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Lawrencel2
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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_{0}\sum\frac{1-cos(\alpha_{n})}{\alpha_{n}(h+cos^{2}\alpha_{n})}e^{-\alpha_{n}x}sin(\alpha_{n}y)
where tan\alpha_{n}=\frac{-\alpha_{n}}{h} (\alpha>0
u_{xx}+u_{yy}=0
u(x,y)= X(x)Y(y)
I Cant seem to even come close to the answer.
Maybe some hints as to how i should approach this?
Y(y)=C(e^{\alpha_{n}y}-e^{-\alpha_{n}y}), Y(0)=0,<br /> X(x)=C_{1}cos(\alpha_{n}x)+C_{2}sin(\alpha_{n}x),<br /> <br /> \frac{-\alpha_{n}}{h}=tanh(\alpha_{n})
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_{0}\sum\frac{1-cos(\alpha_{n})}{\alpha_{n}(h+cos^{2}\alpha_{n})}e^{-\alpha_{n}x}sin(\alpha_{n}y)
where tan\alpha_{n}=\frac{-\alpha_{n}}{h} (\alpha>0
Homework Equations
u_{xx}+u_{yy}=0
u(x,y)= X(x)Y(y)
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?
Y(y)=C(e^{\alpha_{n}y}-e^{-\alpha_{n}y}), Y(0)=0,<br /> X(x)=C_{1}cos(\alpha_{n}x)+C_{2}sin(\alpha_{n}x),<br /> <br /> \frac{-\alpha_{n}}{h}=tanh(\alpha_{n})
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