A fundamental solution and its derivatives

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Discussion Overview

The discussion revolves around the properties of fundamental solutions to partial differential equations (PDEs) and their derivatives. Participants explore the implications of linearity and the commutation of differential operators, particularly in the context of the Laplace equation and other ordinary differential equations (ODEs).

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether the derivatives of a fundamental solution to a PDE remain solutions, suggesting that linearity allows for interchanging derivatives with the differential operator.
  • Another participant argues that the relationship depends on the commutator of the operator with differentiation, providing examples to illustrate when operators commute and when they do not.
  • There is a discussion about specific operators, such as [D^2 + 3D - 7] and [xD - 1], and how their commutation affects the solutions derived from them.
  • Participants clarify the application of the product rule in differentiation and correct earlier misunderstandings about the manipulation of differential expressions.
  • A later post shifts focus to the Laplace equation, questioning whether the gradient of a fundamental solution is also a solution, and seeks a coordinate-free understanding of this property.
  • Another participant notes that while the Laplacian of a vector and a scalar are not identical, the usual vector operators should commute, providing identities related to the Laplacian and gradient.

Areas of Agreement / Disagreement

Participants express differing views on the conditions under which derivatives of fundamental solutions remain solutions to PDEs. There is no consensus on the implications of linearity versus the commutation of operators, and the discussion remains unresolved regarding the broader applicability of these concepts.

Contextual Notes

Participants highlight the importance of understanding the commutation of operators and the application of the product rule, indicating that assumptions about linearity may not universally apply. The discussion also touches on the nuances of vector versus scalar Laplacians.

Who May Find This Useful

This discussion may be useful for students and professionals interested in the properties of differential equations, particularly in understanding the behavior of solutions and the implications of operator commutation in various contexts.

hanson
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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?
 
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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.
 
lurflurf said:
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?
 
hanson said:
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 [itex]D(x)D+ xD(D)- D= D+ xD^2- D= xD^2[/itex]. "[itex]D[xD- 1]= xD^2- D[/itex]" is wrong. You can "factor" constants out of the differential expression but not functions of the variable.
 
HallsofIvy said:
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 [itex]D(x)D+ xD(D)- D= D+ xD^2- D= xD^2[/itex]. "[itex]D[xD- 1]= xD^2- D[/itex]" is wrong. You can "factor" constants out of the differential expression but not functions of the variable.

Thank you very much!
 
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?
 
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
 
lurflurf said:
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.
 

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