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I'd like to find all solutions of $u_t$ = $u_{xx}$ of the form

$$u = \left(\frac{1}{\sqrt{t}}\right)v\left(\frac{x}{2\sqrt{t}}\right).$$

I know that this problem reduces to solving a second order ODE for $v(y)$ with one solution $v(y) = e^{-y^2}$. Thus I must seek other linearly independent solutions of the form $v(y) = h(y)e^{-y^2}$.

All help and suggestions are appreciated


Update: Here is the derivation of the ODE, which was asked for in the comments. Suppose: \begin{align} y(x,t) &\equiv \frac{1}{2}xt^{-1/2}\\\ \\ u(x,y) &\equiv t^{-1/2}v(y)\\\ \\ \left. \frac{\partial u}{\partial t} \right|_x &= \left. \frac{\partial^2 u}{\partial x^2} \right|_t\tag{$\ast$}. \end{align} Then: \begin{align} \left. \frac{\partial u}{\partial t} \right|_x &= -\frac{1}{2}t^{-3/2}v(y) - \frac{1}{4}t^{-2}x\frac{dv}{dy}\\ \left. \frac{\partial^2 u}{\partial x^2} \right|_t &= \frac{1}{4}t^{-3/2} \frac{d^2v}{dy^2}. \end{align} So $(\ast)$ becomes: \begin{equation} \frac{1}{8}t^{-3/2} \frac{d^2v}{dy^2} + \frac{1}{8}t^{-2}x\frac{dv}{dy} + \frac{1}{4}t^{-3/2}v=0. \end{equation} Assuming $x\neq 0$, this implies: \begin{equation} \frac{d^2v}{dy^2} + 2y\frac{dv}{dy} + 2v=0,\tag{$\star$} \end{equation} which is the ODE referred to above.

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    $\begingroup$ What is that second order ODE? $\endgroup$ – Stefan Smith Apr 6 '13 at 1:21
  • $\begingroup$ @Stefan I added the ODE $(\star)$ to the question; my edit is sitting in the queue. rhm52: I hope you don't mind. $\endgroup$ – Douglas B. Staple Apr 6 '13 at 3:17
  • $\begingroup$ Thank you Douglas I should have included that with my original post. $\endgroup$ – rmh52 Apr 6 '13 at 3:23
  • $\begingroup$ Since it's a linear second order ODE, if you have one solution, you should try using variation of parameters to get the other. $\endgroup$ – Ray Yang Apr 6 '13 at 6:23
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    $\begingroup$ This is homework and I've actually solved it. I'll post the solution when I get the chance. $\endgroup$ – rmh52 Apr 6 '13 at 21:56
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If $v(y)=h(y)\mathrm e^{-y^2}$, then $v$ solves the ODE $v''+2yv'+2v=0$ if and only if $h$ solves the ODE $h''-2yh'=0$. Thus, $h'(y)=a\cdot\mathrm e^{y^2}$ for some $a$ and $v=v_{a,b}$ for some $(a,b)$, where $$ v_{a,b}(y)=\mathrm e^{-y^2}\left(a\int_0^y\mathrm e^{z^2}\mathrm dz+b\right). $$ Note that the functions $v_{1,0}$ and $v_{0,1}$ are linearly independent. Hence the set of solutions of the ODE $v''+2yv'+2v=0$ is indeed the two dimensional vector space $\{v_{a,b}\mid(a,b)\in\mathbb R^2\}$. The function $v$ in the question is $v_{0,1}$.

Proof: If $w(y)=\mathrm e^{-y^2}$ then $w'=-2yw$ and $w''=(4y^2-2)w$. Hence, if $v=hw$ then $v'=(h'-2yh)w$, $v''=(h''-4yh'-2h+4y^2h)w$ and $v''+2yv'+2v=(h''-2yh')w$. End of the proof.

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