# Modified heat equation steady state

1. Jul 17, 2011

### dp182

1. The problem statement, all variables and given/known data
determine the steady state equation for the given heat equation and boundary conditions

2. Relevant equations
Ut=Uxx-4(U-T)
U(0,T)=T U(4,T)=0 U(x,0)=f(x)

3. The attempt at a solution
I put Ut=0
so 0=UInf''-4(Uinf-T)
then once I tried to integrate I ended up with a Uinf thats impossible to isolate is there a specific method used for this kind of heat equation?

2. Jul 17, 2011

### hunt_mat

The steady state is usually when the system is time independent, so:
$$\frac{\partial U}{\partial t}=0$$
So this will turn your equation into and ODE, which I see you have already done and your ODE is:
$$\frac{\partial^{2}U}{\partial x^{2}}-4U+4T=0$$
I am not too sure what T is, is it a constant? A function?

Last edited: Jul 17, 2011
3. Jul 17, 2011

### LCKurtz

What do you mean by "impossible to isolate". Calling what you call Uinf u(x) for simplicity the equation you have written is just

u''(x) - 4u(x) = -4T
u(0) = u(T) = 0

A second order constant coefficient ODE.

4. Jul 18, 2011

### dp182

I believe T is constant. so I tried using variation of parameters to solve (Uinf=X) X''-4X=-4T
and came up with a particular solution of
c1e2x+c2e-2x-T then solved for the constants using boundary conditions. is this method right as all the example I have seen involve just integrating it twice.

5. Jul 18, 2011

### LCKurtz

That should work, but I'm guessing it gets a bit messy solving for the constants. Here's a little trick you might like to learn. You probably know that instead of using the pair {e2x,e-2x} for the general solution of the homogeneous equation, you could have used {sinh(2x),cosh(2x)}, which simplifies things a bit when you calculate X(0) = 0. But an even better choice is the pair {sinh(x),sinh(T-x}. Try evaluating the constants when you write your solution

X(x) = Asinh(x)+Bsinh(T-x) + T

and you will see what I mean.