# Solution to diffusion equation in 1d spherical polar coordinates

Ok,

I have been given the steady state diffusion equation in 1d spherical polar coordinates as;

D.1/(r^2).'partial'd/dr(r^2.'partial'dc/dr)=0

I know that the solution comes in the form c(r) = A+B/r where A and B are some constants. I just dont know how to get from here to there. I have tried doing differentiation by parts on the equation then integrating the result, with no success. I can form a second order differential equation of the form;

r^2.'partial'd2c/dr2 +2r'partial'dc/dr = 0

but again dont know where to go from here. Any help greatly appreciated!

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tiny-tim
Homework Helper
D.1/(r^2).'partial'd/dr(r^2.'partial'dc/dr)=0
Hi captainst1985! Welcome to PF!

(btw, if you type alt-d, it prints ∂)

I don't understand what the D is at the beginning of the line.

If I ignore that, the equation says ∂/∂r(r^2.∂c/∂r) = 0;

so r^2.∂c/∂r must be independent of r (-B, say);

so ∂c/∂r = -B/r^2;

so c = A + B/r.

Hi,

Can you explain why one would say that:

so r^2.∂c/∂r must be independent of r (-B, say)?

Why would you equate r^2.∂c/∂r to be -B? What is the basis for this and/or the technique that does this called? Ive gotten a little forgetful on some of these techniques...

The OP also had the multiplier on the front end that is effectively:

D/r^2

How does that come into play when solving?

Thanks!

tiny-tim
Homework Helper
Welcome to PF!

Hi JHZR2! Welcome to PF!
Hi,

Can you explain why one would say that:

so r^2.∂c/∂r must be independent of r (-B, say)?
Because ∂/∂r of that is 0, ie (in words) the derivative of that with respect to r is 0, so changing r doesn't change it, ie it must be independent of r.
The OP also had the multiplier on the front end that is effectively:

D/r^2

How does that come into play when solving?
Because if 1/r2 times something is 0, then the something must also be 0, so we can ignore the 1/r2 !

(The D probably just indicates that it's the fourth exercise in the homework … A. B. C. D. …)

(btw, typing "alt-d" for "∂" only works on a Mac )