Help needed to solve a differential equation

[lavster]

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]

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.

 

[lavster]

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 don't 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 :)

 

[lavster]

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