Laplace equation on a semi infinite slab

[bobred]

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 [tex]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[/tex]

 

[bobred]

Any ideas anyone?