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bobred
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Homework Statement
Could someone check my work plaese.
\frac{\partial^2u}{\partial x^2}(x,y)+\frac{\partial^2u}{\partial y^2}(x,y)=0
(0<x<1, 0<y)
\frac{\partial u}{\partial x}(0,y)=\frac{\partial u}{\partial y}(1,y)=0
u(x,y)\rightarrow k as y\rightarrow\infty
u(x,0)=f(x) (0\leqx\leq1)
Homework Equations
\ddot{X}-\mu X=0 and \ddot{Y}+\mu Y=0, \mu = k^2 and k=\pi r[\tex]<br /> <br /> X(x)=A \cos(kx)+B \sin(kx) so \dot{X}=-Ak\sin(kx)+Bk\cos(kx)<br /> <br /> With the boundary conditions we have X_r(x)=A_r \cos(r\pi x)<br /> <br /> \ddot{Y}+\mu Y=0 gives Y(y)=Ce^{r\pi y}+De^{-r\pi y} with the boundary conditions and setting C=0, Y(y)=De^{-r\pi y}<br /> <h2>The Attempt at a Solution</h2><br /> <br /> So we have u(x,y)=A_0+\sum^\infty _{r=1} {A_r\cos(r\pi x)e^{-r\pi y}}<br /> <br /> u(x,0)=f(x)=A_0+\sum^\infty _{r=1} {A_r\cos(r\pi x)<br /> <br /> A_0 = \int^1 _0 \cos(r\pi x) dx Here r=0 so A_0 = 1<br /> <br /> A_r =2 \int^1 _0 \cos(r\pi x)\cos(r\pi x) dx Here I get all A coefficients as 1, is this right?<br /> <br /> Thanks
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