PDE for temperature distribution in rectangle

[collectedsoul]

Homework Statement

A rectangular chip of dimensions a by b is insulated on all sides and at t=o temperature u=0. The chip produces heat at a constant rate h. Find an expression for u(x,y,t)

Homework Equations

δu/δt = h + D(δ²u/δx² + δ²u/δy²) x∈(0,a), y∈(0,b)

The Attempt at a Solution

I'm just wondering what effect the insulation on all sides will have on the equation. The temperature goes up by h per unit time throughout the chip but will the other term be affected?

 

[pasmith]

"Insulated" means that there is no heat flux into or out of the rectangle. Here this reduces to the requirement that the component of $\nabla u$ normal to the boundary must vanish. This gives you the boundary condition you need in order to solve the PDE.

 

[collectedsoul]

Why just the normal component? If its insulated won't the tangential component also vanish? So δu/δx=δu/δy=0 at x=0,a and y=0,b.

The question actually says that there is no need to solve the pde to get to u(x,y,t) - that there is a way to guess the answer without solving it. I can't think of any such way. Any ideas?

 

[pasmith]

Why just the normal component? If its insulated won't the tangential component also vanish? So δu/δx=δu/δy=0 at x=0,a and y=0,b.

The physical constraint is "no heat flux across the boundary". The heat flux is $D\nabla u$, and the heat flux across the boundary is the component of $D\nabla u$ normal to the boundary. Flux parallel to the boundary is not a problem.

The question actually says that there is no need to solve the pde to get to u(x,y,t) - that there is a way to guess the answer without solving it. I can't think of any such way. Any ideas?

Is there any reason why the temperature should vary with position here, given that the heat source doesn't depend on position and the boundary conditions are consistent with a uniform temperature?

 

[collectedsoul]

Alright...got it.

There's another related PDE: d²u/dx² + d²u/dy² = a*u with u = 0 on all edges for some eigenvalues a. I'm not sure how to go about this. I can start by solving the steady state equation but what do I do about the condition?

 

[pasmith]

Seek a separable solution $u(x,y) = X(x)Y(y)$. You will find that there are particular values of $a$ (eigenvalues) for which $u(x,y) \equiv 0$ is not the only such solution.

 

[collectedsoul]

I'm not clear what you mean by eigenvalues for which u=0 is not the only solution. Could you please elaborate?

 

[pasmith]

Simple example: Let $X'' = kX$ subject to $X(0) = X(\pi) = 0$. The general solution of that ODE with satisfies $X(0) = 0$ is $$X(x) = \begin{cases} A\sin(\sqrt{|k|}x), & k < 0 \\ Ax, & k = 0 \\ A\sinh(\sqrt{k}x), & k > 0\end{cases}$$ In the second and third alternatives, the only way to satisfy $X(\pi) = 0$ is to take $A = 0$, which means that $X(x) = 0$. However in the first case we can set $k = -n^2 > 0$ for integer $n$ and then $X(x) = A \sin(nx)$ satisfies $X(\pi) = 0$ for any value of $A$. Thus $\{-n^2 : n \in \mathbb{N}\}$ are the eigenvalues, and $\sin(nx)$ is the corresponding eigenfunction. We don't need $n < 0$ because $\sin(-|n|x) = -\sin(|n|x)$.

 

[collectedsoul]

Maybe I'm being dense but I'm not able to separate the variables, I get X^''/X + Y^''/Y = a. So how do I get separate equations for x and y?

 

[pasmith]

Maybe I'm being dense but I'm not able to separate the variables, I get X^''/X + Y^''/Y = a. So how do I get separate equations for x and y?

X''/X is a function of x only. Y''/Y is a function of y only. Their sum is a constant. Therefore each of them is constant.

 

[collectedsoul]

So X^''/X=m (a cpnstant) and Y^''/Y = n (another constant) and m + n = a?

 

[pasmith]

So X^''/X=m (a cpnstant) and Y^''/Y = n (another constant) and m + n = a?

Yes.

 

[collectedsoul]

Ok...so do I break the equation down into 2 parts? A steady state one where d2u/dx2 + d2u/dy2 = 0 and another where it's = a*v and then add the 2 solutions? Or do I just do the latter equation?