What is wrong with this integral?

  • Thread starter hiyok
  • Start date
  • #1
109
0

Main Question or Discussion Point

Hi,

Let me describe the question. Suppose you have two distinct solutions, say, [itex]G_{1}(z)[/itex] and [itex]G_{2}(z)[/itex], to such a linear differential equation, [itex](\partial^2_z-q^2)G(z)=\delta(z)[/itex], where [itex]\delta(z)[/itex] denotes the Dirac function and [itex]q^2[/itex] is a constant. Now I'd like to evaluate this integral: [itex]\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)[/itex]. I will use this property: [itex]f(z)\delta^{n}(z)=(-1)^{n}f^{n}(z)\delta(z)[/itex]. Thus, if I in the first place integrate over [itex]z_1[/itex], I should then find
[tex]\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*}[/tex] On other hand, if it is first integrated over [itex]z_2[/itex], one would instead find [itex]G_1(z-z')[/itex], 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:

Answers and Replies

  • #2
Mute
Homework Helper
1,388
10
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
109
0
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
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,371
4,900
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
109
0
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
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,371
4,900
Forgive my ignorance, but what is [itex]\delta^{n}(z)[/itex]?
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
109
0
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
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,371
4,900
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.
So you meant [itex]∂^{n}(z)[/itex], not [itex]\delta^{n}(z)[/itex]?
 
  • #9
Mute
Homework Helper
1,388
10
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
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?

So you meant [itex]∂^{n}(z)[/itex], not [itex]\delta^{n}(z)[/itex]?
No, the OP meant ##\delta^{(n)}(z)##, the nth derivative of the Dirac delta function.
 
  • #10
109
0
@ 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.
 

Related Threads for: What is wrong with this integral?

Replies
3
Views
951
  • Last Post
Replies
7
Views
2K
Replies
1
Views
2K
Replies
6
Views
2K
  • Last Post
Replies
2
Views
691
Replies
9
Views
4K
  • Last Post
Replies
5
Views
2K
Replies
4
Views
2K
Top