1. PDE: Finding the steady-state temp

Hey guys,

I am trying to find the steady state temp for the following Partial Diff. Eqn:

Ut = (alpha)^2 * Uxx - (beta)U

IC: 0<x<1

BCs: U(0,t) = 1 and U(1,t) = 1.

I know how to approach the problem, but I am getting stuck on the Integrating U. Here is what I have:

1.) SETUP OF PROBLEM
steady-state => Ut = 0, so
(alpha)^2 * Uxx - (beta)U = 0
(alpha)^2 * Uxx = (beta)U

2.) INTEGRATION/SOLVE
Integral((alpha)^2 * Uxx)dx = Integral ((beta)U)dx
((alpha)^2 * Ux = (beta) * How do I do [Integral(U) dx] ????

This is where I get stuck, so if you could please show me how to Integrate(U) dx or what I need to do to fix this? Thanks for any and all help.

2. Originally Posted by spearfish
Hey guys,

I am trying to find the steady state temp for the following Partial Diff. Eqn:

Ut = (alpha)^2 * Uxx - (beta)U

IC: 0<x<1

BCs: U(0,t) = 1 and U(1,t) = 1.

I know how to approach the problem, but I am getting stuck on the Integrating U. Here is what I have:

1.) SETUP OF PROBLEM
steady-state => Ut = 0, so
(alpha)^2 * Uxx - (beta)U = 0
(alpha)^2 * Uxx = (beta)U

2.) INTEGRATION/SOLVE
Integral((alpha)^2 * Uxx)dx = Integral ((beta)U)dx
((alpha)^2 * Ux = (beta) * How do I do [Integral(U) dx] ????

This is where I get stuck, so if you could please show me how to Integrate(U) dx or what I need to do to fix this? Thanks for any and all help.
In the steady state you have:

$\alpha^2 u''(x) =\beta u(x)$

If you are familiar with ODE's you should know that the general solution of this second order constant coefficients homogeneous ODE is:

$u(x)=A e^{\sqrt{\beta}/\alpha x}+Be^{-\sqrt{\beta}/\alpha x}$

Now use the boundary conditions to deduce $A$ and $B$

CB

3. Because is $U_{t} =0$ , the $U(*,*)$ is a function of the only $x$ and the PDE becomes the following ordinary second order initial values problem...

$\alpha^{2}\cdot \frac{d^{2}U}{dx^{2}} - \beta\cdot U= 0$ , $U(0)=U(1)=1$

Kind regards

$\chi$ $\sigma$

4. Thanks a bunch CaptainBlack and Chisigma. I had to review some ODE notes from a while back, but I am able to work out the problem now. Thanks again.