# Laplace Eq with Dirichlet boundary conditions in 2D (solution check)

PhysicsMark

## Homework Statement

The steady state temperature distribution, T(x,y), in a flat metal sheet obeys the partial differential equation:

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

Separate the variables and find T everywhere on a square flat plate of sides S with boundary conditions:

$$T(0,y)=T(S,y)=T(x,0)=0$$

$$T(x,S)=T_0$$

## The Attempt at a Solution

For a solution, I get:

$$T(x,y)=\sum_{n=1}^{\infty}\frac{4T_0}{n\pi}sin(\frac{n\pi}{S}x)\frac{sinh(\frac{n\pi}{S}y)}{sinh(n\pi)}$$

I am not sure if I have given a sufficient enough answer for the coefficient of the series. I got the coefficient by doing the following:

$$T(x,s)=\sum_{n=1}^{\infty}A_n{sin(\frac{n\pi}{S}x)}sinh(n\pi)=T_0$$

$$B_n=A_n{sinh(n\pi)}$$

$$B_n=\frac{2}{S}\int_{0}^{S}T_0{sin(\frac{n\pi}{S}x)}dx$$

$$B_n=\frac{4{T_0}}{n\pi}$$ for "n" odd and 0 for "n" even.

Does anyone know if the coefficient is incorrect?

Homework Helper
Gold Member
If even $n$ coefficients are zero, why are you summing over both even and odd $n$ in your final solution?

PhysicsMark
Are you suggesting an index change? As in swapping out all n's to the right of sigma with m, where m = 2n?

Or changing the n's to n(n+1)?

Homework Helper
Gold Member
Well, there are two common ways of doing this:

(1) Replace $$\sum_{n=1}^{\infty}$$ by $$\sum_{n=1,3,5,\ldots}^{\infty}$$

(2) Replace $n$ by $2n+1$...

$$T(x,y)=\sum_{n=1,3,5}^{\infty}\frac{4T_0}{n\pi}sin(\frac{n\pi}{S}x)\frac{sinh(\frac{n\pi}{S}y)}{sinh(n\pi)}=\sum_{n=0}^{\infty}\frac{4T_0}{(2n+1)\pi}sin(\frac{(2n+1)\pi}{S}x)\frac{sinh(\frac{(2n+1)\pi}{S}y)}{sinh((2n+1)\pi)}$$

PhysicsMark
Well, there are two common ways of doing this:

(1) Replace $$\sum_{n=1}^{\infty}$$ by $$\sum_{n=1,3,5,\ldots}^{\infty}$$

(2) Replace $n$ by $2n+1$...

$$T(x,y)=\sum_{n=1,3,5}^{\infty}\frac{4T_0}{n\pi}sin(\frac{n\pi}{S}x)\frac{sinh(\frac{n\pi}{S}y)}{sinh(n\pi)}=\sum_{n=0}^{\infty}\frac{4T_0}{(2n+1)\pi}sin(\frac{(2n+1)\pi}{S}x)\frac{sinh(\frac{(2n+1)\pi}{S}y)}{sinh((2n+1)\pi)}$$

First, I forgot to thank you for taking the time to reply. Thank you.

Gosh, I see. Sorry about the earlier post. I was thinking of only including the even numbers for some reason.

Would you say that what you have posted above is "as good as it gets"?

I was concerned about how I found the coefficient because I don't think I fully comprehend the term T_0. I read T_0 to be initial temperature. I assumed it was a constant, and treated it as such in the integration of it with the sine term.

My expression for T_0, indicates that it is a function of x.

Is it correct to interpret T_0 as the initial temperature?

If T_0 is a function of x, then are we talking about how the temperature changes from the initial temperature to a lesser temperature as we move away from the source of the initial temperature?

Christina2010
Hey, I think either way is correct. you could do from 1 to inf, and assume that the even numbers are 0, also, you could sum odd number. I followed your way to calculate the Bn, but I got the same answer as the PhysicsMark had.

Christina2010
Well, there are two common ways of doing this:

(1) Replace $$\sum_{n=1}^{\infty}$$ by $$\sum_{n=1,3,5,\ldots}^{\infty}$$

(2) Replace $n$ by $2n+1$...

$$T(x,y)=\sum_{n=1,3,5}^{\infty}\frac{4T_0}{n\pi}sin(\frac{n\pi}{S}x)\frac{sinh(\frac{n\pi}{S}y)}{sinh(n\pi)}=\sum_{n=0}^{\infty}\frac{4T_0}{(2n+1)\pi}sin(\frac{(2n+1)\pi}{S}x)\frac{sinh(\frac{(2n+1)\pi}{S}y)}{sinh((2n+1)\pi)}$$

Hey, I think either way is correct. you could do from 1 to inf, and assume that the even numbers are 0, also, you could sum odd number. I followed your way to calculate the Bn, but I got the same answer as the PhysicsMark had.

Homework Helper
Gold Member
Would you say that what you have posted above is "as good as it gets"?

Yes, everything else in your calculations is correct.

I was concerned about how I found the coefficient because I don't think I fully comprehend the term T_0. I read T_0 to be initial temperature. I assumed it was a constant, and treated it as such in the integration of it with the sine term.

My expression for T_0, indicates that it is a function of x.

I assume you are worried about the following expression?

$$T(x,s)=\sum_{n=1}^{\infty}A_n{sin(\frac{n\pi}{S}x)}sinh(n\pi)=T_0$$

Keep in mind, that for $0\leq x\leq S$,

$$\sum_{n=0}^{\infty}\frac{4}{(2n+1)\pi}\sin\left(\frac{(2n+1)\pi}{S}x\right)=1$$

which is indeed a constant, not a function of $x$.

Homework Helper
Gold Member
Hey, I think either way is correct. you could do from 1 to inf, and assume that the even numbers are 0, also, you could sum odd number. I followed your way to calculate the Bn, but I got the same answer as the PhysicsMark had.

The problem is that the expression

$$\sum_{n=1}^{\infty}\frac{4T_0}{n\pi}sin(\frac{n\pi}{S}x)\frac{sinh(\frac{n\pi}{S}y)}{sinh(n\pi)}$$

does not imply that the even terms are zero. You need to explicitly sum over only odd $n$.

PhysicsMark
Keep in mind, that for $0\leq x\leq S$,

$$\sum_{n=0}^{\infty}\frac{4}{(2n+1)\pi}\sin\left(\frac{(2n+1)\pi}{S}x\right)=1$$

which is indeed a constant, not a function of $x$.

Ahh, I see. Thanks again.