Find a solution for differential equations

[ip88]

I have to solve the following differential equation (that I found in an article):

$\frac{1}{r^{5}}\partial_{r}(r^{5}\partial_{r}h(r)) -E\frac{h(r)}{r^{2}} = - \frac{C}{r^{5}}\delta(r-r_{0})$

where E and C are two constants.

The authors of the article first find a solution of the previous equation when $r≠r_{0}$ and the result is:

$h(r)=Ar^{c_{\pm}}$ where $c_{\pm} = -2 \pm \sqrt{E+4}$

Then they have to find the value of the constant A. In order to reach this aim they say that I have to integrate the differential equation and the result should be:

$\frac{C}{2\sqrt{E+4}}\times$ $\frac{1}{r_{0}^{4}}(\frac{r}{r_{0}})^{c_{+}}$ if r ≤ r_{0}

and

$\frac{C}{2\sqrt{E+4}}\times$ $\frac{1}{r^{4}}(\frac{r_{0}}{r})^{c_{+}}$ if r ≥ r_{0}

However I'm not able to recover this result. The point is that when I insert the solution of the homogeneous equation and I integrate the left side of the equation is always equal to zero. Probably I did not understand correctly what is the procedure that I have to follow in order to find the solution to the inhomogeneous equation (once I have the solution of the homogeneous equation), but I do not know how to reach this aim.

 

[vanhees71]

The general solution for $r \neq r_0$ is
$$h(r)=A_1 r^{c_+}+A_2 r^{c_-},$$
where the constants $A_1$ and $A_2$ can be different for $r<r_0$ and $r>r_0$ in such a way that you get the $\delta$-distribution singularity. Since it's a 2nd-order equation you can demand that $h$ is continuous at $r_0$. This gives you one constraint for your four unknowns. You get a 2nd constraint by integrating the differential equation over an infinitesimal interval containing $r_0$ in its interior, leading to a condition for the appropriate jump in $h'(r)$.

Further you need some more conditions to fix the function completely. These should be given from the application treated with this function (Green's function for some physical problem?).

 

[the_wolfman]

Hi ip88,

To expand upon what vanhees71 said.

After solving the homgenous equation you have two different solutions


$h_< = Ar^{c_+} + B r^c{_-}$
and
$h_> = L r^{c_+} + M r^{c_-}$

The solution $h_<$ is only valid for $r < r_0$ and $h_>$ is only valid for $r > r_0$.

The goal is then to find the constants $A,B,L, M$. There are 4 unknowns, and we need 4 equations. Two equations come from the boundary conditions of the ODE. A third equation is obtained by requiring $h_< =h_>$ at $r_0$.

To get a forth equation we integrate the differential equation from $r_0 -\epsilon$ to $r_0 +\epsilon$ and then take the limit that $\epsilon \rightarrow 0$. This last step is where all the magic happen and it gives you a "jump condition".

To see how this works let's take the integral: (I've also multiplied the equation by r^5)
$\int_{r_0-\epsilon}^{r_0+\epsilon} dr \left(\partial_r (r^5 \partial_r h) - Er^3h=-C\delta(r-r0) \right)$

The integral of the first term is:
$\left. r^5 \partial_r h \right|_{r_0-\epsilon}^{r_0+\epsilon}$
and in the limit that $\epsilon \rightarrow 0$ we get $r_0^5 \left(h'_>-h'_< \right)\left. \right|_{r_0}$. The primes are the derivatives in r.

The integral of the second term goes to zero in the limit that $\epsilon \rightarrow 0$.

Finally the integral of the third term is -C

Putting this together gives the jump condition
$r_0^5 \left(h'_>-h'_< \right)\left. \right|_{r_0} =-C$

 

[ip88]

Thank you very much for the explanation