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What does the solution $u(x,t)$ of $u_t=u_{xx}+\frac{1}{x}u_x$ on $[0,1]$ with the following initial condition look like? $u(\frac{1}{2},0)=\delta(\frac{1}{2})$ (i.e. delta function at $x=\frac{1}{2}$)

I tried the separation of variables method as follows but I couldn't derive the precise solution:

Suppose $u(x,t)=X(x)T(t)$, then $\frac{\dot{T}}{T}=c$ and $\frac{X''}{X}+\frac{X'}{xX}=c$. This implies $T(t)=e^{ct}$ and the ODE $xX''+X'-cxX=0$. I guess the solution of the ODE looks like $p(x)e^{\alpha x}$ for some polynomial $p(x)=a_0+a_1x+\ldots$ and some constant $\alpha$. I plugged it into the ODE and I got $(m+1)^2a_{m+1}+(2m+1)\alpha a_m+(\alpha^2-c)a_{m-1}=0$ for any $m=1,2,\ldots.$ I think solutions like $p(x)e^{\alpha x+ct}$ with the above relation for the coefficients of $p$ could form a basis for the space of solutions. But I don't know how to use the initial condition to find the coefficient!

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That ODE has Bessel functions as solutions. Also to have uniqueness some boundary conditions should be given. – Andrew Feb 5 '13 at 17:38
Thanks for your comment. We can assume Neumann boundary conditions. – Soli Feb 5 '13 at 19:36
An homogeneous Neumann condition on the left end $u_x(0,t)=0$ leads to a smooth solution. In the inhomogeneous case existence of a (at least smooth near the boundary) solution doesn't seems to me so evident. – Andrew Feb 5 '13 at 21:20
It might be useful to rewrite the PDE as $xu_t=(xu_x)_x$. – Joonas Ilmavirta Jan 27 '15 at 13:41

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