- #1
sa1988
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Homework Statement
Solve ∇2T(x, y) = 0
with boundary conditions
T(0, y) = T(L, y) = T0
T(x, L/2) = T(x, -L/2) = T0 + T1sin(πx/L)
Homework Equations
The Attempt at a Solution
Set T(x, y) = X(x)Y(y)
Then ∇2T(x,y) = (∂2X/∂x2) Y + (∂2Y/∂y2) X = 0
Rearrange to find two separate ODEs:
d2X/dx2 = k2X
d2Y/dy2 = -k2Y
Two solutions:
X(x) = Aekx + Be-kx
Y(x) = Ceiky + De-iky
Thus
T(x, y) = (Aekx + Be-kx)(Ceiky + De-iky)
Boundary conditions:
T(0, y) = (A+B)(Ceiky + De-iky) = T0
T(L, y) = (AekL + Be-kL)(Ceiky + De-iky) = T0
Equate the two and rearrange to find:
B = AekL
Next boundary condition:
T(x, L/2) = (Aekx + Be-kx)(CeikL/2 + De-ikL/2) = T0 + T1sin(πx/L)
T(x, -L/2) = (Aekx + Be-kx)(Ce-ikL/2 + DeikL/2) = T0 + T1sin(πx/L)
Equate the two and rearrange to find:
C=D
Thus:
T(x, y) = AC(ekx + ekLe-kx)(eiky + e-iky)
Returning to first boundary conditions:
T(0, y) = AC(1 + ekL)(eiky + e-iky) = T0
True for all y, so set y=0, then can find:
AC = T0/2(1+ekL)
So now:
T(x, y) = (T0/2(1+ekL))(ekx + ekLe-kx)(eiky + e-iky)
Returning to second boundary condition:
T(x, L/2) = (T0/2(1+ekL))(ekx + ekLe-kx)(eikL/2 + e-ikL/2) = T0 + T1sin(πx/L)
True for all x, so set x=0:
T(0, L/2) = (T0/2(1+ekL))(1 + ekL)(eikL/2 + e-ikL/2) = T0
Reduces to:
(eikL/2 + e-ikL/2) = 2
Which solves to give:
k = 2πn/L , for n∈ℤ
Thus:
T(x, y) = (T0/2(1+e2πn))(e2πnx/L + e2πne-2πnx/L)(ei2πny/L + e-i2πny/L)
But when I check this back for ∇2T(x, y) = 0 it doesn't work.
I think the problem is where I make the "same for all x so set x=0" assumptions..?
But if I don't make those assumptions I can't find AC or k.
It's all rather confusing..!