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Let's have an equation

$$ u_{t} - (xu)_{x} - \frac{1}{2}u_{xx} = 0, \quad u(x, 0) = g(x), \quad -\infty < x < \infty , \quad 0 < t < \infty . $$ I need to find a Green function for it. So, $$ u_{t} - (xu)_{x} - \frac{1}{2}u_{xx} = \delta (x - x_{0})\delta (t - t_{0}). $$ I tried to use Laplace transform for it, but then I got $$ su(x, s) - g(x) - u(x, s) - xu_{x}(x, s) - \frac{1}{2}u_{xx}(x, s) = \delta (x - x_{0})e^{-t_{0}s}. $$

I don't know what to do with $xu_{x}(x, s)$. Can you help me with this problem or with method?

Maybe, some substitution like $u(x, p) = v(x, p)w(x)$ can help?

doraemonpaul
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John Taylor
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1 Answers1

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The ordinary differential equation in $x$ that you have is a form of the Hermite ODE with an inhomogeneous term, so you can solve it using the standard theory of inhomogeneous second-order linear ODEs.

Note, by the way, that if you do a Fourier transform over $x$, the term $xu_x$ transforms into $$(ik)(i\partial_k) u_k(t),$$ so a Fourier transform will turn the second-order PDE w.r.t. $(x,t)$ into a first-order linear PDE w.r.t. $(k,t)$. You can then use, for example, the method of characteristics to solve it. This might be easier than using the Laplace transform.

Kirill
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  • Thank you! So, after using Fourier transform I got $$ u_{t}(k, t) + ku_{k}(k, t) - u(k, t) + \frac{k^{2}}{2}u(k, t) = \frac{1}{2 \pi}\delta (t - t_{0}). $$ After using characteristics method I got $$ \frac{dt}{1} = \frac{dk}{k} = \frac{du}{u\left(1 - \frac{k^{2}}{2}\right) + \frac{1}{2 \pi}\delta (t - t_{0})}. $$ First of all there will be hard to find system of integrals of this equation. So, maybe, there will be better to use Fourier transform over x, t? – John Taylor May 25 '13 at 23:07
  • "...Note, by the way, that if you do a Fourier transform over x..."

    How to prove it?

     – John Taylor May 25 '13 at 23:07
  • Integration by parts changes powers of $x$ into derivatives w.r.t. $k$, so long as the function decays quickly at infinity. – Kirill May 25 '13 at 23:12
  • Thank you! Method of characteristics gives $$ke^{-t} = C_{1}, \quad ue^{\frac{C_{1}^{2}}{4}e^{2t} - t} - \frac{h(t)}{\sqrt{2 \pi}}e^{\frac{C_{1}^{2}}{4}}, $$ and initial condition gives $$ ue^{\frac{C_{1}^{2}}{4}e^{2t} - t} - \frac{h(t)}{\sqrt{2 \pi}}e^{\frac{C_{1}^{2}}{4}} = 0 \Rightarrow u = \frac{h(t)e^{t}}{\sqrt{2 \pi}}e^{-\frac{k^{2}}{4}(1 - e^{-2t})}. $$ It is easy to take inverse Fourier transform. – John Taylor May 26 '13 at 11:12
  • "...Note, by the way, that if you do a Fourier transform over x..." Maybe, to $-\partial_{k}(k u)$? – John Taylor May 26 '13 at 11:43