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Another Green's function

  • #1
BiGyElLoWhAt
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Homework Statement


Find the green's function for y'' +4y' +3y = 0 with y(0)=y'(0)=0 and use it to solve y'' +4y' +3' =e^-2x

Homework Equations


##y = \int_a^b G*f(z)dz##

The Attempt at a Solution


##\lambda^2 + 4\lambda + 3 = 0 \to \lambda = -1,-3##

##G(x,z) = \left\{ \begin{array}{ll}
Ae^{-x} + Be^{-3x} & z<x \\
Ce^{-x} + De^{-3x} & x<z
\end{array} \right. ##

y(0) = 0 -> A=-B
y'(0) =0 -> A=B=0

continuity:
##Ce^{-z} +De^{-3z}=0 \to C = -De^{-2z}##
##-Ce^{-z} -3De^{-3z} = 1 \to [De^{-2^z}]e^{-z} - 3De^{-3z} = -2De^{-3z} = 1##
##D=-1/2e^{3z}## & ##C= 1/2e^z##

##G(x,z) = \left\{ \begin{array}{ll}
0 & z<x \\
1/2e^ze^{-x} - 1/2e^{3z}e^{-3x} & x<z
\end{array} \right. ##

The problem comes in when I go to integrate to get the solution.
##y = \int_x^{\infty} [ 1/2e^{-(x+z)} - 1/2 e^{z-3x}]dz##
The second integral is divergent in z.
 
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Answers and Replies

  • #2
Orodruin
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Assuming z > 0 you have put the wrong side of the Green's function to zero ...
 
  • #3
Ray Vickson
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Homework Statement


Find the green's function for y'' +4y' +3y = 0 with y(0)=y'(0)=0 and use it to solve y'' +4y' +3' =e^-2x

Homework Equations


##y = \int_a^b G*f(z)dz##

The Attempt at a Solution


##\lambda^2 + 4\lambda + 3 = 0 \to \lambda = -1,-3##

##G(x,z) = \left\{ \begin{array}{ll}
Ae^{-x} + Be^{-3x} & z<x \\
Ce^{-x} + De^{-3x} & x<z
\end{array} \right. ##

y(0) = 0 -> A=-B
y'(0) =0 -> A=B=0

continuity:
##Ce^{-z} +De^{-3z}=0 \to C = -De^{-2z}##
##-Ce^{-z} -3De^{-3z} = 1 \to [De^{-2^z}]e^{-z} - 3De^{-3z} = -2De^{-3z} = 1##
##D=-1/2e^{3z}## & ##C= 1/2e^z##

##G(x,z) = \left\{ \begin{array}{ll}
0 & z<x \\
1/2e^ze^{-x} - 1/2e^{3z}e^{-3x} & x<z
\end{array} \right. ##

The problem comes in when I go to integrate to get the solution.
##y = \int_x^{\infty} [ 1/2e^{-(x+z)} - 1/2 e^{z-3x}]dz##
The second integral is divergent in z.
So, choose the opposite convention: ##G(x,z) = 0## if ##x < z## and ##G(x,z) = A e^{-x} + Be^{-3x}## if ##x > z##. Nevermind the boundary conditions ##y(0) = y'(0) = 0## on the Green's function. You can adjust the homogeneous solution to ensure that ##y(0) = y'(0) = 0## on the actual solution ##y(x)## itself----that is where you want it to occur!

If you work it out you will see that the particular solution(for ##x > 0##) is
$$y_p(x) = \int_0^{\infty} G(x,z) f(z) \, dz = \int_0^x G(x,z) f(z) \, dz$$
which involves only convergent integrations.
 
  • #4
BiGyElLoWhAt
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Assuming z > 0 you have put the wrong side of the Green's function to zero ...
Why? I'm not sure I understand this part, then. So, if I assume z <0, I have put the correct side to zero, yes? Does the rest follow? Because then I end up with the first integral diverging with limits -infty to x. Or is there a flaw here as well? I'm missing something between boundary conditions and the limits of integration, I think.
 
  • #5
BiGyElLoWhAt
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So, choose the opposite convention: ##G(x,z) = 0## if ##x < z## and ##G(x,z) = A e^{-x} + Be^{-3x}## if ##x > z##. Nevermind the boundary conditions ##y(0) = y'(0) = 0## on the Green's function. You can adjust the homogeneous solution to ensure that ##y(0) = y'(0) = 0## on the actual solution ##y(x)## itself----that is where you want it to occur!

If you work it out you will see that the particular solution(for ##x > 0##) is
$$y_p(x) = \int_0^{\infty} G(x,z) f(z) \, dz = \int_0^x G(x,z) f(z) \, dz$$
which involves only convergent integrations.
I wasn't aware you could choose one side to be zero. I know we have had problems where neither side was zero, say one had a sin term and the other a cosine term. The zero on one sade emerged from the boundary conditions. Is this still a viable way to solve these problems?
 
  • #6
Ray Vickson
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I wasn't aware you could choose one side to be zero. I know we have had problems where neither side was zero, say one had a sin term and the other a cosine term. The zero on one sade emerged from the boundary conditions. Is this still a viable way to solve these problems?
Just plug in the final result for ##y_p(x)## into the DE. Does it work? If so, it is correct; if not, it is incorrect. (However, we cooked up ##G(x,z)## in such a way that it does, in fact, work!)

If you have characteristic roots of opposite signs, so for example, you have ##e^x## and ##e^{-2x}## as homogeneous solutions, then, indeed, you may need to take ##G = A e^x## for ##x < z## and ##G = B e^{-2x}## for ##x > z## (or maybe the other way round); that keeps your integrals convergent. However, if both characteristic roots have the same sign, then you are more-or-less forced to keep ##G## bounded (or at least, not too "explosive" at ##\infty##) by choosing a one-sided version.

PS. I did not understand your comment in a reply in another thread, about having to do more integrals if the Green's function does not satisfy the problem's overall boundary condition. You still have to perform only a single integration of the form ##\int_0^{\infty} G(x,z) f(z) dz## (although it may split up into two pieces ##z < x## and ##z > x## in some cases). The problem boundary conditions are fitted by tuning the constants in the homogeneous solution, and so do not need any more integrations.
 
  • #7
BiGyElLoWhAt
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Hmm.. I see what you're saying. That doesn't chang the behavior of G? "Bounding" the limits, that is.
 
  • #8
vela
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Why? I'm not sure I understand this part, then. So, if I assume z <0, I have put the correct side to zero, yes?
Yes. If you assume ##z<0##, the Green's function would be
$$G(x,z) = \begin{cases} 0 & x>z \\ \frac 12 e^{-(x-z)} - \frac 12 e^{-3(x-z)} & x < z \end{cases}.$$ When ##x=0##, you have that ##x>z##, so ##G(0,z)=0## and ##G'(0,z)=0##. The boundary conditions are satisfied.
Does the rest follow? Because then I end up with the first integral diverging with limits -infty to x. Or is there a flaw here as well? I'm missing something between boundary conditions and the limits of integration, I think.
It's easy to get mixed up here. When you're finding ##G##, you typically think of ##z## as fixed and ##x## as the variable. When you do the integral to find ##y##, you want to consider ##x## fixed and ##z## as changing (because it's the variable of integration).

The solution for ##x<0## is given by
$$y(x) = \int_{-\infty}^0 G(x,z)e^{-2z}\,dz.$$

Look at the integral while considering ##x## fixed. From ##z=-\infty## to ##z=x##, you have ##z<x##, so ##G(x,z)## vanishes. You only get contributions when ##z>x##, so the limits of the integral become
$$y(x) = \int_x^0 G(x,z)e^{-2z}\,dz.$$ No infinities anymore, so everything converges.
 
  • #9
vela
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I wasn't aware you could choose one side to be zero.
Ray's not saying you can arbitrarily set one side to 0. You chose the wrong end of ##G## and consequently found ##A=B=0## and kept ##C## and ##D## around. If you assume ##z>0##, then when ##x=0##, you have ##x<z##. So in your original work, you should have found ##C=D=0## and kept the ##A## and ##B## terms around.

If you assume ##z<0##, however, your work above was correct. So the Green's function you found is good for ##z<0##, but not good for ##z>0##.
 
  • #10
BiGyElLoWhAt
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Ahhh, Ok, that makes sense. I've caught a few careless mistakes similar to this one before. I just wasn't seeing the inconsistency here. Thanks everyone.
 

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