- #1
BiGyElLoWhAt
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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 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).