Author Topic: Halloween Challenge 7  (Read 167 times)

Victor Ivrii

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XinYu Zheng

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Re: Halloween Challenge 7
« Reply #1 on: October 30, 2016, 11:47:02 PM »
Making Fourier Transform $u(x,y)\to \hat{u}(k,y)$ we can rewrite the ODE as
$$\begin{cases}
-k^2\hat{u}+\hat{u}_{yy}=0\\
(\hat{u}_y+\alpha\hat{u})_{y=0}=\hat{f}(k)
\end{cases}$$
This ODE has solution $\hat{u}(k,y)=A(k)e^{-|k|y}+B(k)e^{|k|y}$. We will discard the second term because it goes unbounded as $y\to\infty$. Now using the boundary condition we have
$$(\hat{u}_y+\alpha\hat{u})_{y=0}=\hat{f}(k)\implies (-|k|+\alpha)A(k)=\hat{f}(k)\implies A(k)=\frac{\hat{f}(k)}{\alpha-|k|}$$
Now, if $\alpha<0$, then the RHS function has no singularities and we are good. To write the solution as fourier integral, we want IFT of $e^{-|k|y}/(\alpha-|k|)$, which we cannot find in closed form. So for now we can just write
$$u(x,y)=\frac{1}{2\pi}\int_{-\infty}^\infty f(x-z) \left(\int_{-\infty}^\infty \frac{e^{-|k|y+ikz}}{\alpha-|k|}\,dk\right)\,dz$$
If $\alpha>0$, then we have a problem as $\frac{\hat{f}(k)}{\alpha-|k|}$ could be singular. In this case, we need to require $\hat{f}(\alpha)=0$, which means
$$\int_{-\infty}^\infty f(x)e^{-i\alpha x}\,dx=0$$
This is a necessary condition, but not sufficient. $\hat{f}(k)$ needs to approach 0 faster than $\alpha-|k|$ does as $|k|\to\alpha$. Otherwise, we may still have problems with convergence. Assuming $\hat{f}(k)$ vanishes quickly enough at $\alpha$, the solution is given by the same equation.

Victor Ivrii

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Re: Halloween Challenge 7
« Reply #2 on: October 31, 2016, 09:14:13 AM »
 :D