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Consider $\frac{1}{r}\partial_{r}u(r,t)=u_{t}(r,t)$ with $u(0,t)=g(t),u(R,t)=f(t) $ and $u(r,0)=h(r)$.

Attempt

1)Power series

$$\partial_{r}u_{n}(r,t)\frac{r^{n-1}}{n!}+u_{n}(r,t)\frac{r^{n-2}}{(n-1)!}=\frac{r^{n}}{n!}\partial_{t}u_{n}(r,t)\Rightarrow$$

$$\partial_{r}u_{n}(r,t)\frac{r}{n}+u_{n}(r,t)=\frac{r^{2}}{n}\partial_{t}u_{n}(r,t)\Rightarrow $$

$$u_{n}(r,t)=ae^{n/r+\lambda( r^{2}+t)b}.$$

Same issue

2)Difference method doesn't work because for zero boundary conditions $\frac{1}{r}u_{r}=\lambda u\Rightarrow u=ae^{\lambda r^{2}/2+b}\equiv 0$.

3)setting u=vw s.t. $v(0,t)=1$ and $v(R,t)=2$ e.g. for

$$w=g(t)^{\frac{R-r}{R}}(f(t)/2)^{\frac{r}{R}}$$

but gives $\frac{1}{r}v_{r}+\frac{1}{r}wln(r/R)=v_{t}+w_{t}$.

Motivation

The $u(r,t):=\frac{1}{area(\partial B_{r}(0))}\int_{\partial B_{r}(0)}\phi(x,t)dx$ where $\Delta \phi=\phi_{t}$ satisfies the above equation.

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  • $\begingroup$ You wrote : "Consider $\frac{1}{r}\partial_{r}u(r,t)=u_{t}(r,t)$ with $u(0,t)=g(t),u(R,t)=f(t) $ and $u(r,0)=h(x)$." What is this $x$ which appears totally unexpected ? $\endgroup$
    – JJacquelin
    Feb 25, 2017 at 21:31

1 Answer 1

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$$\frac{1}{r}\partial_{r}u(r,t)=\partial_{t}u(r,t)$$ The calculus below concerns continuous and differentiable functions. Any method to solve this PDE (especially the method of characteristics) leads to the general solution : $$u(r,t)=F\left( t+\frac{r^2}{2}\right)$$ where $F$ is any differentiable function.

The function $F$ has to be determined according to some boundary and/or initial condition(s).

If those conditions are sufficient, $F$ can be determined. If those conditions are overabundant and not consistent one to another, there is no solution.

The specified conditions $u(0,t)=g(t),\quad u(R,t)=f(t) $ and $u(r,0)=h(r)$ involve three given functions : $g,f,h$ which is too much for a first order PDE. The conditions are overabundant. So the question is : are they consistent one to another ?

Consider the first condition $\quad u(0,t)=g(t) \quad\to\quad F(t-0)=F(t)=g(t)\quad$ determines the function $F=g$. So the solution of the PDE according to the first condition is : $$u(r,t)=g\left( t+\frac{r^2}{2}\right)$$

Then, let apply the second condition $\quad u(R,t)=f(t)=g\left( t+\frac{R^2}{2}\right)$

So, they are two cases : if the given functions $g$ and $f$ are not related according to the above relationship, the problem (given PDE and given conditions) has no solution. If the given functions $g$ and $f$ are related according to the above relationship, the problem (given PDE and given conditions) has a solution : $$u(r,t)=g\left( t+\frac{r^2}{2}\right)=f\left( t+\frac{r^2}{2}-\frac{R^2}{2}\right)$$

Then, let apply the third condition $\quad u(r,0)=h(r)=g\left(\frac{r^2}{2}\right)$

Again, they are two cases : if the given functions $g$ and $h$ are not related according to the above relationship, the problem (given PDE and given conditions) has no solution. If the given functions $g$ and $h$ are related according to the above relationship, the problem (given PDE and given conditions) has a solution : $$u(r,t)=g\left( t+\frac{r^2}{2}\right)=f\left( t+\frac{r^2}{2}-\frac{R^2}{2}\right)=h\left(\sqrt{2t+r^2}\right)$$

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