Consider the PDE $$\frac{\partial u}{\partial t}=\frac{D}{r}\frac{\partial ^2}{\partial r^2}(ru),$$ where $D>0$ and we have the auxiliary conditions \begin{align*} u(r,0)&=0, \\ u(a,0)&=u_0.\end{align*}

I want to solve this problem using Laplace transforms. Defining the Laplace transform of a function $f:(0,\infty) \to \mathbb{R}$ as $$F(p)=\int_0^{\infty} e^{-pt}f(t) \, dt, $$ it follows that the PDE becomes $$pU(r,p)=\frac{D}{r}\frac{\partial ^2}{\partial r^2}(rU(r,p)).$$

After applying the boundary condition and requiring that the Laplace transform be bounded, I find that $$U(r,p)=\frac{au_0}{pr}e^{(a-r)\sqrt{p/D}}.$$

However, I'm told that the correct answer should be $$U(r,p)=\frac{au_0}{pr}\frac{\sinh\left(r\sqrt{\frac{p}{D}}\right)}{\sinh\left(a\sqrt{\frac{p}{D}}\right)}.$$

Are these answers equivalent or have I made an error?

  • $\begingroup$ Where does $k$ come from? $\endgroup$ – Kyle Mar 31 '18 at 15:57
  • $\begingroup$ @Kyle Sorry, it was supposed to be $D$, not $k$. I've changed this now. $\endgroup$ – Si.0788 Mar 31 '18 at 16:29

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