Question 6.5: (a) Determine the potential function for the region inside t...

(a) The potential V in this case depends on x and y. Laplace’s equation becomes

\nabla^2 V=\frac{\partial^{2} V}{\partial x^{2}}+\frac{\partial^{2} V}{\partial y^{2}}=0                                                                                               (6.5.1)

We have to solve this equation subject to the following boundary conditions:

V(x=0,0\leq y\lt a)=0                                                                                               (6.5.2a)

V(x=b,0\leq y\lt a)=0                                                                                               (6.5.2b)

V(0\leq x\leq b,y=0)=0                                                                                               (6.5.2c)

V(0\lt x\lt b,y=a)=V_{o}                                                                                               (6.5.2d)

We solve eq. (6.5.1) by the method of separation of variables; that is, we seek a product solution of V. Let

V(x,y)=X(x)Y(y)                                                                                               (6.5.3)

where X is a function of x only and Y is a function of y only. Substituting eq. (6.5.3) into eq. (6.5.1) yields

X^{"}Y+Y^{"}X=0

Dividing through by XY and separating X from Y gives

-\frac{X^{"}}{X}=\frac{Y^{"}}{Y}                                                                                               (6.5.4a)

Since the left-hand side of this equation is a function of x only and the right-hand side is a function of y only, for the equality to hold, both sides must be equal to a constant \lambda; that is

-\frac{X^{"}}{X}=\frac{Y^{"}}{Y}=\lambda                                                                                               (6.5.4b)

The constant \lambda is known as the separation constant. From eq. (6.5.4b), we obtain

X^{"}+\lambda X=0                                                                                               (6.5.5a)

and

Y^{"}-\lambda Y=0                                                                                               (6.5.5b)

Thus the variables have been separated at this point and we refer to eq. (6.5.5) as separated equations. We can solve for X(x) and Y(y) separately and then substitute our solutions into eq. (6.5.3). To do this requires that the boundary conditions in eq. (6.5.2) be separated, if possible. We separate them as follows:

V(0,y)=X(0)Y(y)=0\rightarrow X(0)=0                                                                                  (6.5.6a)

V(b,y)=X(b)Y(y)=0\rightarrow X(b)=0                                                                                  (6.5.6b)

V(x,0)=X(x)Y(0)=0\rightarrow Y(0)=0                                                                                  (6.5.6c)

V(x,a)=X(x)Y(a)=V_{o}  (inseparable)                                   (6.5.6d)

To solve for X(x) and Y(y) in eq. (6.5.5), we impose the boundary conditions in eq. (6.5.6). We consider possible values of \lambda that will satisfy both the separated equations in eq. (6.5.5) and the conditions in eq. (6.5.6).

CASE 1.
If \lambda=0, then eq. (6.5.5a) becomes

X^{"}=0   or  \frac{d^{2}X}{dx^{2}}=0

which, upon integrating twice, yields

X=Ax+B                                                                                               (6.5.7)

The boundary conditions in eqs. (6.5.6a) and (6.5.6b) imply that

X(x=0)=0\rightarrow 0=0+B   or  B=0

and

X(x=b)=0\rightarrow 0=A\cdot b+0   or  A=0

 because b\neq 0. Hence our solution for X in eq. (6.5.7) becomes

X(x)=0

which makes V=0 in eq. (6.5.3). Thus we regard X(x)=0 as a trivial solution and we conclude that \lambda\neq 0.

CASE 2.
If \lambda\lt 0, say \lambda=-\alpha^{2}, then eq. (6.5.5a) becomes

X^{"}-\alpha^{2}X=0   or  (D^{2}-\alpha^{2})X=0

where D=\frac{d}{dx}, that is

DX=\pm\alpha X                                                                                               (6.5.8)

showing that we have two possible solutions corresponding to the plus and minus signs. For the plus sign, eq. (6.5.8) becomes

\frac{dX}{dx}\alpha X   or   \frac{dX}{X}=\alpha dx

Hence

\int\frac{dX}{X}=\int\alpha dx   or  \ln X=\alpha x+\ln A_{1}

where \ln A_{1} is a constant of integration. Thus

X=A_{1}e^{\alpha x}                                                                                               (6.5.9a)

Similarly, for the minus sign, solving eq. (6.5.8) gives

X=A_{2}e^{-\alpha x}                                                                                               (6.5.9b)

The total solution consists of what we have in eqs (6.5.9a) and (6.5.9b); that is

X(x)=A_{1}e^{\alpha x}+A_{2}e^{-\alpha x}                                                                                               (6.5.10)

Since \cosh\alpha x=(e^{\alpha x}+e^{-\alpha x})/2   and   \sinh\alpha x=(e^{\alpha x}-e^{-\alpha x})/2

or

e^{\alpha x}=\cosh\alpha x+\sinh\alpha x   and   e^{-\alpha x}=\cosh\alpha x-\sinh\alpha x , eq. (6.5.10) can be written as

X(x)=B_{1}\cosh\alpha x+B_{2}\sinh\alpha x                                                                                               (6.5.11)

where B_{1}=A_{1}+A_{2} and B_{2}=A_{1}-A_{2}. In view of the given boundary conditions, we prefer eq. (6.5.11) to eq. (6.5.10) as the solution. Again, eqs. (6.5.6a) and (6.5.6b) require that

X(x=0)=0\rightarrow 0=B_{1}\cdot(1)+B_{2}\cdot(0)   or  B_{1}=0

and

X(x=b)=0\rightarrow 0=0+B_{2}\sinh \alpha b

Since \alpha\neq 0 and b\neq 0,  \sinh \alpha b,  cannot be zero. This is due to the fact that  \sinh x=0 if and only if x=0 as shown in Figure 6.8. Hence B_{2}=0 and

X(x)=0

This is also a trivial solution and we conclude that \lambda cannot be less than zero.

CASE 3.
If \lambda\gt 0, say \lambda=\beta^{2}, then eq. (6.5.5a) becomes

X^{"}+\beta^{2}X=0

that is

(D^{2}+\beta^{2})X=0  or   DX=\pm j\beta X                         (6.5.12)

where j=\sqrt{-1}. From eqs. (6.5.8) and (6.5.12), we notice that the difference between Cases 2 and 3 is the replacement of \alpha by j\beta. By taking the same procedure as in Case 2, we obtain the solution as

X(x)=C_{o}e^{j\beta x}+C_{1}e^{-j\beta x}                                                                                               (6.5.13a)

Since e^{j\beta x}=\cos\beta x+j\sin\beta x and e^{-j\beta x}=\cos\beta x-j\sin\beta x, eq. (6.5.13a) can be written as

X(x)=g_{o}\cos\beta x+g_{1}\sin\beta x                                                                                               (6.5.13b)

where g_{o}=C_{o}+C_{1} and g_{1}=j(C_{o}-C_{1}).

In view of the given boundary conditions, we prefer to use eq. (6.5.13b). Imposing the conditions in eqs (6.5.6a) and (6.5.6b) yields

X(x=0)=0\rightarrow 0=g_{o}\cdot(1)+0   or  g_{o}=0

and

X(x=b)=0\rightarrow 0=0+g_{1}\sin\beta b

Suppose g_{1}\neq 0 (otherwise we get a trivial solution), then

\sin\beta b=0=\sin n\pi\rightarrow \beta b=n\pi

\beta=\frac{n\pi}{b},       n=1,2,3,4,…                                                                                               (6.5.14)

Note that, unlike \sinh x, which is zero only when x=0, \sin x is zero at an infinite number of points as shown in Figure 6.9. It should also be noted that n\neq 0 because \beta\neq 0; we have already considered the possibility \beta=0 in Case 1, where we ended up with a trivial solution. Also we do not need to consider n=-1,-2,-3,-4,… because \lambda=\beta^{2} would remain the same for positive and negative integer values of n. Thus for a given n, eq. (6.5.13b) becomes

X_{n}(x)=g_{n}\sin\frac{n\pi x}{b}                                                                                               (6.5.15)

Having found X(x) and

\lambda=\beta^{2}=\frac{n^{2}\pi^{2}}{b^{2}}                                                                                               (6.5.16)

we solve eq. (6.5.5b), which is now

Y^{"}-\beta^{2}Y=0

The solution to this is similar to eq. (6.5.11) obtained in Case 2; that is,

Y(y)=h_{o}\cosh\beta y+h_{1}\sinh\beta y

The boundary condition in eq. (6.5.6c) implies that

Y(y=0)=0\rightarrow 0=h_{o}\cdot(1)+0   or  h_{o}=0

Hence our solution for Y(y) becomes

Y_{n}(y)=h_{n}\sinh\frac{n\pi y}{b}                                                                                               (6.5.17)

Substituting eqs. (6.5.15) and (6.5.17), which are the solutions to the separated equations in eq. (6.5.5), into the product solution in eq. (6.5.3) gives

V_{n}(x,y)=g_{n}h_{n}\sin\frac{n\pi x}{b}\sinh\frac{n\pi y}{b}

This shows that there are many possible solutions V_{1},  V_{2},  V_{3},  V_{4} and so on, for n=1,  2,  3,  4, and so on.

By the superposition theorem, if V_{1},  V_{2},  V_{3},…,  V_{n} are solutions of Laplace’s equation, the linear combination

V=c_{1}V_{1}+c_{2}V_{2}+c_{3}V_{3}+…+c_{n}V_{n}

(where c_{1},  c_{2},  c_{3},…,  c_{n} are constants) is also a solution of Laplace’s equation. Thus the solution to eq. (6.5.1) is

V(x,y)=\sum\limits_{n=1}^{\infty}c_{n}\sin\frac{n\pi x}{b}\sinh\frac{n\pi y}{b}                                                                                     (6.5.18)

where c_{n}=g_{n}h_{n} are the coefficients to be determined from the boundary condition in eq. (6.5.6d). Imposing this condition gives

V(x,y=a)=V_{o}=\sum\limits_{n=1}^{\infty}c_{n}\sin\frac{n\pi x}{b}\sinh\frac{n\pi a}{b}                                                                                     (6.5.19)

which is a Fourier series expansion of V_{o}. Multiplying both sides of eq. (6.5.19) by \sin m \pi x/b and integrating over 0\lt x\lt b gives

\int_{0}^{b}V_{o}\sin\frac{m\pi x}{b}dx=\sum\limits_{n=1}^{\infty}c_{n}\sinh\frac{n\pi a}{b}\int_{0}^{b}\sin\frac{m\pi x}{b}\sin\frac{n\pi x}{b}dx                                       (6.5.20)

By the orthogonality property of the sine or cosine function (see Appendix A.9).

\int_{0}^{\pi}\sin mx\sin nx dx= \begin{cases} 0, & m\neq n \quad \\ \pi/2, & m=n \end{cases}

Incorporating this property in eq. (6.5.20) means that all terms on the right-hand side of eq. (6.5.20) will vanish except one term in which m=n. Thus eq (6.5.20) reduces to

\int_{0}^{b}V_{o}\sin\frac{n\pi x}{b}dx=c_{n}\sinh\frac{n\pi a}{b}\int_{0}^{b}\sin^{2}\frac{n\pi x}{b}dx-V_{o}\frac{b}{n\pi}\cos\frac{n\pi x}{b}|_{0}^{b}

=c_{n}\sinh\frac{n\pi a}{b}\frac{1}{2}\int_{0}^{b}\left(1-\cos\frac{2n\pi x}{b}\right)dx

\frac{V_{o}b}{n\pi}(1-\cos n\pi)=c_{n}\sinh\frac{n\pi a}{b}\cdot\frac{b}{2}

or

c_{n}\sinh\frac{n\pi a}{b}=\frac{2V_{o}}{n\pi}(1-\cos n\pi)= \begin{cases} \frac{4V_{o}}{n\pi}, & n=1,3,5,… \\ 0, & n=2,4,6,… \end{cases}

that is

c_{n}= \begin{cases} \frac{4V_{o}}{n\pi\sinh\frac{n\pi a}{b}}, & n=odd \\ 0, & n=even \end{cases}                                                                                     (6.5.21)

Substituting this into eq. (6.5.18) gives the complete solution as

V(x,y)=\frac{4V_{o}}{\pi}\sum\limits_{n=1,3,5,…}^{\infty} \frac{\sin\frac{n\pi x}{b}\sinh\frac{n\pi y}{b}}{n\sinh\frac{n\pi a}{b}}                                                                                     (6.5.22)

Check:

\nabla^2 V=0,    V(x=0,y)=0=V(x=b,y)=V(x,y=0),   V(x,y=a)=V_{o}

The solution in eq. (6.5.22) should not be a surprise; it can be guessed by mere observation of the potential system in Figure 6.7. From this figure, we notice that along x, V varies from 0(at    x=0) to 0(at   x=b) and only a sine function can satisfy this requirement. Similarly, along y, V varies from 0(at   y=0) to V_{o}(at   y=a) and only a hyperbolic sine function can satisfy this. Thus we should expect the solution as in eq. (6.5.22).

To determine the potential for each point (x,y) in the trough, we take the first few terms of the convergent infinite series in eq. (6.5.22). Taking four or five terms may be sufficient.

(b) For x=a/2 and y=3a/4, where b=2a, we have

V\left(\frac{a}{2},\frac{3a}{4}\right)=\frac{4V_{o}}{\pi}\sum\limits_{n=1,3,5,..}^{\infty}\frac{\sin n\pi/4 \sinh 3n\pi/8}{n \sinh n\pi/2}

=\frac{4V_{o}}{\pi}\left[\frac{\sin\pi/4 \sinh 3\pi/8}{\sinh\pi/2}+\frac{\sin 3\pi/4 \sinh 9\pi/8}{3\sinh 3\pi/2}+\frac{\sin 5\pi/4\sinh 15\pi/8}{5\sinh 5\pi/2}+…\right]

=\frac{4V_{o}}{\pi}(0.4517+0.0725-0.01985-0.00645+0.00229+…)=0.6374V_{o}

It is instructive to consider a special case of a=b=1m and V_{o}=100V. The potentials at some specific points are calculated by using eq. (6.5.22), and the result is displayed in Figure 6.10(a). The corresponding flux lines and equipotential lines are shown in Figure 6.10(b). A simple MATLAB program based on eq (6.5.22) is displayed in Figure 6.11. This self explanatory program can be used to calculate V(x,y) at any point within the trough. In Figure 6.11, V(x=b/4,  y=3a/4) is typically calculated and found to be 43.2V.