Solving Laplace Equations using this boundary conditions?

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astrodeva
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The equation is Uxx + Uyy = 0
And domain of solution is 0 < x < a, 0 < y < b
Boundary conditions:
Ux(0,y) = Ux(a,y) = 0
U(x,0) = 1
U(x,b) = 2

What I've done is that I did separation of variables:
U(x,y)=X(x)Y(y)

Plugging into the equation gives:
X''Y + XY'' = 0

Rearranging:
X''/X = -Y''/Y = k

For case k > 0, I saw that it gives no non-trivial solutions.
For case k = 0, I solved it and found U(x,y) = y/b + 1

For case k < 0, I'm slightly lost.
X'' + kX = 0
Y'' - kX = 0

upload_2016-4-9_12-32-57.png


Using the X boundary conditions:
upload_2016-4-9_12-33-41.png


upload_2016-4-9_12-34-11.png

Using the Y boundary condition:
upload_2016-4-9_12-34-47.png

Using Fourier Series to find the coefficient:
upload_2016-4-9_12-34-59.png


But the integral just gives Dn = 0, and this doesn't satisfy u(x,0) = 1.

Can someone explain where I went wrong?

Thanks!
 

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Your working is confusing.

If you start with [itex]X'' = kX[/itex] and want [itex]k < 0[/itex], you should then be writing [itex]\sin(\sqrt{|k|}x)[/itex] and so forth, or defining [itex]k = -c^2[/itex] where [itex]c \geq 0[/itex].

You want to be using cosines, which in fact you end up doing, but only after expressly stating that [itex]X_n = A_n\sin(n\pi x/a)[/itex].

Now the cosine expansion of [itex]1[/itex] on [itex][0,a][/itex] is [itex]\cos(0x)[/itex]. Thus you have correctly determined that all the other coefficients vanish.