Solving the 2D laplace equation. Quick question about the final stage

[Gregg]

Homework Statement

$\frac{\partial^2 V}{\partial x^2} +\frac{\partial^2 V}{\partial y^2} = 0$

it's defined for $V=0$ for

$x=0, 0<y<b$,
$x=a, 0<y<b$,
$y=0, 0<x<a$

$V=V_0 x/a$ for $y=b, 0<x<a$

Homework Equations

None.

The Attempt at a Solution

I've got

$V(x,b) = \displaystyle V_0 x/a = \sum_{n=1}^{\infty} K_n \sin ({n \pi x \over a}) \sinh ({n \pi b \over a})$

At first I thought I could set K to 0 for n>1 and do a simple division, but I don't know what K1 is. Do I always need to find Fourier coefficients. I have seen a problem www.robots.ox.ac.uk/~jmb/lectures/pdelecture5.pdf here where in the final stage they do something like this. I'm guessing I need to do the former, but I am interested to know if the other method is possible.

As for the coefficients, the function is odd so the only integral I need is the sine integral.

$K_n = \frac{1}{a} \int_0^{2a} \frac{V x}{a} \sin ({n \pi x \over a}) dx =-\frac{2V}{n\pi}$

I think this is right:

$V(x,y) = \sum_{n=1}^{\infty} -\frac{2V}{n\pi} \sin ({n \pi x \over a}) \sinh ({n \pi y \over a})$

 

[LCKurtz]

Homework Statement

$\frac{\partial V}{\partial x} +\frac{\partial V}{\partial y} = 0$

Presumably those are second order partials.

it's defined for $V=0$ for

$x=0, 0<y<b$,
$x=a, 0<y<b$,
$y=0, 0<x<a$

$V=V_0 x/a$ for $y=b, 0<x<a$

Homework Equations

None.

The Attempt at a Solution

I've got

Presumably you got for your proposed solution$$
V(x,y) = \sum_{n=1}^\infty K_n\sin\frac{n\pi x}{a}\sinh\frac{n\pi y}{a}$$It is when you put in $y=b$ you get the next equation, and the left side of that equation should be $V(x,b)$, not $V(x,y)$ since you are trying to make the last boundary condition work.

$V(x,y) = \displaystyle V_0 x/a = \sum_{n=1}^{\infty} K_n \sin ({n \pi x \over a}) \sinh ({n \pi b \over a})$

At first I thought I could set K to 0 for n>1 and do a simple division, but I don't know what K1 is. Do I always need to find Fourier coefficients. I have seen a problem www.robots.ox.ac.uk/~jmb/lectures/pdelecture5.pdf here where in the final stage they do something like this. I'm guessing I need to do the former, but I am interested to know if the other method is possible.

As for the coefficients, the function is odd so the only integral I need is the sine integral.

$K_n = \frac{1}{a} \int_0^{2a} \frac{V x}{a} \sin ({n \pi x \over a}) dx =-\frac{2V}{n\pi}$

I think this is right:

$V(x,y) = \sum_{n=1}^{\infty} -\frac{2V}{n\pi} \sin ({n \pi x \over a}) \sinh ({n \pi y \over a})$

No, you've lost your way here. Of course you need a Fourier Series. How else can the sum of sine terms equal $V_0x/a$? First, it doesn't matter whether that function is odd because whatever it might be, you would use the odd half-range expansion since all you have to work with is sine terms. What you need is for$$K_n\sinh\frac {n\pi b} a$$ to be the half range Fourier coefficient usually called $b_n$. So set$$
K_n\sinh\frac {n\pi b} a = \frac 2 a\int_0^a \frac {V_0x} a \sin\frac{n\pi x}a dx$$
Solve that for $K_n$ and put in in the formula for $V(x,y)$.

 

[Gregg]

I get $K_n = -\frac{2V(-1)^n}{n \pi \sinh ({n \pi b \over a } )}$