How to prove that solution to ODE in spherical coordinate is equivalent to the ODE in cartesian coordinates if it is a thin shell

Question

Solving a diffusion-type ODE across a spherical shell, the equation is: $$\frac{d}{dr}\left(r^2\frac{df}{dr}\right)=0\tag{1}$$ with boundary conditions $f(r_1)=f_1$ and $f(r_2)=f_2$.

The solution is: $$f(r)=\frac{f_1-f_2}{\frac{r}{r_1}-\frac{r}{r_2}}+\frac{f_1r_1-f_2r_2}{r_1-r_2}\tag{2}$$ Now I know intuitively that if the overall thickness of the shell $\Delta r=r_2-r_1\ll r_1$, then this can be reduced to Cartesian coordinates. In that case the ODE is: $$\frac{d^2f}{dr^2}=0\tag{3}$$ With the same boundary conditions, then the solution is: $$f(r)=f_1+(f_2-f_1)\frac{r-r_1}{r_2-r_1}\tag{4}$$ All fine so far. What I want to do though is prove that Equation 2 reduces to Equation 4 when $\Delta r\ll r_1$, but I can't seem to figure out how it should be done.

My approach has been to substitute $r_1 + \Delta r$ for $r_2$ and substitute $r_1+\xi$ for $r$, and then use the geometric series $\frac{1}{1-x}\approx 1+x$ (for $x\ll 1$) type of substitution to get $r$ up into the numerator, but the terms don't cancel and I can't get it to work out.

Is there another approach I should be using? Some trick or technique I'm not remembering?

Answer 1

Your idea is correct. As the thickness of the shell diminishes -- the effect of the curvature of the domain becomes negligible. The solution in rectangular coordinates becomes a close approximation to the curvilinear solution.

As you suggested, let $\Delta = r_2 - r_1$ and $\xi = r - r_1$.

The spherical solution is

$$f(r) = \frac{f_2r_2-f_1r_1}{r_2-r_1} + \frac{f_1-f_2}{r(1/r_1 - 1/r_2)}=\frac{f_2(r_1 + \Delta)-f_1r_1}{\Delta} + \frac{f_1-f_2}{\Delta}\frac{r_1(r_1+ \Delta)}{r}\\=f_2(1 + \frac{r_1}{\Delta})-f_1\frac{r_1}{\Delta} + \frac{f_1-f_2}{\Delta}\frac{(r_1+ \Delta)}{1+\frac{\xi}{r_1}}.$$

With $\frac{\xi}{r_1} \ll 1$, we have

$$\frac1{{1+\frac{\xi}{r_1}}} \approx 1 - \frac{\xi}{r_1}.$$

Then

$$f(r) \approx f_2 +\frac{f_2r_1}{\Delta}-\frac{f_1r_1}{\Delta}+ \frac{f_1-f_2}{\Delta}(r_1 + \Delta) + \frac{f_2-f_1}{\Delta}(r_1+\Delta)\frac{\xi}{r_1} \\ = f_1 + \frac{f_2-f_1}{\Delta}\left(1+\frac{\Delta}{r_1}\right)\xi$$

Finally, with $\frac{\Delta}{r_1} \ll 1$ we have

$$f(r) \approx f_1 + \frac{f_2-f_1}{\Delta}\xi= f_1 + \frac{f_2-f_1}{r_2-r_1}(r-r_1)$$

Answer 2

I did the change of variables on the original ODE to see if RRL was right in saying that it would also lead to transforming the spherical ODE to the Cartesian one.

Original ODE: $$\frac{d}{dr}\left(r^2\frac{df}{dr}\right)=0$$ and B.C.'s: $f(r_1)=f_1$ and $f(r_2)=f_2$. I did a change of variables $\xi=\frac{r-r_1}{\Delta r}$, where $\Delta r=r_2-r_1$. Substituting this in we get: $$\frac{1}{\Delta r}\frac{1}{d\xi}\left( \frac{\left( r_1+\Delta r\xi\right)^2 }{\Delta r} \frac{df}{d\xi} \right)=0$$ And boundary conditions change to $f(0)=f_1$ and $f(1)=f_2$. Expanding out the derivative and the squared term: $$\frac{1}{\Delta r^2}\left( {r_1}^2+2r_1 \Delta r \xi +\Delta r^2 \xi^2 \right)\frac{d^2f}{d\xi^2}+\left( 2\frac{r_1}{\Delta r}+2\xi \right)\frac{df}{d\xi}=0$$ Multiplying the entire equation by $\left(\frac{\Delta r}{r_1}\right)^2$, then performing one more substitution $\varepsilon = \frac{\Delta r}{r_1}$: $$\frac{d^2 f}{d\xi^2}+2\varepsilon \left( \xi \frac{d^2 f}{d\xi^2} +\frac{df}{d\xi} \right)+\varepsilon ^2 \left( \xi^2 \frac{d^2 f}{d\xi^2}+\xi\frac{df}{d\xi}\right)=0$$ So since $\varepsilon\ll 1$, this essentially becomes a regular perturbation problem. Take only the $\mathcal{O}(1)$ term and you get the Cartesian equation, take all the terms and you get the full spherical coordinate equation. I was kind of hoping that if you took only the $\mathcal{O}(1)$ and $\mathcal{O}(\varepsilon)$ terms that it would then be equivalent to cylindrical coordinates, but it doesn't quite work out: the same expansion done from the cylindrical equation is $$\frac{d^2 f}{d\xi^2}+\varepsilon \left( \xi \frac{d^2 f}{d\xi^2} +\frac{df}{d\xi} \right)=0$$ so it's off by a factor of $2$ in the $\mathcal{O}(\varepsilon)$ term.