Diffusion Equation on a plate - 2 dimensions

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Homework Help Overview

The problem involves determining the steady-state temperature distribution in a thin plate situated in the x-y plane, with specific boundary conditions. The edges of the plate are held at various temperatures, including a sinusoidal temperature along one edge, while the other edges are maintained at 0°C. The context is rooted in the diffusion equation, which governs heat distribution in the plate.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the transition from solving problems for a rod to a two-dimensional plate, questioning how to apply boundary conditions in this new context. There are attempts to apply separation of variables and concerns about the complexity introduced by the additional dimension. Some participants suggest modifications to the governing equations and explore the implications of dropping certain terms.

Discussion Status

The discussion is ongoing, with participants providing hints and suggestions for approaching the problem. Some participants express confusion regarding the application of separation of variables in two dimensions, while others offer guidance on how to set up the equations correctly. There is no clear consensus yet, as various interpretations and methods are being explored.

Contextual Notes

Participants note the importance of boundary conditions and the potential simplifications that could arise from specific assumptions about the temperature distribution. There is mention of a reference to a similar problem in literature, which may provide additional context for solving the current problem.

samee
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Homework Statement



The edges of a thin plate are held at the temperature described below. Determine the steady-state temperature distribution in the plate. Assume the large flat surfaces to be insulated.

If the plate is lying along the x-y plane, then one corner would be at the origin. The height of the plate would be 1m along the y-axis and the length would be 2m along the x-axis. The edge along the y-axis is being held at 0 C. The edge along the x-axis is being held at 0 C. The edge parallel to the x-axis is being held at 0 C. The edge parallel to the y-axis is being held at 50sin(pi*y) C.

Homework Equations



So this question is actually just a diffusion equation. Alpha2uxx=ut
and
u(x,t)=X(x)T(t)=(C1coskx+C2sinkx)e-K2alpha2t+C3+C4x

The Attempt at a Solution



So the problem I'm having is previously everything was done on a rod. Now I have a plate. I know that for a rod, my boundary conditions would be the temperature at the ends, for example, if this were a rod I would say that u(0,t)=0 and u(2,t)=50sin(piy). BUT! How can I say that if I have corners and a second dimension to deal with?
 
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samee said:

Homework Statement



The edges of a thin plate are held at the temperature described below. Determine the steady-state temperature distribution in the plate. Assume the large flat surfaces to be insulated.

If the plate is lying along the x-y plane, then one corner would be at the origin. The height of the plate would be 1m along the y-axis and the length would be 2m along the x-axis. The edge along the y-axis is being held at 0 C. The edge along the x-axis is being held at 0 C. The edge parallel to the x-axis is being held at 0 C. The edge parallel to the y-axis is being held at 50sin(pi*y) C.

Homework Equations



So this question is actually just a diffusion equation. Alpha2uxx=ut
and
u(x,t)=X(x)T(t)=(C1coskx+C2sinkx)e-K2alpha2t+C3+C4x
No, it isn't. There are two space variables, x and y. The equation is \alpha^2(u_{xx}+ u_{yy})= u_t.
(Corrected thanks to LCKurtz.)

The Attempt at a Solution



So the problem I'm having is previously everything was done on a rod. Now I have a plate. I know that for a rod, my boundary conditions would be the temperature at the ends, for example, if this were a rod I would say that u(0,t)=0 and u(2,t)=50sin(piy). BUT! How can I say that if I have corners and a second dimension to deal with?
 
Last edited by a moderator:
Halls has a typo there, he means$$
\alpha^2(u_{xx} + u_{yy})=u_t$$
in case it isn't obvious.
 
Okay, so I'm going to expand the formula from alpha2uxx to alpha2(uxx+uyy)=ut

Then, u(x,y,t)=(X(x)+Y(y))T(t)

and

X'"X+Y"/Y=T'/alpha2T=λ and λ=0

so, X(x)=ax+b, Y(y)=cy+d and T(t)=e

Then u(x,y,t)=c(ax+cy+b+d)

right?
 
LCKurtz said:
Halls has a typo there, he means$$
\alpha^2(u_{xx} + u_{yy})=u_t$$
in case it isn't obvious.

samee said:
Okay, so I'm going to expand the formula from alpha2uxx to alpha2(uxx+uyy)=ut

Then, u(x,y,t)=(X(x)+Y(y))T(t)

and

X'"X+Y"/Y=T'/alpha2T=λ and λ=0

Assuming that X''' is a typo meaning X'', please show us how you got that equation with that substitution for u(x,y,t). Or are you just guessing?
 
I was just guessing. Since u(x,t)=X(x)T(t), I assumed that U(x,y,t)=[X(x)+Y(y)]T(t) because x and y are vectors and T isn't. And yes, I meant X"/X, not X'''/X
 
samee said:
I was just guessing. Since u(x,t)=X(x)T(t), I assumed that U(x,y,t)=[X(x)+Y(y)]T(t) because x and y are vectors and T isn't. And yes, I meant X"/X, not X'''/X

X, Y, and T are not vectors, they are ordinary functions. And it is one thing to guess what the answer might be but surely if you are doing mathematics you would check your guess to see if you got lucky. If you try it, you will not be able to verify your guess.

I will give you a hint: Try u(x,y,t) = X(x)Y(y)T(t). No guessing from here. Plug it in and see what you get. Work it out.
 
I'm confused because this was supposed to be solved with separation of variables. I don't know how to do that with 3 variables.
 
alpha2(X"/X+Y"/Y)=T'/T

X"/X+Y"/Y=T'/(Talpha2)=λ=0

If I were just doing X"/X=T'/(Talpha2), I would use separation of variables method to solve and get that X(x)=ax+b and T(t)=c

Since this is not just X"/X, but X"/X+Y"/Y, I don't think I know how to solve the problem.
 
  • #10
samee said:
alpha2(X"/X+Y"/Y)=T'/T

X"/X+Y"/Y=T'/(Talpha2)=λ=0

If I were just doing X"/X=T'/(Talpha2), I would use separation of variables method to solve and get that X(x)=ax+b and T(t)=c

Since this is not just X"/X, but X"/X+Y"/Y, I don't think I know how to solve the problem.

There is no reason to assume ##\lambda=0##. Since each fraction uses a different variable you can assume each fraction is constant. So you might start with$$
\frac{X''} X = \lambda,\ \frac{Y''} Y = \mu,\ \frac{T''}{\alpha^2 T}=\lambda +\mu$$
Start with the ##Y## equation. Apply u=XYT to the boundary conditions to see what the boundary conditions for Y must be. Then solve the Y eigenvalue problem just like you did for a 1D rod. After that, do the X equation, followed by the T equation.

I have to leave for the rest of the afternoon but that should get you started.
 
  • #11
Okay, I'm looking more carefully through my book, and I found a problem with a temperature difference on a thin disk. They simplify the problem and explain, "since u does not vary with θ, we dropped the uθθ term and solved the reduced equation. If you have doubts about this step, observe that equation 79 does satisfy the boundary conditions."

So can I drop the uyy term since the plate have a temperature gradient along x and not y? That would make the whole problem much simpler.
 
  • #12
May I suggest, and I know LCKurtz is doing a fine job helping you, that you find "Basic Partial Differential Equations" by Bleecker and Csordas in the Library and just work through the example? And the reason I suggest this is because you seem to be having trouble with just the concepts, for example separation of variables and it's always nice to first just work through an example in the book, or two or three, then attempt to work on a new one yourself.
 
  • #13
samee said:
... Determine the steady-state temperature distribution in the plate...

Steady-state means u_t=0 so the equation to solve is u_{xx}+u_{yy}=0 and u=u(x,y)
 

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