp\left(-\left(\frac{n^2}{L^2}+\frac{m^2}{K^2}\right)
\pi^2 k t\right),
$

where this time

\begin{displaymath}\begin{split}c^3_{p,q} &= -\left(\frac{b_p}{\sinh(\pi n K/L)}...
...t)  dy  &=\frac{2L^2qb_n}{\pi(K^2p^2 + L^2 q^2)} \end{split}\end{displaymath}    

So putting it all together we have

\begin{displaymath}\begin{split}u(x,y,t) &= U_1(x,y,t) + U_2(x,y,t) + v(x,y,t) \...
...t) + \sinh\left(\frac{\pi n}{L}(K-y)\right)\right]. \end{split}\end{displaymath}    

See Figure 2 for some graphs. $ \qedsymbol$

Figure 2: Snapshots of the graph of $ u(x,y,t)$ from Example 2. We have set $ L=K=\pi $ again. You can see that the homogenous part $ U(x,y,t)$ approaches zero and the steady state solution is $ v$.
\begin{figure}\begin{center}
\epsfig{file=u-t0.00.eps,angle=-90,width=2in} \hfil...
...sfig{file=u-t2.50.eps,angle=-90,width=2in} \ \medskip
\end{center}
\end{figure}


Matthew Leingang 2002-11-08