# What is wrong with this integral?

1. Sep 16, 2012

### hiyok

Hi,

Let me describe the question. Suppose you have two distinct solutions, say, $G_{1}(z)$ and $G_{2}(z)$, to such a linear differential equation, $(\partial^2_z-q^2)G(z)=\delta(z)$, where $\delta(z)$ denotes the Dirac function and $q^2$ is a constant. Now I'd like to evaluate this integral: $\int^{\infty}_0\int^{\infty}_0dz_1dz_2G_1(z-z_1)G_2(z'-z_2)(\partial^2_{z_1}-q^2)\delta(z_1-z_2)$. I will use this property: $f(z)\delta^{n}(z)=(-1)^{n}f^{n}(z)\delta(z)$. Thus, if I in the first place integrate over $z_1$, I should then find
$$\begin{eqnarray*} \int^{\infty}_0\int^{\infty}_0dz_1dz_2G_1(z-z_1)G_2(z'-z_2)(\partial^2_{z_1}-q^2)\delta(z_1-z_2) & = &\int^{\infty}_0dz_2G_2(z'-z_2)\int^{\infty}_0dz_1G_1(z-z_1)(\partial^2_{z_1}-q^2)\delta(z_1-z_2)\\ & = & \int^{\infty}_0dz_2G_2(z'-z_2)\int^{\infty}_0dz_1(\partial^2_{z_1}-q^2)G_1(z-z_1)\delta(z_1-z_2)\\ & = &\int^{\infty}_0dz_2G_2(z'-z_2)\int^{\infty}_0dz_1\delta(z-z_1)\delta(z_1-z_2)\\ & = & G_2(z-z') \end{eqnarray*}$$ On other hand, if it is first integrated over $z_2$, one would instead find $G_1(z-z')$, which differs from previous result. I don't know how to make sense of such ambiguity. Can anybody come to help ?!

Thank you in advance.

hiyok

Last edited by a moderator: Sep 16, 2012
2. Sep 16, 2012

### Mute

If your calculations are all correct, it looks to me like it demonstrates that your two distinct solutions are actually not distinct - they are the same.

(Which I would expect - I am not aware of any differential equation having more than one Green's function for a given set of initial/boundary conditions.)

3. Sep 16, 2012

### hiyok

Dear Mute,
Thanks for your reply.
Please note we are dealing with a second order linear inhomogeneous differential equation, and therefore there must be infinite number of solutions. The equation itself can not specify a unique solution. Boundary conditions must be imposed to this end. So, one can easily find two different solutions to my equation.
hy

4. Sep 18, 2012

### haruspex

A few things i don't follow in your algebra.
Surely ∂ is an operator, so you can't go swapping the order like that.
Secondly, your integral is not symmetric between z1 and z2 (specifically, ∂z1).
Thirdly, there is a change of order of integration. That is not always valid. Needs to be justified.

5. Sep 19, 2012

### hiyok

Dear haruspex,
thanks for your reply.
1. The reason why I swapped the order of acting the differential operator is because I made of use a property of Dirac function, as mentioned in the original post.
2. The integral is actually symmetric between z1 and z2, because the Dirac function is even.
3. Changing the integration order should be allowed in this case, because the integration domain is rectangular.

hy

6. Sep 20, 2012

### haruspex

Forgive my ignorance, but what is $\delta^{n}(z)$?
My problem with the symmetry is that if I swap z1 with z2, G1 with G2, and z with z' in the original expression it does not leave it quite the same. The difference looks crucial, so I doubt that you can arrive at those two different results by essentially the same procedure.

Re order of integration, a rectangular domain is not enough. Integration is a limit process, and changing the order of limits has pitfalls. However, I doubt that's the problem here. Changing the operator order is the most suspect part.

7. Sep 20, 2012

### hiyok

Sorry that I did not make that symbol clear. The superscript n indicates differentiating n times. I find that it is possible to eliminate the ambiguity via Fourier transform.

8. Sep 21, 2012

### haruspex

So you meant $∂^{n}(z)$, not $\delta^{n}(z)$?

9. Sep 21, 2012

### Mute

In order to use the property $G_i(z_1-z_2)\partial^2_{z_1} \delta(z_1-z_2) = [\partial^2_{z_1}G_i(z_1-z_2)]\delta(z_1-z_2)$, the G's must be technically vanish at the boundaries, no? So are you not then implicitly assuming both G's have the same boundary conditions in using this identity?

No, the OP meant $\delta^{(n)}(z)$, the nth derivative of the Dirac delta function.

10. Sep 22, 2012

### hiyok

@ haruspex: please see Mute's post.

@ Mute: to use that property, I don't think the G's have to vanish at the boundary. The Dirac function automatically does that. The G's are not with the same boundary conditions, or they would be the same.

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