Help needed to solve a differential equation

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Homework Statement



solve for u:

[tex]u_t=G + \mu(u_{rr}+\frac{1}{r}u_r)[/tex]

with boundary conditions u=0 on r=a and [tex]u_r=0[/tex] 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:

[tex]\frac{G_0 a^2}{4\mu}(1-\frac{r^2}{a^2}[/tex]

The Attempt at a Solution



u is independent of t so [tex]u_t=0[/tex].
it is an inhomogeneous differential equation so i thought youd solve [tex](u_rr+\frac{1}{r}u_r)=0[/tex] first.
I then thought ud let [tex]p=u_r[/tex] to get [tex]p_r+\frac{1}{r}p[/tex] 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 : [tex]u_r=r+c[/tex] to get [tex]u=\frac{r^2}{2}+rc+d[/tex]. This looks wrong and i have no idea how to introduce the \frac{G}{/mu} term.
any help will be much appreciated!
 

Answers and Replies

  • #2
HallsofIvy
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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 [itex]p(r)= cr^{-1}= c/r[/itex], where c= [itex]e^C[/itex], not r+ c.

Then u'= c/r so [itex]u''= cln(r)+ d[/itex]. 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 [itex]u(r)= v(r)ln(r)+ w(r)[/itex] for two unknown function v(r) and w(r).

Differentiating, [itex]u'= v'ln(r)+ v/r + w'[/itex]. 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 [itex]v'ln(r)+ w'= 0[/itex]. That means that [itex]u'= v/r[/itex] and so [itex]u''= v'/r- v/r^2[/itex]. Putting those into the original euation,
[tex]u''+ (1/r)u'= v'/r- v/r^2+ (1/r)(v/r)= v'/r= -G[/tex]

So v'= -Gr and [itex]v(x)= -(1/2)Gr^2[/itex]. 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.
 
  • #3
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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 [itex]v'ln(r)+ w'= 0[/itex].

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

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

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

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 :)
 
  • #4
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also when im doing the integration by parts should i be leaving it as indefinite integrals or doing it between 0 and a?
thanks
 

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