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:

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

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

Homework Equations

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?

 

[gabbagabbahey]

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)?

 

[gabbagabbahey]

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.

 

[gabbagabbahey]

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 $x$.

 

[gabbagabbahey]

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.