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I was following this example from Griffiths (example 3.3). The solution to the following problem

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

$$V(y=0,a)=0$$

$$V(x=0)=V_0$$

is

$$V(x,y) = \sum_{n=1,\text{odd}}^\infty \frac{4V_0}{n\pi} e^{-\frac{n\pi x}{a}} \sin\left(\frac{n\pi y}{a}\right)$$

I am not sure how it satisfies the boundary condition $V(x=0)=V_0$. When I try to put the result in my calculator with $V_0=47,x=0,a=5,N=100$:

$$V(0,y) = \sum_{n=1}^{100} \frac{4 (47)}{(2n+1)\pi} \sin\left(\frac{(2n+1)\pi y}{5}\right)$$

It gives zero when $y=0,5$ and when $y=3$, it gives $-9.7875...$. I expected the answer to be $47$ for all y values. What is the problem?

Jimmy Yang
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  • Like you said, it's not possible. – Chee Han Dec 30 '22 at 18:40
  • @CheeHan What do you mean? How do I know whether the BC is satisfied? – Jimmy Yang Dec 30 '22 at 18:48
  • Of course you only get equality in the limit when you sum up to infinity (and it's only for $0<y<a$, not at the endpoints $y=0$ and $y=a$). You can't expect to get the right answer from just a partial sum. I suggest that you plot some partial sums for $V(0,y)$ as functions of $y \in [0,a]$, so that you can see what happens when you keep adding more terms. – Hans Lundmark Dec 30 '22 at 22:15
  • @HansLundmark I still got -9.9 with N=1000 and N=100000 – Jimmy Yang Dec 30 '22 at 22:21
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    Oh, I see the mistake now. You should start the sum on $n=0$ instead of $n=1$, so that $2n+1$ starts from $1$ as it should (rather than from $3$). – Hans Lundmark Dec 30 '22 at 22:25
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    @HansLundmark Thank you, I think it answered my question ;-; – Jimmy Yang Dec 30 '22 at 22:29

0 Answers0