1
$\begingroup$

I am trying to solve the following problem.

I have the two ODEs \begin{align} T'+\alpha\lambda T&=0 \\ \frac{d}{dr}\left(r\frac{dR}{dr}\right)+\lambda rR&=0, \end{align} with boundary conditions $R(b)=0, \ R'(0)=0$ and $T(0)=f(r)$. I am trying to solve the ODE for $T$ and hence write down the general solution of $u(r,t)=R(r)T(t)$ as an infinite series.

Transforming the ODE for $R$ into a Bessel equation of order $0$ when $\lambda=k^2>0$ and $p=kr$ gives the solution $$R_n(p)=A_nJ_0(p)+B_nY_0(p)\implies R_n(r)=A_nJ_0(kr)+B_nY_0(kr).$$ Solving the ODE for $T$ gives $$T(t)=Ce^{-\alpha k^2 t},$$ as the characteristic equation is $\beta+\alpha k^2=0\implies\beta=-\alpha k^2$. Does this then mean that the general solution for $u$ is,

\begin{align} u(r,t)&=\sum_{n=1}^{\infty}R_n(r)T_n(t) \\ u(r,t)&=\sum_{n=1}^{\infty}e^{-\alpha k^2 t}\left(A_nJ_0(kr)+B_nY_0(kr)\right)? \end{align} Is this methodology correct? Thank you kindly for you help.

$\endgroup$
  • $\begingroup$ @Mattos Does the $Y_0$ term disappear? If a boundary condition was that $R$ is finite as $r\rightarrow 0^+$ I would understand that is does vanish, but these boundary conditions don't seem to indicate this. $\endgroup$ – user654924 Mar 17 at 2:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy