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Find a solution for differential equations

  1. Jan 23, 2014 #1
    I have to solve the following differential equation (that I found in an article):

    [itex]\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})[/itex]

    where E and C are two constants.

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

    [itex]h(r)=Ar^{c_{\pm}}[/itex] where [itex]c_{\pm} = -2 \pm \sqrt{E+4} [/itex]

    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:

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


    [itex]\frac{C}{2\sqrt{E+4}}\times [/itex] [itex]\frac{1}{r^{4}}(\frac{r_{0}}{r})^{c_{+}}[/itex] 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.
  2. jcsd
  3. Jan 23, 2014 #2


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

    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?).
  4. Jan 23, 2014 #3
    Hi ip88,

    To expand upon what vanhees71 said.

    After solving the homgenous equation you have two different solutions

    [itex] h_< = Ar^{c_+} + B r^c{_-}[/itex]
    [itex] h_> = L r^{c_+} + M r^{c_-}[/itex]

    The solution [itex] h_< [/itex] is only valid for [itex] r < r_0[/itex] and [itex] h_>[/itex] is only valid for [itex] r > r_0[/itex].

    The goal is then to find the constants [itex]A,B,L, M[/itex]. 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 [itex] h_< =h_> [/itex] at [itex] r_0 [/itex].

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

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

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

    The integral of the second term goes to zero in the limit that [itex] \epsilon \rightarrow 0 [/itex].

    Finally the integral of the third term is -C

    Putting this together gives the jump condition
    [itex] r_0^5 \left(h'_>-h'_< \right)\left. \right|_{r_0} =-C[/itex]
  5. Jan 23, 2014 #4
    Thank you very much for the explanation
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