# A question about boundary conditions in Green's functions

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• BiGyElLoWhAt
In summary, the conversation discusses finding the Green's function for a given differential equation and boundary conditions. There is confusion about whether y(x)=0 should be considered as a solution and how to apply the boundary conditions in solving for the coefficients. The conversation also mentions issues with setting up the integrals and determining the limits. The question is left unresolved and the speaker mentions seeking guidance from their professor.
BiGyElLoWhAt
Gold Member
I have a couple homework questions, and I'm getting caught up in boundary applications. For the first one, I have y'' - 4y' + 3y = f(x) and I need to find the Green's function.
I also have the boundary conditions y(x)=y'(0)=0. Is this possible? Wouldn't y(x)=0 be of the form of a solution? Should this be y(x_0)? I assumed that it was correct, but then getting into the boundary condition application for G(x,z), I'm not sure what to do. Can I limit this boundary condition to one half of my G? So when I have
...{Ae^3x + Be^x for z<x
G = {
...{Ce^3x + De^x for x<z

Can I use y(x) = 0 to say A=B=0? Or do I need to treat y(x) as a single point, and evaluate at that point, then set it equal to zero (In which case A and B would be functions of (x_0) ).

For another problem, I have y'' + 4y' +3x = e^-2x and y(0)=y'(0)=0.
I think I'm either not applying the boundaries correctly in solving for my coefficients, or in setting up the integral, as I ended up with a nonconvergant integral. I'm relatively certain that I solved for the constants correctly. I think it's arbitrary which set I use in setting y and y' = 0. At least I can't think of a good reason why It should matter. I chose A=B=0 (that's what you get when you set y and y' = 0 for z<x)
Then when you apply the continuity conditions (##G(z,z)_+ - G(z,z)_- = 0\ \text{and} \ G'(z,z)_+ - G'(z,z)_- = 1##) You end up with C=1/2 e^z and D = -1/2 e^3z. Now, when I multiply both of these by f(z) = e^-2z I effectively have an integral of the form e^-z -e^z, the first part converges, but the second doesn't. This is of course assuming my limits (x, ifty) are correct. I'm not sure I understand exactly how to get my limits. My reasoning for this is that since in my initial assumption I chose z<x to contain my y(0) boundary and we don't have another boundary in y (only in y'), so therefore I need to integrate from 0 to x (which is 0) and from x to infinity (since there is a discontinuity at x).

Can anyone shed some light? My book seems to just get them from somewhere. We assume a solution of the form int [G(x,z)f(z)] from a to b with a and b being gotten from the boundary conditions, but in examples where you have one y boundary and one y' boundary, it seems as though one of the limits is infinity. This should have a fairly simple solution (as I did one earlier of the same form without using green's functions).

I got ahold of my professor, and y(x)=y'(0)=0 is a typo, and should be y(0)=y'(0) = 0. The second question still stands, however.

## 1. What are boundary conditions in Green's functions?

Boundary conditions in Green's functions refer to the conditions that must be satisfied at the boundaries of a physical system in order for the Green's function to accurately describe the behavior of the system. These conditions can include constraints on the values of the function or its derivatives at the boundaries.

## 2. How are boundary conditions incorporated into Green's functions?

Boundary conditions are typically incorporated into Green's functions through the use of delta functions or other mathematical techniques. The boundary conditions are used to determine the coefficients of the delta functions, which then become part of the overall Green's function solution.

## 3. Why are boundary conditions important in Green's functions?

Boundary conditions are crucial in Green's functions because they allow us to accurately model physical systems by taking into account the specific conditions at the boundaries. Without incorporating these conditions, the Green's function solution may not accurately reflect the behavior of the system.

## 4. What happens if the boundary conditions are not satisfied in Green's functions?

If the boundary conditions are not satisfied in Green's functions, the resulting solution may be incorrect or invalid. This can lead to inaccurate predictions and unreliable results when using the Green's function to model physical systems.

## 5. Can boundary conditions change the behavior of Green's functions?

Yes, boundary conditions can significantly affect the behavior of Green's functions. By changing the constraints at the boundaries, we can alter the overall shape and characteristics of the Green's function, leading to different solutions and predictions for the physical system being modeled.

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