Author Topic: Lecture 21  (Read 2645 times)

Miranda Jarvis

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Lecture 21
« on: December 19, 2012, 01:23:30 PM »
I think that where you state the Schrödinger equation in the beginning of lecture 21 'consider problem: . . .' it is supposed to be $u_t=iku_{xx}$ what is posted now doesn't have the $i$ and makes no sense with the subsequent solution . . .

Victor Ivrii

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Re: Lecture 21
« Reply #1 on: December 19, 2012, 02:16:00 PM »
I think that where you state the Schrödinger equation in the beginning of lecture 21 'consider problem: . . .' it is supposed to be $u_t=iku_{xx}$ what is posted now doesn't have the $i$ and makes no sense with the subsequent solution . . .

Yes, fixed.

Miranda Jarvis

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Re: Lecture 21
« Reply #2 on: December 19, 2012, 02:34:26 PM »
for the Laplace equation in half-strip where it says "with g(x)=0 we could reduce it by the method of continuation to the problem in the whole strip and solve it by Fourier transform" would that be using a Fourier transform with respect to y this time . . . if yes then would the fact that we would be using an odd continuation influence the answer compared to if we had $u_y|_{y=0}=0$ I don't see how this would influence the solution . . .

Victor Ivrii

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Re: Lecture 21
« Reply #3 on: December 19, 2012, 03:08:56 PM »
for the Laplace equation in half-strip where it says "with g(x)=0 we could reduce it by the method of continuation to the problem in the whole strip and solve it by Fourier transform" would that be using a Fourier transform with respect to y this time . . . if yes then would the fact that we would be using an odd continuation influence the answer compared to if we had $u_y|_{y=0}=0$ I don't see how this would influence the solution . . .

Yes, for $u|_{y=0}=0$ we use an odd continuation and for $u_y|_{y=0}=0$ an even continuation. F.e. consider
\begin{align*}
& u_{xx}+u_{yy}=0,\qquad 0<x<1, y>0\\
& u|_{x=0}=0,\\
& u|_{x=1}=e^{-y},\\
&u|_{y=0}=0.
\end{align*}
Then taking an odd continuation $\Phi(y)$ of $\phi(y)=e^{-y}$ which will be $\pm e^{-|y|}$ as $\pm y>0$, we find its FT $ \hat{\Phi}(\eta)=\frac{i\eta}{\pi(1+\eta^2)}$. If instead we had $u_y|_{y=0}=0$ we would take an even continuation and FT of $\Phi$ would be $ \hat{\Phi}(\eta)=\frac{1}{\pi(1+\eta^2)} $.

The rest of solution would be the same (but with different $\hat{\Phi}(\eta)$ !):
\begin{gather*}
\hat{u}_{xx}-\eta^2 \hat{u}=0 \implies \hat{u}= A(\eta)\cosh (x\eta) + B(\eta)\sinh(x\eta),\\
\hat{u}|_{x=0}=0\implies A(\eta)=0 ,\qquad \hat{u}|_{x=1}=\hat{\Phi}(\eta)\implies B(\eta)= \hat{\Phi}(\eta)/\sinh (\eta),\\
\hat{u}= \hat{\Phi}(\eta)\sin(x\eta)/\sinh (\eta)\implies \\
u(x,y)= \int_{-\infty}^\infty \frac{\hat{\Phi}(\eta)\sin(x\eta)}{\sinh (\eta)} e^{iy\eta}\,d\eta.
\end{gather*}

« Last Edit: December 19, 2012, 05:20:24 PM by Victor Ivrii »

Miranda Jarvis

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Re: Lecture 21
« Reply #4 on: December 19, 2012, 04:06:17 PM »
thank you for the clarification