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

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Answer 1

Textbook Question

Chapter 4, Problem 4.1P

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

Expert Solution & Answer

To determine

Whether a negative or zero value of $λ2$ 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 $λ2$ will result in solutions that cannot satisfy the prescribed boundary conditions.

Explanation of Solution

Formula Used:

The temperature distribution equation is given by,

$θ(x,y)=X(x)⋅Y(x)$ .......(I)

The two dimensional steady state conduction equations is given by,

$∂2θ∂x2+∂2θ∂y2=0$ .......(II)

Calculation:

Substitute Equation (I) in Equation (II).

$∂2θ∂x2+∂2θ∂y2=0∂2( X⋅Y)∂x2+∂2( X⋅Y)∂y2=0−Y∂2X∂x2=X∂2Y∂y2$ .......(III)

Divide Equation (III) by $XY$ .

$−YXY∂2X∂x2=XXY∂2Y∂y2−1X∂2X∂x2=1Y∂2Y∂y2$ .......(IV)

Use separation constant $λ2$ in Equation (IV).

$−1X∂2X∂x2=λ2∂2X∂x2+Xλ2=0$ .......(V)

$1Y∂2Y∂y2=λ2∂2Y∂y2−Yλ2=0$ .......(VI)

Case 1:

Assume $λ2=−k2$ .

Substitute $λ2=−k2$ in Equation (V).

$∂2X∂x2−Xk2=0$

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

$m2−k2=0m=±k$

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

The general solution for the auxiliary equation is given by,

$X(x)=C1e−kx+C2ekx$ .......(VII)

Substitute $λ2=−k2$ in Equation (VI).

$∂2X∂x2+Yk2=0$

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

$m2+k2=0m=±ik$

The general solution for the auxiliary equation is given by,

$Y(y)C3cosky+C4sinky$ .......(VIII)

Substitute $(C1e−kx+C2ekx)$ for $X(x)$ and $(C3cosky+C4sinky)$ for $Y(y)$ in Equation (I).

$θ(x,y)=(C1e−kx+C2ekx)⋅(C3cosky+C4sinky)$ .......(IX)

Apply boundary condition 1: at $y=0$ , $θ(x,0)=0$ in Equation (IX).

$θ(x,0)=(C1e −kx+C2e kx)⋅[C3cos(k×0)+C4sin(k×0)]0=(C1e −kx+C2e kx)⋅(C3)C3=0$

Apply boundary condition 2: at $x=0$ , $θ(0,y)=0$ in Equation (IX).

$θ(x,y)=(C1e −kx+C2e kx)⋅(C3cosky+C4sinky)θ(0,y)=(C1e −k×0+C2e k×0)⋅(C3cosky+C4sinky)0=(C1+C2)⋅(0×cosky+C4sinky)C4=0$

A trivial solutions results as the constant $C4=0$ .

Case 2:

Assume $λ2=0$ .

Substitute $0$ for $λ2$ in Equation (V).

$∂2Y∂y2−Y(0)=0∂2Y∂y2=0$

Integrate the above obtained equation.

$dYdy=C7$

Integrate the above equation.

$Y(y)=C7y+C8$ .......(X)

Substitute $(C5x+C6)$ for $X(x)$ and $(C7y+C8)$ for $Y(y)$ in Equation (I).

$θ(x,y)=(C5xC6)⋅(C7y+C8)$ .......(XI)

Apply boundary condition 2: at $x=0$ , $θ(0,y)=0$ in Equation (XI).

$θ(x,y)=(C5x+C6)⋅(C7y+C8)θ(0,y)=[C5×(0)+C6]⋅(C7y+C8)0=(C6)⋅(C7y+C8)C6=0$

Apply boundary condition 1: at $y=0$ , $θ(x,0)=0$ in Equation (XI).

$θ(x,y)=(C5x+C6)⋅(C7y+C8)θ(x,0)=(C5x+0)⋅[C7(0)+C8]0=(C5x)⋅(C8)C8=0$

Apply boundary condition 3: at $x=L$ , $θ(L,0)=0$ in Equation (XI).

$θ(x,y)=(C5x+C6)⋅(C7y+C8)θ(L,y)=[( C 5×L)+0]⋅(C7y+0)0=(C5L)⋅(C7y)C5=0$

Apply boundary condition 4: at $y=W$ , $θ(x,W)=1$ in Equation (XI).

$θ(x,y)=(C5x+C6)⋅(C7y+C8)θ(x,W)=[(0×x)+0]⋅[( C 7×W)+0]1=C7×W1≠0$

Conclusion:

The boundary condition 4 results in a trivial solution so a negative or zero value of $λ2$ will result in solutions that cannot satisfy the prescribed boundary conditions.

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