# Help needed to solve a differential equation

## Homework Statement

solve for u:

$$u_t=G + \mu(u_{rr}+\frac{1}{r}u_r)$$

with boundary conditions u=0 on r=a and $$u_r=0$$ on r=0

where G is a constant, u is a function of r only and u_r is the derivative of u with respect to r etc

## Homework Equations

the solution is:

$$\frac{G_0 a^2}{4\mu}(1-\frac{r^2}{a^2}$$

## The Attempt at a Solution

u is independent of t so $$u_t=0$$.
it is an inhomogeneous differential equation so i thought youd solve $$(u_rr+\frac{1}{r}u_r)=0$$ first.
I then thought ud let $$p=u_r$$ to get $$p_r+\frac{1}{r}p$$ and then use separation of variables integral thing to get p = r+c where c is a constant. (initially i got logs but i took the exponential).
then i converted back to u : $$u_r=r+c$$ to get $$u=\frac{r^2}{2}+rc+d$$. This looks wrong and i have no idea how to introduce the \frac{G}{/mu} term.
any help will be much appreciated!

HallsofIvy
Homework Helper
Part of your difficulty is that you lost a sign and another is that you took the exponential incorrectly. Yes, the differential equation reduces to the first order equation p'+ p/r= 0 or p'= -p/r, so that dp/p= -dr/r and, integrating, ln(p)= -ln(r)+ C. But taking the exponential gives $p(r)= cr^{-1}= c/r$, where c= $e^C$, not r+ c.

Then u'= c/r so $u''= cln(r)+ d$. That is essentially saying that two independent solutions to the equation are ln(r) and 1.

To find a solution to the entire equation, use "variation of parameters". Look for a solution of the form $u(r)= v(r)ln(r)+ w(r)$ for two unknown function v(r) and w(r).

Differentiating, $u'= v'ln(r)+ v/r + w'$. There are, in fact, many solutions of that form, and we need only one, so to simplify the problem we will look only for solutions such that $v'ln(r)+ w'= 0$. That means that $u'= v/r$ and so $u''= v'/r- v/r^2$. Putting those into the original euation,
$$u''+ (1/r)u'= v'/r- v/r^2+ (1/r)(v/r)= v'/r= -G$$

So v'= -Gr and $v(x)= -(1/2)Gr^2$. From v'ln(r)+ w'= 0 and v'= -Gr, we have w'= Grln(r). That can be integrated by parts: let u= ln(r) and dv= r dr so that du= (1/r)dr and v= (1/2)r^2.

The terms involving ln(r) will turn out to have coefficient 0 in order that u(0) exist.

Thanks!

There are, in fact, many solutions of that form, and we need only one, so to simplify the problem we will look only for solutions such that $v'ln(r)+ w'= 0$.

how do you know this if you dont know the solution already?

ive got $$v=-\frac{1}{2}Gr^2+d$$ and $$w=Inr\frac{1}{2}r^2-\frac{Gr^2}{4}+f$$

giving $$u=-(\frac{1}{2}Gr^2+d) ln r +GInr\frac{1}{2}r^2-\frac{Gr^2}{4}+f$$

from the u(0) bc d=0 and from the (the other two In cancels) from the u_r(a) bc the f vanishes and so i do not know how to proceed. Also i think i have a sign error perhaps...

thanks :)

also when im doing the integration by parts should i be leaving it as indefinite integrals or doing it between 0 and a?
thanks