Subsection 7.3.2 Two Dimensional Equation of Heat Flow
Consider the flow of heat in a metal plate of uniform thickness in the directions parallel to the breadth of the plate. The surface of the plate is insulated as shown in figure Figure 7.3.3. Let \(u(x,y,t)\) be the temperature at any point \((x,y)\) of the plate at any time \(t\) is given by
\begin{equation}
\frac{\partial u}{\partial t}= h^{2} \left( \frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}\right) \tag{7.3.48}
\end{equation}
If the edges are kept at zero temperature then the boundary conditions are
\begin{equation*}
u(0,y,t) = u(a,y,t) =u(x,0,t) = u(x,b,t)= 0
\end{equation*}
and the initial condition is given as \(u(x,y,0) = f(x,y)\text{.}\) Let us assume that
\begin{equation}
u(x,y,t) = X(x)Y(y)T(t)\tag{7.3.49}
\end{equation}
\begin{equation*}
YT\frac{\partial^{2}X}{\partial x^{2}}+XT\frac{\partial^{2}Y}{\partial y^{2}} =\frac{XY}{h^{2}}\frac{\partial T}{\partial t}
\end{equation*}
or,
\begin{equation}
h^{2}\left[\frac{1}{X}\frac{\partial^{2}X}{\partial x^{2}}+\frac{1}{Y}\frac{\partial^{2}Y}{\partial y^{2}}\right] = \frac{1}{T}\frac{\partial T}{\partial t} = -m \quad (\text{or,} -\lambda^{2})\tag{7.3.50}
\end{equation}
or,
\begin{equation}
\frac{\partial T}{\partial t} =-mT \text{or,} \frac{\partial T}{\partial t}+mT=0 \Rightarrow T=Ae^{-mt}\tag{7.3.51}
\end{equation}
and
\begin{equation*}
\frac{1}{X}\frac{\partial^{2}X}{\partial x^{2}} +\frac{m}{h^{2}}=-\frac{1}{Y}\frac{\partial^{2}Y}{\partial y^{2}}=k^{2}\quad \text{(say)}
\end{equation*}
or,
\begin{equation}
\frac{\partial^{2}Y}{\partial y^{2}}+k^{2}Y=0 \quad \Rightarrow Y=B\cos ky+C\sin ky\tag{7.3.52}
\end{equation}
also,
\begin{equation*}
\frac{\partial^{2}X}{\partial x^{2}}+\left(\frac{m}{h^{2}}-k^{2}\right)X = 0
\end{equation*}
\begin{equation}
\frac{\partial^{2}X}{\partial x^{2}}+p^{2}X = 0 \quad \Rightarrow X=D\cos px+E\sin px\tag{7.3.53}
\end{equation}
where,
\begin{equation}
p^{2}= \frac{m}{h^{2}}-k^{2}\tag{7.3.54}
\end{equation}
- \(u(x,y,t) = 0\) at \(x=0\) give \(D=0\text{,}\)
- \(u(x,y,t)= 0\) at \(y=0\) gives \(B=0\text{,}\)
- \(u(x,y,t)= 0 \) at \(x=a\) i.e., \(0=Ae^{-mt}[C\sin ky][E\sin pa] \) \(\Rightarrow \sin pa =0=\sin r\pi \text{,}\) \(\therefore p=\frac{r\pi}{a}\text{,}\) \(r=1,2,3,\cdots\text{,}\)
- \(u(x,y,t) = 0\) at \(y=b\) i.e., \(0=Ae^{-mt}[C\sin kb][E\sin px]\) \(\Rightarrow \sin kb =0=\sin s\pi \text{.}\) \(\therefore k=\frac{s\pi}{b}\text{,}\) \(s=1,2,3,\cdots\text{.}\)
Putting these values in equation (7.3.54), we get -
\begin{equation*}
\frac{r^{2}\pi^{2}}{a^{2}} =\frac{m}{h^{2}}- \frac{s^{2}\pi^{2}}{b^{2}}
\end{equation*}
or,
\begin{equation*}
\frac{m}{h^{2}}=\left[\frac{r^{2}}{a^{2}}+ \frac{s^{2}}{b^{2}}\right]\pi^{2}
\end{equation*}
\begin{equation}
\therefore m= h^{2}\pi^{2}\left[\frac{r^{2}}{a^{2}}+ \frac{s^{2}}{b^{2}}\right] = m_{rs} \quad (say)\tag{7.3.55}
\end{equation}
The solution (7.3.49) now becomes -
\begin{equation*}
u=\sum\limits_{r=1}^{\infty}\sum\limits_{s=1}^{\infty}A_{rs}e^{-m_{rs}t}\left(E_{r}\sin\frac{r\pi x}{a}\right)\left(C_{s}\sin\frac{s\pi y}{b}\right)
\end{equation*}
\begin{equation}
=\sum\limits_{r,s=1}^{\infty}a_{rs}e^{-m_{rs}t}\sin\frac{r\pi x}{a}\sin\frac{s\pi y}{b}\tag{7.3.56}
\end{equation}
To determine the value of constant \(a_{rs}\) imposing the initial condition, \(u(x,y,0)=f(x,y)\text{,}\) we get -
\begin{equation}
f(x,y)=\sum\limits_{r,s=1}^{\infty}a_{rs}\sin\frac{r\pi x}{a}\sin\frac{s\pi y}{b}\tag{7.3.57}
\end{equation}
multiplying both sides by \(\sin\frac{r\pi x}{a}\cdot \sin\frac{s\pi y}{b} \) and integrating from 0 to \(a\) and \(b\text{,}\) we get -
\begin{equation}
a_{rs} =\frac{4}{ab}\int\limits_{0}^{a}\int\limits_{0}^{b} f(x,y)\sin\frac{r\pi x}{a}\cdot \sin\frac{s\pi y}{b}\,dx \,dy \tag{7.3.58}
\end{equation}
