1
$\begingroup$

Suppose we have an infinite fluid surrounding a ball. I want to solve the following PDE, which describes the heat diffusion in the fluid (simplifications have been made): $$\partial_t\theta=\alpha\Delta\theta$$ with boundary conditions $\theta(r=R,t)=0$ and $\lim_{r\to\infty}\theta(r,t)=0$, and initial condition $\theta(r,t=0)=f(r)$. Using separation of variables $\theta(r,t)=F(r)G(t)$. So for $G$ I obtain the ODE $$ G'+\alpha\lambda^2 G=0 $$ which integrates into $G(t)=A\exp\left(-\alpha\lambda^2 t\right)$, where $-\lambda^2$ is the separation constant. Now for $F(r)$ I obtain the ODE $$ F''+\frac{2}{r}F'+\lambda^2 F=0 $$ which integrates into $F(r)=B\frac{\sin(\lambda r)}{r}+C\frac{\cos(\lambda r)}{r}$. So the second boundary condition is automatically satisfied, and the first one yields $$ B\sin(\lambda R)+C\cos(\lambda R)=0. $$ But I don't know how to go further. If I put $B=C=0$, then I obtain the trivial solution, which isn't feasible. I could put $B=0$ and $C\neq 0$, or $B\neq0$ and $C=0$, but then the corresponding values of $\lambda$ would yield two different solutions. Finally if I keep both $B\neq 0$ and $C\neq 0$ then I obtain $D\tan(\lambda R)=1$ where $D=-\frac{C}{B}$, from where I cannot go further.

Thanks for any help, or any textbook that deals with that kind of problem.

$\endgroup$
  • $\begingroup$ Are you sure you want $r\to \infty$ and not $t\to\infty$? Moreover, are you assuming radial symmetry to get to such simplification? $\endgroup$ – b00n heT Feb 2 '17 at 11:01
  • $\begingroup$ @b00nheT: Yes I want $r\to\infty$ and I can assume radial symmetry, essentially because the boundary/initial conditions don't make use of the other variables $\theta$ and $\varphi$. $\endgroup$ – xuanphong Feb 2 '17 at 11:12
1
$\begingroup$

The $r$ solution would be $$ \frac{\sin(\lambda(r-R))}{r}=\cos(\lambda R)\frac{\sin(\lambda r)}{r}-\sin(\lambda R)\frac{\cos(\lambda r)}{r}. $$ So the heat equation solution is $$ \int_{0}^{\infty}e^{-\alpha\lambda^2 t}c(\lambda)\frac{\sin(\lambda(r-R))}{r}d\lambda, $$ where $c(\lambda)$ is a coefficient function determined by the initial condition $$ rf(r)=\int_{0}^{\infty}c(\lambda)\sin(\lambda(r-R))d\lambda \\ (r+R)f(r+R) = \int_{0}^{\infty}c(\lambda)\sin(\lambda r)d\lambda. $$ The coefficient function is then determined by the inverse Fourier sine transform: $$ c(\lambda) = \frac{2}{\pi}\int_{0}^{\infty}(s+R)f(s+R)\sin(\lambda s)ds $$

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.