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Inhomogeneous PDE

by Lacero
Tags: inhomogeneous
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Lacero
#1
Apr7-08, 08:37 PM
P: 32
Hi all,

1. The problem statement, all variables and given/known data

Determine the equilibrium temperature distribution (if it exists). For what values of B, are there solutions.

2. Relevant equations

a) Ut = Uxx + 1, U(x,0) = f(x), Ux(0,t) = 1, U(L,t) = B

b) Ut = Uxx + X - B, U(x,0) = f(x), Ux(0,t) = 0, U(L,t) = 0

3. The attempt at a solution

a) Assume solution U(x,t) = V(x,t) + K

=> Ut = Vt
=> Uxx = Vxx -- Sub into Ut = Uxx + 1

Vt = Vxx + 1

Now to get homogeneous boundary conditions,

U(0,t) = V(0,t) + K = 1 ?
but K = 1 => V(0,t) = 1?

// Trouble at the BC

b)

Assume solution U(x,t) = V(x,t) + W(x)

=> Ut = Vt
=> Uxx = Vxx + W'' -- Sub into Ut = Uxx + Q(x)

Vt = Vxx + W'' + Q

Now, let W'' + Q = 0 or W'' = -Q
=> Vt = Vxx

V(0,t) = 0
V(L,t) = 0

U(x,0) = V(x,0) + W(x) = f(x)
=> V(x,0) = f(x) - W(x)

Now Solve transient solution, v(x,t)
. .
. .
. .
V(x,t) = (2/L)sum(n=1 to inf)(int((f(x)-W(x))sin(npi/L)xdx)... etc

Now Solve steady state solution, W" = -Q B - X
W" + X + B = 0
. .
. .
W(x) = Scos(zx) + Tsin(zx) + Yp... etc

Now U(x,t) = W(x) + V(x,t)

Am I right at all???
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