Thread: BVP

**ux0** · October 29th 2009, 07:16 AM

For what constant value of q does the boundary value problem

$u_t=u_{xx} + q$ . . . $0<x<2, t>0$
$u_x(0,t)=2$ . . . . . $t \geq 0$
$u_x(2,t)=0$ . . . . . $t \geq 0$
$u(x,o)=f(x)$ . . . . . $0 \leq x \leq 2$

have a steady state solution? Find the complete solution when

$ f(x)=\left\{\begin{matrix} 2x \mapsto 0 \leq x \leq 1 \\ x \mapsto 1 \leq x \leq 2 \end{matrix}\right.$

So to find the steady state solution i did the following steps. (The notation i use is the notation we use in class)

$0=U''+q \Rightarrow U''=-q$

$U'(x)=-qx + A$

$U(x)= \frac{-q}{2}x^2 + Ax + B$

using my endpoint conditions i get

$U'(2) = 0 = -q2 + A$

$A=2q$

$U'(0)=2=2q \rightarrow 1=q \rightarrow A=2$

So my steady state solution is

$U(x)=\frac{-1}{2}x^2+2x+B$

which is only valid when $q = 1$ ... If this part is right, I can't seem to find the complete solution when i have the Constant B...

**Rebesques** · January 21st 2010, 06:12 AM

Take B=0, and it is a steady state solution.

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