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Dirac Delta function - as a solution to Modified Helmholtz equation

Posted 9 years ago

So i am solving a differential equation which has a delta function. When i solve the differential equation, the solution comes up with a Heaviside Function for the particular solution of the DiracDelta. But in the book Arfken, we see that the solution for a modified Helmholtz equation with a delta function is

Equation : p''[z] - k^2*p[z] = DiracDelta(z-z2)

Solution in Mathematica : constansts + Heaviside Function

Solution in Arfken : Exp[-kAbs[z-z2]]/2k

How do i make sure Mathematica gives me the Arfken result instead of a Heaviside function????

4 Replies

I don't know of any systematic Mathematica procedure for solving this equation. You apparently can't put in f[Infinity] == 0 as a subsidiary condition. (Machines have not yet completely replaced humans!)

POSTED BY: S M Blinder

I agree with solution you have given, but how do i implement the same in Mathematica. Using the solution from this expression, I need to solve more Differential Equations.

DSolve[{C''[z] - q^2*C[z] == -DiracDelta[z - z2], C[Infinity] == 0, C[-Infinity] == 0}, C[z], z]

I tried giving the boundary conditions as the above mentioned, I ended getting an error message as shown below:

DSolve::bvlim: For some branches of the general solution, unable to compute the limit at the given points. Some of the solutions may be lost. >>

If you would be kind enough to give me the solution for this error.


G(Z) = E^(-k|Z|)/(2k).

POSTED BY: S M Blinder

The solution is a Green's function G{z,z2) = G{z-z2) = G{Z}, with Z = z-z2, which obeys the differential equation G''{Z)-k^2 G(Z) = -delta(Z). The minus sign is conventional.

In[45]:= DSolve[G''[Z] - k^2*G[Z] == -DiracDelta[Z], G[Z], Z]

Out[45]= {{G[Z] -> 
   E^(k Z) C[1] + E^(-k Z) C[2] - (
    E^(-k Z) (-1 + E^(2 k Z)) HeavisideTheta[Z])/(2 k)}}

Now we need to determine C[1] and C[2] by the boundary conditions G[Infinity]=G[-Infinity]=0. From the first condition, as Z->+Infinity, E^(-k Z) becomes negligible, so the asymptotic solution approaches G(Z) = E^(k Z) C[1] - E^(k Z)/{2k).This implies that C[1] = 1/(2k). Setting C[2]= 0, we find G(Z) = E^(-k Z)/(2k) for Z>0. The second boundary condition leads to E^(k Z)/(2k) for Z<0. Thus the final result is G(Z) = E^(k|Z|)/(2k).

POSTED BY: S M Blinder
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