Chapter 4, Problem 4.1P

In the method of separation of variables (Section 4.2) for two-dimensional, steady-state conduction, the separation constant ${\lambda }^2$ in Equations 4.6 and 4.7 must be a positive constant. Show that a negative or zero value of will result in ${\lambda }^2$ solutions that cannot satisfy the prescribed boundary conditions.

To determine

To show: The value of ${\lambda }^2$ does not support the given boundary condition.

Explanation of Solution

From the textbook,

Constant of separation is given as,

  $\begin{array}{l}\frac{d^{2}X}{d{x}^2}+{\lambda }^{2}X=0\mathrm{.....}\left(1\right)\\ \frac{d^{2}Y}{d{y}^2}+{\lambda }^{2}Y=0\mathrm{.....}\left(2\right)\end{array}$

The distribution of temperature can be given as

  $\theta \left(x,y\right)=X\left(x\right)Y\left(y\right)\mathrm{.....}\left(3\right)$

Now,

  $\begin{array}{l}\text{When}{\lambda }^2=0;\\ \text{Using}\text{equation}\left(\text{1}\right)\text{,}\left(\text{2}\right)\text{and}\left(\text{3}\right)\\ X=C_1+C_{2}x\\ Y=C_3+C_{4}y\\ \theta \left(x,y\right)=\left(C_1+C_{2}x\right)\left(C_3+C_{4}y\right)\mathrm{......}\left(4\right)\\ \text{Applying}\text{boundary}\text{condition:}\\ At\\ x=0;\\ \theta \left(0,y\right)=\left(C_1+C_2\cdot 0\right)\left(C_3+C_{4}y\right)=0\\ C_1=0\\ Aty=0;\\ \theta \left(x,0\right)=\left(0+C_{2}X\right)\left(C_3+C_4\cdot 0\right)=0\\ C_3=0\\ Atx=L;\\ \theta \left(L,0\right)=\left(0+C_{2}L\right)\left(0+C_4\cdot y\right)=0\\ C_2=0\\ Aty=W;\\ \theta \left(x,W\right)=\left(0+0\cdot x\right)\left(0+C_4\cdot W\right)=1\\ 0\ne 1\end{array}$

  $\text{When}{\lambda }^2<0;$

The solution of equation (1) and (2) can be expressed as,

  $\begin{array}{l}X=C_5{e}^{-\lambda x}+C_6{e}^{+\lambda x}and\\ Y=C_7\cos \lambda y+C_8\sin \lambda y\\ and\\ \theta \left(x,y\right)=\left[C_5{e}^{-\lambda x}+C_6{e}^{+\lambda x}\right]\left[C_7\cos \lambda y+C_8\sin \lambda y\right]\\ \text{Again}\text{applying}\text{boundary}\text{condition:}\\ Aty=0;\\ \theta \left(x,0\right)=\left[C_5{e}^{-\lambda x}+C_6{e}^{+\lambda x}\right]\left[C_7\cos 0+C_8\sin 0\right]=0\\ \Rightarrow C_7=0\\ Atx=0;\\ \theta \left(0,y\right)=\left[C_5{e}^0+C_6{e}^0\right]\left[0+C_8\sin \lambda y\right]=0\\ \Rightarrow C_8=0\end{array}$

  $\begin{array}{l}Now,\\ If{C}_8=0;\\ \text{A}\text{trivial}\text{solution}\text{results}\text{.}\\ or,\\ C_5=-C_6\\ x=L;\\ \theta \left(L,y\right)=C_5\left[e^{-xL}-e^{+xL}\right]C_8\sin \lambda y=0\end{array}$

Therefore,

We obtain the value of C₈ or C₅ is zero. So, it will generate a trivial solution that does not depend on the value of x and y.