Graduate Square of an integral containing a Green's Function

[Celeritas]

TL;DR

Square of an integral containing a Green's Function.

Let's say you have a tensor u with the following components:
$$u_{ij}=\nabla_i\nabla_j\int_{r'}G(r,r')g(r')dr'$$
Where G is a Green function, and g is just a normal well behaved function. My question is what is the square of this component? is it
$${u_{ij}}^2=\bigg[\nabla_i\nabla_j\int_{r'}G(r,r')g(r')dr'\bigg]\bigg[\nabla_i\nabla_j\int_{r''}G(r,r'')g(r'')dr''\bigg]$$
or is it
$${u_{ij}}^2=\bigg[\nabla_i\nabla_j\int_{r'}G(r,r')g(r')dr'\bigg]\bigg[{\nabla_i}'{\nabla_j}'\int_{r''}G(r',r'')g(r'')dr''\bigg]$$

The first one makes much more sense to me, but I'm trying to reproduce results from a paper and it appears that they have used the second one. Any help is appreciated.

EDIT:
I'm sorry but it appears I have written the second one wrong. What I mean is:
$${u_{ij}}^2=\nabla_i\nabla_j\int_{r'}G(r,r')g(r')\bigg[{\nabla_i}'{\nabla_j}'\int_{r''}G(r',r'')g(r'')dr''\bigg]dr'$$
I understand that r and r'' are dummy variables, but my problem is writing G in the second integral as G(r',r'') and not G(r,r''). To me it should be the latter because after integrating I should have something depending on r and not r'. If it depends on r' then the result of the second integral will contribute to the first integral (which is over r').

 

[PeroK]

I'm sorry but it appears I have written the second one wrong. What I mean is:
$${u_{ij}}^2=\nabla_i\nabla_j\int_{r'}G(r,r')g(r')\bigg[{\nabla_i}'{\nabla_j}'\int_{r''}G(r',r'')g(r'')dr''\bigg]dr'$$
I understand that r and r'' are dummy variables, but my problem is writing G in the second integral as G(r',r'') and not G(r,r''). To me it should be the latter because after integrating I should have something depending on r and not r'. If it depends on r' then the result of the second integral will contribute to the first integral (which is over r').

Yes, I wondered what happened!

There's clearly something missing here. What is $u_{ij}^2$? Is it really $(u^2)_{ij}$?

 

[Celeritas]

Yes, I wondered what happened!

There's clearly something missing here. What is $u_{ij}^2$? Is it really $(u^2)_{ij}$?

${u_{ij}}^2$ is the square of the element $u_{ij}$. Afterwards one sums over i and j.
In physics, this would give (along with some additional unimportant terms) the free energy contribution of pure shear on a material.

 

[PeroK]

${u_{ij}}^2$ is the square of the element $u_{ij}$. Afterwards one sums over i and j.
In physics, this would give (along with some additional unimportant terms) the free energy contribution of pure shear on a material.

I wonder if it's an identity that depends on the properties of the Green's function?

 

[Celeritas]

I wonder if it's an identity that depends on the properties of the Green's function?

Maybe, but after integrating over dr'', you essentially lose the second green function and are left with G(r,r')g(r')f(r') where f(r') is the result of the integral over dr''. I don't see how any property of Green's function will do anything about that!
But my main concern was that is my way of looking at this term correct? i.e. is this one correct?
$${u_{ij}}^2=\bigg[\nabla_i\nabla_j\int_{r'}G(r,r')g(r')dr'\bigg]\bigg[\nabla_i\nabla_j\int_{r''}G(r,r'')g(r'')dr''\bigg]$$
I will also try to contact the authors but the paper is 22 years old!

 

[PeroK]

Maybe, but after integrating over dr'', you essentially lose the second green function and are left with G(r,r')g(r')f(r') where f(r') is the result of the integral over dr''. I don't see how any property of Green's function will do anything about that!
But my main concern was that is my way of looking at this term correct? i.e. is this one correct?
$${u_{ij}}^2=\bigg[\nabla_i\nabla_j\int_{r'}G(r,r')g(r')dr'\bigg]\bigg[\nabla_i\nabla_j\int_{r''}G(r,r'')g(r'')dr''\bigg]$$
I will also try to contact the authors but the paper is 22 years old!

That expression is correct by definition. There must be some way to equate it to the other. Unless we are missing something there must be a Green's function property behind this.

I haven't see this before, so I can't help any further. Sorry.

 

[Celeritas]

That expression is correct by definition. There must be some way to equate it to the other. Unless we are missing something there must be a Green's function property behind this.

I haven't see this before, so I can't help any further. Sorry.

Thank you. You've helped plenty.

 

[Fred Wright]

I assume that $\nabla_i=\frac{\partial}{\partial x_i}$. For the Cartesian coordinate system$$x_1=x\\
x_2=y\\
x_3=z\\
r=\sqrt{x_1^2+x_2^2+x_3^2}\\
r'=\sqrt{x_1^{'2}+x_2^{'2}+x_3^{'2}}$$I would think that, for example$$
u_{xz}=\int_{r'}\frac{\partial^2}{\partial x \partial z}G(r,r')g(r')dr'$$ and that,$$u_{xz}^2=[\int_{r'}\frac{\partial^2}{\partial x \partial z}G(r,r')g(r')dr'][\int_{r'}\frac{\partial^2}{\partial x \partial z}G(r,r')g(r')dr']$$