Green's Function with Neumann Boundary Conditions

In summary: I think I finally understand what I'm doing wrong!In summary, the student is trying to solve a two-point boundary value problem but is stuck because they do not understand how to construct the Green's function. They are stuck because they are trying to use the functions to approximate a boundary value problem using matrices, but they are unable to find a nonlinear term in the solution.
  • #1
Mattbringssoda
16
1

Homework Statement

[/B]
Determine the Green's functions for the two-point boundary value problem u''(x) = f(x) on 0 < x < 1 with a Neumann boundary condition at x = 0 and a Dirichlet condition at x = 1, i.e, find the function G(x; x) solving

u''(x) = delta(x - xbar) (the Dirac delta function); u'(0) = 0; u(1) = 0

and the functions G_0 (x) solving

u''(x) = 0; u'(0) = 1; u(1) = 0

and G_1(x) solving

u''(x) = 0; u'(0) = 0; u(1) = 1:

Homework Equations

The Attempt at a Solution


Right now I'm most concerned with the first part of the problem, the main Green's function G(x,x) solving: u''(x) = delta(x - xbar) (the dirac delta function); u'(0) = 0; u(1) = 0.

I'm stuck because we're using these functions to make discreet approximations to boundary value problems via matrices, so I assume the functions have to be linear.

However, u'(0) = 0 would have to be of some form u'(x) = x, so the solution would have to be a non-linear term, along the lines of u(x) = x^c. That doesn't seem right in light of what we've been learning...so I don't know what to do here...

Any help is much appreciated!
 
Physics news on Phys.org
  • #2
Mattbringssoda said:
However, u'(0) = 0 would have to be of some form u'(x) = x, so the solution would have to be a non-linear term, along the lines of u(x) = x^c. That doesn't seem right in light of what we've been learning...so I don't know what to do here...
It is not clear what you are trying to do here. Your problem is a relatively simple differential equation with given boundary conditions and so should have a rather straightforward solution. Constructing the general solution to ##G'' = \delta(x-\bar x)##, you should obtain a solution with two undetermined constants that you will have to fix to satisfy the boundary conditions. How are you used to construct the Green's function to ODEs of this type?
 
  • #3
Well, we're only doing a very surface introduction to Greens functions in class, and were using them to construct inverse matrices in which to numerically solve boundary value problems and to look at the stability of those methods.

The only other Greens function we encountered was for one that satisfied G" = δ(x-x), x(0) = 1 and x(1) = 0. The answer was just given and then shown to be the answer. The answers were linear.

In my posted problem, I guess I'm not sure how to get started with the u' as the boundary conditions. I'm guessing something like uc, then u'(0) will give us 0... but I was under the impression that the terms here has to be linear since were using them in matrices to numerically solve problems... I'm clearly not understanding something...
 
  • #4
Start by considering the differential equation for ##x < \bar x##. How does the solution look in this region?
 
  • #5
Hmm, well that's the area I thought would be of form xc for x < x_bar

(PS Thank you for your responses thus far..)
 
  • #6
Why? How does the differential equation look in that region?
 
  • #7
Well, the G" equals a Dirac function with its jump at xBar,

So below xBar: G" = 0,

Meaning G' = a constant ,

Meaning G is generally some form of Cx, where C is a constant...?
 
  • #8
Mattbringssoda said:
Meaning G is generally some form of Cx, where C is a constant...?
No, you forgot one integration constant. You also need to adapt the solution to the boundary condition at ##x = 0##.
 
  • #9
I think I have it now; u" = 0 means that u MUST beinn linear. u' = some constant, and go from there.

Thanks SO much for your help...
 

Related to Green's Function with Neumann Boundary Conditions

1. What is a Green's function with Neumann boundary conditions?

A Green's function with Neumann boundary conditions is a mathematical tool used in solving partial differential equations with boundary conditions that involve the normal derivative of the unknown function. It is a type of Green's function that satisfies both the differential equation and the boundary conditions.

2. How is a Green's function with Neumann boundary conditions used in solving PDEs?

A Green's function with Neumann boundary conditions can be used to express the solution of a PDE as an integral involving the Green's function and the given boundary conditions. This allows for a more efficient and systematic way of solving PDEs with complicated boundary conditions.

3. What is the difference between Green's function with Neumann boundary conditions and Green's function with Dirichlet boundary conditions?

The main difference between these two types of Green's functions is the type of boundary conditions they satisfy. Green's function with Neumann boundary conditions satisfies boundary conditions that involve the normal derivative of the unknown function, while Green's function with Dirichlet boundary conditions satisfies boundary conditions that involve the value of the unknown function itself.

4. Can a Green's function with Neumann boundary conditions be used for any type of PDE?

Yes, a Green's function with Neumann boundary conditions can be used for any type of linear PDE. However, it is not applicable for nonlinear PDEs.

5. Are there any limitations or drawbacks to using a Green's function with Neumann boundary conditions?

One limitation of using a Green's function with Neumann boundary conditions is that it can only be used for linear PDEs. Additionally, the computation of the Green's function can be complex and time-consuming, making it difficult to use in certain practical applications.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
560
  • Calculus and Beyond Homework Help
Replies
1
Views
820
  • Calculus and Beyond Homework Help
Replies
6
Views
551
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
899
  • Calculus and Beyond Homework Help
Replies
3
Views
562
  • Calculus and Beyond Homework Help
Replies
8
Views
452
  • Calculus and Beyond Homework Help
Replies
1
Views
789
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Back
Top