Homogeneous Laplace's Equation

[rizardon]

Homework Statement

u_xx+u_yy=0
u(x,0)=u(x,pie)=0
u(0,y)=0
u_x(5,y)=3siny-5sin4y

Homework Equations

The Attempt at a Solution

Using separable method I get
Y"-kY= 0 and X"+kX=0

For Case 1 and Case 2 where k>0 and k=0 there are no eigenvalues

So Case 3 k<0 gives
Y=ccos(sqrk x y) + dsin(sqrk x y)
y=0, then c=0
y=pie, then dsin(pie x sqrk)=0
k= -n², Y_n=sin(ny)
X"-n²X=0, X=ccosh(nx)+dsinh(nx)

u(x,y)= summation (from n=1 to n=infinity) sin(ny)[c_ncosh(nx)+d_nsinh(nx)]

u(0,y)= summation (from n=1 to n=infinity) c_nsin(ny)

Using Fourier's Series to expand 0, I got 0 so c_n is 0

u_x(x,y)= summation (from n=1 to n=infinity) sin(ny)[d_nncosh(nx)]
u_x(x,0)= summation (from n=1 to n=infinity) sin(ny)[d_nn]

Using Fourier's Series to expand 3siny-5sin4y, I also got 0 so d_n is 0 as well

Therefore u(x,y)=0

However the answer given in the book is

u(x,y)=(3/cosh5)(siny)(sinhx) - (5/cosh20)(sin4y)(sinh4x)

I realized that the book didn't used Fourier's Series but compared the coefficients of the two equations instead. But my question is what happened to the n because if we compare the coefficients aren't we supposed to get

[3/ncosh(5n)] n=1 so we get [3/cosh(5)]
[-5/nsinh(5n)] n=4 so we get [-5/4sinh(20)] so u(x,y) should be

[3/cosh(5)](siny)(sinhx) - [5/4sinh(20)](sin4y)(sinh4x)

Did I do it correctly or is it just a printing error?

 

[LCKurtz]

However the answer given in the book is

u(x,y)=(3/cosh5)(siny)(sinhx) - (5/cosh20)(sin4y)(sinh4x)

They left out a 4 multiplying the cosh(20) in the denominator

[3/cosh(5)](siny)(sinhx) - [5/4sinh(20)](sin4y)(sinh4x)

And you meant cosh(20) instead of sinh(20). They just dropped a 4.