Homework Help: Another Green's function

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1. Dec 21, 2016

BiGyElLoWhAt

1. The problem statement, all variables and given/known data
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

2. Relevant equations
$y = \int_a^b G*f(z)dz$

3. 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.

Last edited by a moderator: Dec 21, 2016
2. Dec 21, 2016

Orodruin

Staff Emeritus
Assuming z > 0 you have put the wrong side of the Green's function to zero ...

3. Dec 21, 2016

Ray Vickson

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. Dec 21, 2016

BiGyElLoWhAt

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. Dec 21, 2016

BiGyElLoWhAt

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. Dec 21, 2016

Ray Vickson

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. Dec 21, 2016

BiGyElLoWhAt

Hmm.. I see what you're saying. That doesn't chang the behavior of G? "Bounding" the limits, that is.

8. Dec 22, 2016

vela

Staff Emeritus
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.
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. Dec 22, 2016

vela

Staff Emeritus
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. Dec 22, 2016

BiGyElLoWhAt

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.