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Question

Find the solution for $u_t = iu_{xx}$, for $ x \in \mathbb R$ and for $t \gt 0$, with the initial condition $u(x,0) = exp(-x^2)$.

Solution

For the assignment I get to use any transformation method (Laplace, Fourier and possibly separation of variables). I've seen other people solving similar problems with separation but I want to try to solve it with all methods and I am currently stuck trying to find a particular solution for the ODE after using Laplace transformation. Any help is appreciated! Thanks!

$$\mathcal L \{u(x,t) \} = U(x,s) = \int_0^\infty u(x,t)e^{-st}dt,$$ $$\mathcal L\{u_t(x,t)\} = sU(x,s) - u(x,0),$$ $$\mathcal L \{u_{xx}(x,t) \} = \frac{d^2}{dx^2}U(x,s) = U''(x,s).$$ Inserting into the original equation yields

$$ sU(x,s) - u(x,0) - iU''(x,s) = 0,$$ $$ sU(x,s) - iU''(x,s) = exp(-x^2).$$

Homogenous solution

For the homogenous solution I get $$U_h(x,s) = k_1e^{\sqrt{(s/i)}x}+k_2e^{-\sqrt{(s/i)}x}$$

Particular solution

Using variation of parameters I get that the Wronskian should be $W = -2\sqrt{\frac{s}{i}}$ and

$$ c_1(s) = -\frac{1}{2}\int_0^\infty \frac{e^{-\sqrt{(s/i)}x-x^2}}{\sqrt{\frac{s}{i}}} ds = \frac{ie^{-x^2}}{x}e^{-x\sqrt{-is}}, $$ $$c_2(s) = \frac{1}{2}\int_0^\infty \frac{{e^{\sqrt{(s/i)}x-x^2}}}{\sqrt{\frac{s}{i}}} ds = \frac{ie^{-x^2}}{x}e^{x\sqrt{-is}}.$$ According to WolframAlpha the particular solution can be written as $U_p(x,s) = \frac{2ie^{-x^2}}{x}$, but I don't know how. Inserting into $U(x,s)$ yields $$U(x,s) = k_1e^{\sqrt{(s/i)}x}+k_2e^{-\sqrt{(s/i)}x}+\frac{2ie^{-x^2}}{x}.$$ Inserting the solution into $sU-iU''= e^{-x^2}$ I get $$s\frac{2i}{x}+8x+\frac{4}{x}+\frac{4}{x^3}=1$$ so it doesn't add up. Is there some way to simplify the problem or the expressions?

hehan123
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    Presumably you applied the Laplace transform to $t\mapsto u_t(x,t)$ and to $t\mapsto u_{xx}(x,t)$ keeping $x$ fixed ? If so I do not see how you get this ODE for $U(s)$. Please elaborate. You may also want to look at this. – Kurt G. Oct 20 '22 at 15:15
  • Thank you for your reply. I made some rushed mistakes and edited the solution. Let me know if you want me to elaborate more. – hehan123 Oct 20 '22 at 20:03
  • I looked at the post you linked and the answer, but I don't quite follow. – hehan123 Oct 20 '22 at 20:15
  • The ODE now looks correct to me. To find the particular solution you can use this method. – Kurt G. Oct 21 '22 at 04:49

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