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

  • #1

Homework Statement



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

[tex]\frac{\partial^2{T}}{\partial{x}^2}+{\frac{\partial^2{T}}{\partial{y}^2}}=0[/tex]

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

[tex]T(0,y)=T(S,y)=T(x,0)=0[/tex]

[tex]T(x,S)=T_0[/tex]


Homework Equations





The Attempt at a Solution



For a solution, I get:

[tex]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)}[/tex]

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:

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

[tex]B_n=A_n{sinh(n\pi)}[/tex]

[tex]B_n=\frac{2}{S}\int_{0}^{S}T_0{sin(\frac{n\pi}{S}x)}dx[/tex]

[tex]B_n=\frac{4{T_0}}{n\pi}[/tex] for "n" odd and 0 for "n" even.

Does anyone know if the coefficient is incorrect?
 

Answers and Replies

  • #2
gabbagabbahey
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If even [itex]n[/itex] coefficients are zero, why are you summing over both even and odd [itex]n[/itex] in your final solution?
 
  • #3
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)?
 
  • #4
gabbagabbahey
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Well, there are two common ways of doing this:

(1) Replace [tex]\sum_{n=1}^{\infty}[/tex] by [tex]\sum_{n=1,3,5,\ldots}^{\infty}[/tex]

(2) Replace [itex]n[/itex] by [itex]2n+1[/itex]....

[tex]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)}[/tex]
 
  • #5
Well, there are two common ways of doing this:

(1) Replace [tex]\sum_{n=1}^{\infty}[/tex] by [tex]\sum_{n=1,3,5,\ldots}^{\infty}[/tex]

(2) Replace [itex]n[/itex] by [itex]2n+1[/itex]....

[tex]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)}[/tex]

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?
 
  • #6
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.
 
  • #7
Well, there are two common ways of doing this:

(1) Replace [tex]\sum_{n=1}^{\infty}[/tex] by [tex]\sum_{n=1,3,5,\ldots}^{\infty}[/tex]

(2) Replace [itex]n[/itex] by [itex]2n+1[/itex]....

[tex]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)}[/tex]

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.
 
  • #8
gabbagabbahey
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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?

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

Keep in mind, that for [itex]0\leq x\leq S[/itex],

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

which is indeed a constant, not a function of [itex]x[/itex].
 
  • #9
gabbagabbahey
Homework Helper
Gold Member
5,002
7
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

[tex]\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)}[/tex]

does not imply that the even terms are zero. You need to explicitly sum over only odd [itex]n[/itex].
 
  • #10
Keep in mind, that for [itex]0\leq x\leq S[/itex],

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

which is indeed a constant, not a function of [itex]x[/itex].

Ahh, I see. Thanks again.
 

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