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Consider the heat equation in polar coordinates, $$\frac{\partial u}{\partial t}=h^2\left(\frac{\partial^2 u}{\partial r^2}+\frac{2}{r}\frac{\partial u}{\partial r}\right), t>0, 0<r<R.$$ A sphere of radius $R$ is initially at constant temperature $u_0$ throughout, then has surface temperature $u_1$ for $t>0$. Find the temperature throughout the sphere for $t>0$ and in particular in the center $u_c$.

I can solve the heat equation when in cartesian form, however polar coordinates has always been my weakness and i have been stumped for a while, i have let $U(r,t)=F(r)G(t)$ then substituted into our original equation to get

$$G'(t)+(\lambda^2)G(t)=0 \text{ where } \lambda=kh$$

$$F"(r) + F'(r)/r + k^2F(r)=0$$

i have the initial condition that $U(r,0)=u_0$ and the boundary conditions that $u(R,t)=u_1$

however i do not know where to go from here, any help is appreciated.

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  • $\begingroup$ Welcome to MSE! Please see meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference for a quick overview of how to input math formulas. $\endgroup$
    – KittyL
    Dec 17, 2016 at 12:04
  • $\begingroup$ Hint: try letting $F=f/\sqrt{r}$. It will reduce the second ODE. $\endgroup$
    – KittyL
    Dec 17, 2016 at 12:38

1 Answer 1

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Hint: Let $u(r,t) = \frac{U(r,t)}{r}$. Then $$\frac{\partial u}{\partial t} =\frac{1}{r}\frac{\partial U}{\partial t} $$ $$\frac{\partial u}{\partial r} = \frac{1}{r}\frac{\partial U}{\partial r} - \frac{U}{r^2}$$ and $$\frac{\partial^2 u}{\partial r^2} = -\frac{1}{r^2}\frac{\partial U}{\partial r} + \frac{1}{r} \frac{\partial^2 U}{\partial r^2} + \frac{2 U}{r^3} -\frac{1}{r^2} \frac{\partial U}{\partial r} = \frac{1}{r} \frac{\partial^2 U}{\partial r^2} - \frac{2}{r^2} \frac{\partial U}{\partial r} + \frac{2 U}{r^3}$$ So the heat equation becomes

$$\begin{align}\frac{1}{r}\frac{\partial U}{\partial t} &= h^2\left(\frac{1}{r} \frac{\partial^2 U}{\partial r^2} - \frac{2}{r^2} \frac{\partial U}{\partial r} + \frac{2 U}{r^3} + \frac{2}{r}\left(\frac{1}{r}\frac{\partial U}{\partial r} - \frac{U}{r^2}\right)\right)\\& =h^2\left(\frac{1}{r} \frac{\partial^2 U}{\partial r^2} - \frac{2}{r^2} \frac{\partial U}{\partial r} + \frac{2 U}{r^3} + \frac{2}{r^2}\frac{\partial U}{\partial r} - \frac{2U}{r^3}\right)\\& = \frac{h^2}{r} \frac{\partial^2 U}{\partial r^2}\end{align}$$ which implies $$\frac{\partial U}{\partial t} = h^2 \frac{\partial U}{\partial r^2}$$ Do you think you take can it from here?

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  • $\begingroup$ Wow never even crossed my mind to do it like this, i reckon i can finish it of from here thank you $\endgroup$ Dec 17, 2016 at 15:36
  • $\begingroup$ Glad I could help. $\endgroup$ Dec 17, 2016 at 16:29

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