Chapter 4, Problem 4.1P

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

To determine

Whether a negative or zero value of ${\lambda }^2$ will result in solutions that cannot satisfy the prescribed boundary conditions.

Answer to Problem 4.1P

The boundary condition 4 results in a trivial solution so a negative or zero value of ${\lambda }^2$ will result in solutions that cannot satisfy the prescribed boundary conditions.

Explanation of Solution

Formula Used:

The temperature distribution equation is given by,

  $\theta \left(x,y\right)=X\left(x\right)\cdot Y\left(x\right)$  .......(I)

The two dimensional steady state conduction equations is given by,

  $\frac{{\partial }^2\theta }{\partial x^2}+\frac{{\partial }^2\theta }{\partial y^2}=0$  .......(II)

Calculation:

Substitute Equation (I) in Equation (II).

  $\begin{array}{c}\frac{{\partial }^2\theta }{\partial x^2}+\frac{{\partial }^2\theta }{\partial y^2}=0\\ \frac{{\partial }^2\left(X\cdot Y\right)}{\partial x^2}+\frac{{\partial }^2\left(X\cdot Y\right)}{\partial y^2}=0\\ -Y\frac{{\partial }^{2}X}{\partial x^2}=X\frac{{\partial }^{2}Y}{\partial y^2}\end{array}$  .......(III)

Divide Equation (III) by $XY$ .

  $\begin{array}{c}-\frac{Y}{XY}\frac{{\partial }^{2}X}{\partial x^2}=\frac{X}{XY}\frac{{\partial }^{2}Y}{\partial y^2}\\ -\frac{1}{X}\frac{{\partial }^{2}X}{\partial x^2}=\frac{1}{Y}\frac{{\partial }^{2}Y}{\partial y^2}\end{array}$  .......(IV)

Use separation constant ${\lambda }^2$ in Equation (IV).

  $\begin{array}{l}-\frac{1}{X}\frac{{\partial }^{2}X}{\partial x^2}={\lambda }^2\\ \frac{{\partial }^{2}X}{\partial x^2}+X{\lambda }^2=0\end{array}$  .......(V)

  $\begin{array}{l}\frac{1}{Y}\frac{{\partial }^{2}Y}{\partial y^2}={\lambda }^2\\ \frac{{\partial }^{2}Y}{\partial y^2}-Y{\lambda }^2=0\end{array}$  .......(VI)

Case 1:

Assume ${\lambda }^2=-k^2$ .

Substitute ${\lambda }^2=-k^2$ in Equation (V).

  $\frac{{\partial }^{2}X}{\partial x^2}-X{k}^2=0$

From the above equation, the auxiliary equation is written as,

  $\begin{array}{c}{m}^2-k^2=0\\ m=\pm k\end{array}$

Here, the roots for the auxiliary equation are unequal and real.

The general solution for the auxiliary equation is given by,

  $X\left(x\right)=C{1}_{}{e}^{-kx}+C_2{e}^{kx}$  .......(VII)

Substitute ${\lambda }^2=-k^2$ in Equation (VI).

  $\frac{{\partial }^{2}X}{\partial x^2}+Y{k}^2=0$

From the above equation, the auxiliary equation is written as,

  $\begin{array}{c}{m}^2+k^2=0\\ m=\pm ik\end{array}$

The general solution for the auxiliary equation is given by,

  $Y\left(y\right)C_3\cos ky+C_4\sin ky$  .......(VIII)

Substitute $\left(C_1{e}^{-kx}+C_2{e}^{kx}\right)$ for $X\left(x\right)$ and $\left(C_3\cos ky+C_4\sin ky\right)$ for $Y\left(y\right)$ in Equation (I).

  $\theta \left(x,y\right)=\left(C_1{e}^{-kx}+C_2{e}^{kx}\right)\cdot \left(C_3\cos ky+C_4\sin ky\right)$  .......(IX)

Apply boundary condition 1: at $y=0$ , $\theta \left(x,0\right)=0$ in Equation (IX).

  $\begin{array}{c}\theta \left(x,0\right)=\left(C_1{e}^{-kx}+C_2{e}^{kx}\right)\cdot \left[C_3\cos \left(k\times 0\right)+C_4\sin \left(k\times 0\right)\right]\\ 0=\left(C_1{e}^{-kx}+C_2{e}^{kx}\right)\cdot \left(C_3\right)\\ C_3=0\end{array}$

Apply boundary condition 2: at $x=0$ , $\theta \left(0,y\right)=0$ in Equation (IX).

  $\begin{array}{c}\theta \left(x,y\right)=\left(C_1{e}^{-kx}+C_2{e}^{kx}\right)\cdot \left(C_3\cos ky+C_4\sin ky\right)\\ \theta \left(0,y\right)=\left(C_1{e}^{-k\times 0}+C_2{e}^{k\times 0}\right)\cdot \left(C_3\cos ky+C_4\sin ky\right)\\ 0=\left(C_1+C_2\right)\cdot \left(0\times \cos ky+C_4\sin ky\right)\\ C_4=0\end{array}$

A trivial solutions results as the constant $C_4=0$ .

Case 2:

Assume ${\lambda }^2=0$ .

Substitute $0$ for ${\lambda }^2$ in Equation (V).

  $\begin{array}{c}\frac{{\partial }^{2}Y}{\partial y^2}-Y\left(0\right)=0\\ \frac{{\partial }^{2}Y}{\partial y^2}=0\end{array}$

Integrate the above obtained equation.

  $\frac{dY}{dy}=C_7$

Integrate the above equation.

  $Y\left(y\right)=C_{7}y+C_8$  .......(X)

Substitute $\left(C_{5}x+C_6\right)$ for $X\left(x\right)$ and $\left(C_{7}y+C_8\right)$ for $Y\left(y\right)$ in Equation (I).

  $\theta \left(x,y\right)=\left(C_{5}x{C}_6\right)\cdot \left(C_{7}y+C_8\right)$  .......(XI)

Apply boundary condition 2: at $x=0$ , $\theta \left(0,y\right)=0$ in Equation (XI).

  $\begin{array}{c}\theta \left(x,y\right)=\left(C_{5}x+C_6\right)\cdot \left(C_{7}y+C_8\right)\\ \theta \left(0,y\right)=\left[C_5\times \left(0\right)+C_6\right]\cdot \left(C_{7}y+C_8\right)\\ 0=\left(C_6\right)\cdot \left(C_{7}y+C_8\right)\\ C_6=0\end{array}$

Apply boundary condition 1: at $y=0$ , $\theta \left(x,0\right)=0$ in Equation (XI).

  $\begin{array}{c}\theta \left(x,y\right)=\left(C_{5}x+C_6\right)\cdot \left(C_{7}y+C_8\right)\\ \theta \left(x,0\right)=\left(C_{5}x+0\right)\cdot \left[C_7\left(0\right)+C_8\right]\\ 0=\left(C_{5}x\right)\cdot \left(C_8\right)\\ C_8=0\end{array}$

Apply boundary condition 3: at $x=L$ , $\theta \left(L,0\right)=0$ in Equation (XI).

  $\begin{array}{c}\theta \left(x,y\right)=\left(C_{5}x+C_6\right)\cdot \left(C_{7}y+C_8\right)\\ \theta \left(L,y\right)=\left[\left(C_5\times L\right)+0\right]\cdot \left(C_{7}y+0\right)\\ 0=\left(C_{5}L\right)\cdot \left(C_{7}y\right)\\ C_5=0\end{array}$

Apply boundary condition 4: at $y=W$ , $\theta \left(x,W\right)=1$ in Equation (XI).

  $\begin{array}{c}\theta \left(x,y\right)=\left(C_{5}x+C_6\right)\cdot \left(C_{7}y+C_8\right)\\ \theta \left(x,W\right)=\left[\left(0\times x\right)+0\right]\cdot \left[\left(C_7\times W\right)+0\right]\\ 1=C_7\times W\\ 1\ne 0\end{array}$

Conclusion:

The boundary condition 4 results in a trivial solution so a negative or zero value of ${\lambda }^2$ will result in solutions that cannot satisfy the prescribed boundary conditions.