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Homogeneous Laplace's Equation

  • Thread starter rizardon
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  • #1
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Homework Statement


uxx+uyy=0
u(x,0)=u(x,pie)=0
u(0,y)=0
ux(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, Yn=sin(ny)
X"-n2X=0, X=ccosh(nx)+dsinh(nx)

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

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

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

ux(x,y)= summation (from n=1 to n=infinity) sin(ny)[dnncosh(nx)]
ux(x,0)= summation (from n=1 to n=infinity) sin(ny)[dnn]

Using Fourier's Series to expand 3siny-5sin4y, I also got 0 so dn 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?
 

Answers and Replies

  • #2
LCKurtz
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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.
 

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