A fundamental solution and its derivatives

[hanson]

Hello, if I have a fundamental solution, ,f, to a partial differential equation L(f)=0, where L is the differential operator, is that true that the derivatives of the fundamental solution, like D(f), will also be solution to the partial differential equation?

Intuitively, is it because things are linear, so I can always interchange the derivatives with the original differential operator of the PDE: L(D(f))=D(L(f))=0?

 

[lurflurf]

It is not for the most part to do with linearity. As you say it depends on the commutator of the operator with differentiation.
Is [L,D]=LD-DL=0?
Consider some examples from ODE.
D[D^2+3D-7]=[D^2+3D-7]D=[D^3+3D^2-7D]
They commute so if f is a solution Df is as well.
what about xD-1?
D[xD-1]=xD^2+D
[xD-1]D=xD^2-D
These are not equal, the operators do not commute.
f=x is a solution of [xD-1]f=0
but Dx=1 is not.

 

[hanson]

It is not for the most part to do with linearity. As you say it depends on the commutator of the operator with differentiation.
Is [L,D]=LD-DL=0?
Consider some examples from ODE.
D[D^2+3D-7]=[D^2+3D-7]D=[D^3+3D^2-7D]
They commute so if f is a solution Df is as well.
what about xD-1?
D[xD-1]=xD^2+D
[xD-1]D=xD^2-D
These are not equal, the operators do not commute.
f=x is a solution of [xD-1]f=0
but Dx=1 is not.

Thanks for your reply. I think I get it, but why
D[D^2+3D-7]=[D^3+3D^2-7D],
but
D[xD-1]=xD^2+D ?
Should D[xD-1]=xD^2+D-D = xD^2?

 

[HallsofIvy]

Thanks for your reply. I think I get it, but why
D[D^2+3D-7]=[D^3+3D^2-7D],
but
D[xD-1]=xD^2+D ?
Should D[xD-1]=xD^2+D-D = xD^2?

I had written out a long response until I realized I had read it backwards! You are completely correct.

D[xD- 1]= D(xD)- D and, using the product rule, that is $D(x)D+ xD(D)- D= D+ xD^2- D= xD^2$. "$D[xD- 1]= xD^2- D$" is wrong. You can "factor" constants out of the differential expression but not functions of the variable.

 

[hanson]

I had written out a long response until I realized I had read it backwards! You are completely correct.

D[xD- 1]= D(xD)- D and, using the product rule, that is $D(x)D+ xD(D)- D= D+ xD^2- D= xD^2$. "$D[xD- 1]= xD^2- D$" is wrong. You can "factor" constants out of the differential expression but not functions of the variable.

Thank you very much!

 

[hanson]

Let's focus on the Laplace equation. Is there a good way to understand the following?
If I have a fundamental solution f to the Laplace equation, then the gradient of f is also a solution to the Laplace equation. I can see why if I express Laplace equation in Cartesian coordinates, and see they the operators commutes in Cartesian coordinates. But is there a coordinate-free way to see this?

 

[lurflurf]

Are you working in 3-space? Strictly speaking the Laplacian of a vector and of Laplacian of a scalar are not identical, but it is a minor concern. All the usual vector operators like Laplacian, gradient, curl and divergence should commute since they do not have spatial dependence.
Recall the following basic identities
Laplacian(vector)=grad(div(vector))-curl(curl(vector))
Laplacian(scalar)=div(grad(scalar))
curl(grad(scalar))=0

so if we have for scalar f
Laplacian(f)=
Laplacian(grad(f))=grad(div(grad(f)))-curl(curl(grad(f)))=grad(0)-curl(0)=0

 

[hanson]

Are you working in 3-space? Strictly speaking the Laplacian of a vector and of Laplacian of a scalar are not identical, but it is a minor concern. All the usual vector operators like Laplacian, gradient, curl and divergence should commute since they do not have spatial dependence.
Recall the following basic identities
Laplacian(vector)=grad(div(vector))-curl(curl(vector))
Laplacian(scalar)=div(grad(scalar))
curl(grad(scalar))=0

so if we have for scalar f
Laplacian(f)=
Laplacian(grad(f))=grad(div(grad(f)))-curl(curl(grad(f)))=grad(0)-curl(0)=0

Thanks lurflurf. Let me read it in detail.