Section 7.6 Heat Equation in two spatial dimensions
The Heat equation in two spatial dimensions can be written:
\begin{equation}
\frac{1}{\kappa} \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial {x}^2} + \frac{\partial^2 u}{\partial {y}^2}\tag{7.6.1}
\end{equation}
In this case, let’s say that we have the following boundary conditions:
\begin{align*}
u\bigr\vert_{x=0} \amp = 0 \amp \frac{\partial u}{\partial x}\biggr\vert_{x=L} \amp = 0 \\
\frac{\partial u}{\partial y} \biggr\vert_{y=0} \amp = 0 \amp \frac{\partial u}{\partial y}\biggr\vert_{y = \ell} \amp = 0
\end{align*}
which means that along the edge
\(x=0\text{,}\) the temperature is 0 and the other three edges are insulated. In addition, assume that the initial condition is
\begin{equation*}
u(x,y,0) = f(x,y)
\end{equation*}
In this section, we will examine how to solve this problem using the separation of variables. Since there are 3 variables, let’s assume that the solution
\(u\) can be written:
\begin{equation*}
u = X(x) Y(y) T(t)
\end{equation*}
and substituting this into the heat equation, we get:
\begin{align*}
\frac{1}{\kappa} \frac{\partial XYT}{\partial t} \amp = \frac{\partial^2 XYT}{\partial {x}^2} + \frac{\partial^2 XYT}{\partial {y}^2} \\
\frac{1}{\kappa} XYT' \amp = X'' YT + XY''T
\end{align*}
dividing through by
\(XYT\)
\begin{align*}
\frac{1}{\kappa} \frac{T'}{T} \amp = \frac{X''}{X} + \frac{Y''}{Y}
\end{align*}
Since
\(T'/T\) is only a function of
\(t\text{,}\) \(X''/X\) is only a function of
\(x\) and
\(Y''/Y\) is only a function of
\(y\text{,}\) the only option for allowing the above to be true is to assume that
\begin{align*}
\frac{X''}{X} \amp = -k_1 \amp \frac{Y''}{Y} \amp = -k_2
\end{align*}
The boundary condition can also be written in terms of
\(X\) and
\(Y\) as
\(X(0)=0, X'(L)=0, Y'(0)=0, Y'(\ell)=0\text{.}\) Thus, in this case, we have two Sturm-Liouville problems,
\begin{align*}
X''+k_1 X \amp = 0 \amp X(0)=0, \quad X'(L) = 0 \\
Y''+k_1 Y \amp = 0 \amp Y'(0)=0, \quad Y'(\ell) = 0
\end{align*}
The solution to the first is
\begin{align*}
k_1 \amp = \frac{(2n-1)^2\pi^2}{4L^2} \amp \amp \text{for $n=1,2,3, \ldots$}
\end{align*}
\begin{equation*}
X = \sin \frac{(2n-1)\pi}{2L} x
\end{equation*}
The solution to the second is
\begin{align*}
k_2 \amp = \frac{m^2\pi^2}{\ell^2} \amp \amp \text{for $m=0,1,2,3, \ldots$}
\end{align*}
\begin{equation*}
Y = \cos \frac{m\pi}{\ell} y
\end{equation*}
Next, then we need to solve
\begin{align*}
\frac{1}{\kappa} \frac{T'}{T} \amp = -k_1 -k_2 \qquad \text{or} \\
T' + \kappa(k_1+k_2) T \amp = 0
\end{align*}
\begin{equation*}
T_{n,m} = e^{-\kappa(k_1+k_2)t}
\end{equation*}
Then put the solutions together:
\begin{align*}
u_{n,m} \amp = T_{n,m}X_n Y_m \\
\amp = e^{-\kappa(k_1+k_2)t}\sin \frac{(2n-1) \pi x}{2L}\cos \frac{m \pi x}{\ell}
\end{align*}
and the solution that satisfies the boundary conditions is:
\begin{equation*}
u = \sum_{n=1}^{\infty} \sum_{m=0}^{\infty} c_{n,m} e^{-\kappa(k_1+k_2)t}\sin \frac{(2n-1) \pi x}{2L}\cos \frac{m \pi y}{\ell} \,dy \, dx
\end{equation*}
Finally, we apply the initial condition.
\begin{equation*}
u(x,y,0) = f(x,y) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} c_{n,m} \sin \frac{(2n-1) \pi x}{2L}\cos \frac{m \pi y}{\ell}
\end{equation*}
which results in the generalized Fourier Series:
\begin{equation*}
c_{n,m} = \frac{4}{L \ell} \int_0^L \int_0^{\ell} f(x,y) \sin \frac{(2n-1) \pi x}{2L}\cos \frac{m \pi y}{\ell} \,dy \, dx
\end{equation*}
Example 7.6.1.
Find the full solution if
\(L=\ell=\pi\) and
\begin{equation*}
f(x,y) = x y (\pi-y)
\end{equation*}
Solution.
We only need to find
\(c_{m,n}\)
\begin{align*}
c_{n,m} \amp =\frac{4}{\pi^2} \int_0^{\pi}\int_0^{\pi} \sin (n+1/2) x \cos my \, dx \, dy \\
\amp = 16 (-1)^{n+1} {\frac {1+(-1)^m}{{\pi } (2n+1)^2 {m}^{2}}}
\end{align*}
\begin{equation*}
u = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} 16 (-1)^{n+1} {\frac {1+(-1)^m}{{\pi } (2n+1)^2 {m}^{2}}} e^{-\kappa(k_1+k_2)t}\sin (n+1/2) x\cos my
\end{equation*}
where
\(k_1= (n+1/2)^2\) and
\(k_2=m^2\text{.}\)
