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