1. ## Boundary value problem

I have the following problems to solve:
The temperature u(r) in the circular ring shown in the image is determined from the boundary-value problem
where u0 and u1 are constants. Show that Not sure where to start. I need to show and explain every step.
#2 is similar but is sphere with equation of
r(d^2u/dr^2) + 2(du/dr) = 0
u0 and u1 are constants

I know the answer is u(r) = ((u0-u1)/(b-a))*ab/r +(u1b-u0a)/(b-a)
not sure how to get here though.

Any help to start would be great.
Thanks

2. ## Re: Boundary value problem

Substitute $\displaystyle \frac{du}{dr}=p,$ (say). That gets you a first order separable.

Post again if you need further help.

3. ## Re: Boundary value problem

I got that far to get rp'' + p' =0
my issues is with how the variables a, b, u0, u1 get brought in to the equation.

Thanks

4. ## Re: Boundary value problem

If $\displaystyle \frac{du}{dr}=p,$ then $\displaystyle \frac{d^{2}u}{dr^{2}}=\frac{dp}{dr}$ and your equation becomes

$\displaystyle r\frac{dp}{dr}+p=0,$

which is of the first order and is separable.

Solve this for $\displaystyle p$ in terms of $\displaystyle r$ and substitute into your original substitution. That gets you an equation that you can integrate.

It contains two arbitrary constants (of integration) which you evaluate by substituting in the boundary conditions.

Post again if you need further help.

5. ## Re: Boundary value problem

Thank you for your help. Hope you are having a good holiday. Seperables was one of my not so good topics so let me know if this is correct:

r dp/dr+p = 0 separates out to 1/pdp = -1/rdr. then integrate each side and end up with p = e^-lnr+c.

is this right so far? now I take this and sub it in to r dp/dr+p = 0 to get r dp/dr+e^-lnr+c = 0.

now what as I'm left with only r's and a dp?

thanks again

6. ## Re: Boundary value problem

Go back a step and tidy up the RHS.

Remember that

$\displaystyle -\ln(M)=\ln(1/M),$ and that $\displaystyle e^{ln(N)}=N.$

Also, e raised to the power of a constant is just a constant.