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PDE Harmonic Function help!
A bounded harmonic function u(x,y) in a semi-infinite strip x>0, 0<y<1 is to satisfy the boudary conditions:
u(x,0)=0, uy(x,1)=-hu(x,1), u(0,y)=u0,
Where h (h>0) and u0 are constants. Derive the expression:
u(x,y)=2hu[itex]_{0}[/itex][itex]\sum[/itex][itex]\frac{1-cos(\alpha_{n})}{\alpha_{n}(h+cos^{2}\alpha_{n})}e^{-\alpha_{n}x}sin(\alpha_{n}y)[/itex]
where [itex]tan\alpha_{n}=\frac{-\alpha_{n}}{h}[/itex] ([itex]\alpha>0[/itex]
[itex]u_{xx}+u_{yy}=0[/itex]
[itex] u(x,y)= X(x)Y(y) [/itex]
I Cant seem to even come close to the answer.
Maybe some hints as to how i should approach this?
[itex]Y(y)=C(e^{\alpha_{n}y}-e^{-\alpha_{n}y}), Y(0)=0,
X(x)=C_{1}cos(\alpha_{n}x)+C_{2}sin(\alpha_{n}x),
\frac{-\alpha_{n}}{h}=tanh(\alpha_{n}) [/itex]
Homework Statement
A bounded harmonic function u(x,y) in a semi-infinite strip x>0, 0<y<1 is to satisfy the boudary conditions:
u(x,0)=0, uy(x,1)=-hu(x,1), u(0,y)=u0,
Where h (h>0) and u0 are constants. Derive the expression:
u(x,y)=2hu[itex]_{0}[/itex][itex]\sum[/itex][itex]\frac{1-cos(\alpha_{n})}{\alpha_{n}(h+cos^{2}\alpha_{n})}e^{-\alpha_{n}x}sin(\alpha_{n}y)[/itex]
where [itex]tan\alpha_{n}=\frac{-\alpha_{n}}{h}[/itex] ([itex]\alpha>0[/itex]
Homework Equations
[itex]u_{xx}+u_{yy}=0[/itex]
[itex] u(x,y)= X(x)Y(y) [/itex]
The Attempt at a Solution
I Cant seem to even come close to the answer.
Maybe some hints as to how i should approach this?
[itex]Y(y)=C(e^{\alpha_{n}y}-e^{-\alpha_{n}y}), Y(0)=0,
X(x)=C_{1}cos(\alpha_{n}x)+C_{2}sin(\alpha_{n}x),
\frac{-\alpha_{n}}{h}=tanh(\alpha_{n}) [/itex]
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