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I want to solve the following problem:

$$u_{xx}(x,y)+u_{yy}(x,y)=0, 0<x<\pi, y>0 \\ u(0,y)=u(\pi, y)=0, y>0 \\ u(x,0)=\sin x +\sin^3 x, 0<x<\pi$$

$u$ bounded

I have done the following:

$$u(x,y)=X(x)Y(y)$$

We get the following two problems:

$$X''(x)+\lambda X(x)=0 \ \ \ \ \ (1) \ X(0)=X(\pi)=0$$

$$Y''(y)-\lambda Y(y)=0 \ \ \ \ \ (2)$$

To solve the problem $(1)$ we do the following:

The characteristic polynomial is $\mu +\lambda =0$.

  • $\lambda <0$: $\mu=\pm \sqrt{-\lambda}$

    $X(x)=c_1e^{\sqrt{-\lambda}x}+c_2e^{-\sqrt{-\lambda}x}$

    $X(0)=0 \Rightarrow c_1+c_2=0 \Rightarrow c_1=-c_2$

    $X(\pi)=0 \Rightarrow c_1e^{\sqrt{-\lambda}\pi}+c_2 e^{-\sqrt{-\lambda}\pi}=0 \Rightarrow c_2(-e^{\sqrt{-\lambda}\pi}+e^{-\sqrt{-\lambda}\pi})=0 \Rightarrow c_1=c_2=0$

  • $\lambda=0$:

    $X(x)=c_1 x+c_2$

    $X(0)=0 \Rightarrow c_2=0 \Rightarrow X(x)=c_1x$

    $X(\pi)=0 \Rightarrow c_1 \pi=0 \Rightarrow c_1=0$

  • $\lambda >0$ :

    $X(x)=c_1 \cos (\sqrt{\lambda} x)+c_2 \sin (\sqrt{\lambda}x)$

    $X(0)=0 \Rightarrow c_1=0 \Rightarrow X(x)=c_2 \sin (\sqrt{\lambda}x)$

    $X(\pi)=0 \Rightarrow \sin (\sqrt{\lambda}\pi)=0 \Rightarrow \sqrt{\lambda}\pi=k\pi \Rightarrow \lambda=k^2$

For the problem $(2)$ we have the following:

$Y(y)=c_1 e^{ky}+c_2 e^{-ky}$

The general solution is the following:

$$u(x,y)=\sum_{k=0}^{\infty}a_n( e^{ky}+ e^{-ky}) \sin (kx) $$

$$u(x,0)=\sin x+\sin^3 x=\sin x+\frac{3}{4}\sin x-\frac{1}{4}\sin (3x)=\frac{7}{4}\sin x-\frac{1}{4}\sin (3x) \\ \Rightarrow \frac{7}{4}\sin x-\frac{1}{4}\sin (3x)=\sum_{k=0}^{\infty}2a_n\sin (kx) \\ \Rightarrow 2a_1=\frac{7}{4} \Rightarrow a_1=\frac{7}{8}, 2a_3=-\frac{1}{4}=-\frac{1}{8}, a_k=0 \text{ for } k=2,4,5,6,7, 8, \dots $$

Is this correct??

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by separation of variable, you verify that $$e^{\pm ky}\sin kx$$ which satisfies laplace equation and the boundary conditions $u(0,y)=u(\pi,y)$ for any positive integer $k.$ requiring that the solution be bounded for $y$ at $\infty$ gives you that $$e^{-ky}\sin kx$$ is a solution.

we can decompose $$\sin^3 x = \frac 12 \sin x (1-\cos 2x)=\frac12\sin x - \frac 14(\sin 3x - \sin x)=\frac 34 \sin x - \frac 14\sin 3x$$ by principle of superposition the solution is $$u = \frac34e^{-y}\sin x - \frac14e^{-3y}\sin 3x.$$

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  • $\begingroup$ Isn't it as follows?? $$u(x,0)=\sin x+\sin^3 x =\sin x +\frac{3}{4} \sin x -\frac{1}{4} \sin 3x \Rightarrow \dots \Rightarrow u=\frac{7}{4}e^{-y} \sin x-\frac{1}{4}e^{-3y} \sin 3x$$ @abel $\endgroup$ – Mary Star Apr 29 '15 at 16:46
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    $\begingroup$ @MaryStar, that is correct. $\endgroup$ – abel Apr 29 '15 at 16:55
  • $\begingroup$ Ok... Thank you very much!! @abel $\endgroup$ – Mary Star Apr 29 '15 at 18:25
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    $\begingroup$ @MaryStar, you are welcome. $\endgroup$ – abel Apr 29 '15 at 18:48
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BIt might be easier using k all throughout your solutions. For your solution to 1) you have to solve c2 Sin(kπ) = 0. This happens every n π therefore k= n for n = 1,2,... and then lambda = n^2

With solution 2 -> you've established that λ > 0 i.e.λ = k^2 therefore substituting k^2 in place of λ and then using the characteristic equation on part 2) you get λ^2 = k^2. Therefore λ = plus/minus k. The general solution here will be exponentials, not hyperbolic sines/cosines. This should make things easier.

--> Now you have both Solns:

1) X(x) = BnSin(nπx)

2)Y(y) = c1e^(ny) + c2e^(−ny)

If you were told to use Fourier to solve this, then I'm assuming you will probably be able to use an odd expansion of the fourier series at the end to find Kn. This is unlikely though and you will probably have to use the general eigenfunction expansion (the one with all the inner products) to find your Kn. This problem looks similar to the wave equation so it's probably best you revise any lecture notes on that topic from here on out

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  • $\begingroup$ I edited my initial post and now I use exponentials instead of hyperbolic sines/cosines. Could you take a look?? Also why do we have to use an odd expansion of the fourier series?? @user235242 $\endgroup$ – Mary Star Apr 28 '15 at 20:39
  • $\begingroup$ So, $a_n$ comes from the $X(x)$ problem, right?? I changed again my initial post. Could you take a look at it ?? @Maharero $\endgroup$ – Mary Star Apr 28 '15 at 23:48

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