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I encountered this problem when trying to solve a linear PDE, of the form $$ \frac{\partial f}{\partial t}= F\left(f,\frac{\partial f}{\partial r},\frac{\partial^2 f}{\partial r^2}, \cdots\right), $$ for $0\leq r\leq 1$ and $t\geq0$. The initial condition is $f(t=0,r)=0$, and the boundary condition is $f(t,r=1)=0$ and $$ \lim_{r\to0}\left(\frac{\partial f}{\partial r}-\frac{f}{r}\right)=0 $$ The last equation implies $\lim_{r\to0}f(t,r)=c(t)r$. The quantity I'm interested in is $\partial f/\partial r$ at $r=0$ for all $t$, which is essentially $c(t)$ here.

While I could solve this numerically, I'm wondering if it is actually allowed to expand the PDE near small $r$ and then solve the resulting ODE. More specifically, due to the boundary conditions, we know that $f(t,r)=c(t)r$ near $r=0$. Would it then make sense to substitute this into the PDE, transform it into $$ \dot c=\lim_{r\to 0}F(cr,c,0,\cdots) $$ and look for $c(t)$ only? I'm wondering would the boundary condition at $r=1$ have any influence if I actually do this.

I also wonder if it works, would it be extendable to a set of PDEs? e.g. $$ \frac{\partial f}{\partial t}= F\left(f,\frac{\partial f}{\partial r},\frac{\partial^2 f}{\partial r^2}, \cdots,g,\frac{\partial g}{\partial r},\frac{\partial^2 g}{\partial r^2}, \cdots\right),\\ \frac{\partial g}{\partial t}= G\left(f,\frac{\partial f}{\partial r},\frac{\partial^2 f}{\partial r^2}, \cdots,g,\frac{\partial g}{\partial r},\frac{\partial^2 g}{\partial r^2}, \cdots\right), $$ where $F$ and $G$ are nonlinear functions. If I could determine the behaviors of $f$ and $g$ at $r=0$, can I just substitute these ``local'' forms into the PDEs if I'm only interested in $f$ and $g$ at $r=0$?

H. Zhou
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1 Answers1

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It is more natural to use the boundary condition $f(t,0) = 0 \, \forall t > 0$, since then $\partial_r f(t,0) = \lim_{r \to 0} f/r$ follows.

More seriously, however, it is in general not correct to assume that $\partial_r^2 f \vert_{r = 0} = 0$. Note that this would imply for $r = 0$ and all $t$ $$ 0 = \partial_t f = F(0,c(t),0) \, , \quad \text{not} \quad \dot c = F(0,c(t),0) $$ This might give a contradiction, e.g. if $F(0,c,0) = 1$ for all $c$, or it might mean nothing at all if $F(0,c,0) = 0$ for all $c$, or it might imply that $c(t)$ is constant.

So this approach quickly leads to a dead end.

To understand the problem better, it is important to know the type of the pde, its order (highest $r$ deriavtive), and so on.

Hans Engler
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