Green's function of coupled ODEs

For functions $a(x)$ and $b(x)$ and "sources" $S_1(f,g)$, $S_2(f,g)$ and $S_3(f,g)$ lets say one has the differential equations for functions $f(x)$ and $g(x)$,

$f' + af + bg = S_1(f,g) + S_2(f,g)$

$g' + f = S_3(f,g)$

Now if $S_1= S_2 = S_3 = 0$ i.e all the sources are turned off then the system reduces to the differential equations, $g'' + ag' - bg = 0$ (by eliminating $f$ as $f = - g'$). For this 2nd order ODE for g let $G_g$ be the Green's function.

And let $f_{13}(x)$ and $g_{13}(x)$ be the solutions to the system when $S_1$ and $S_3$ are on but $S_2 = 0$.

(..if $f_0(x)$ and $g_0(x)$ are the solutions with all sources turned off then $f_{13}$ and $g_{13}$ can be thought to be given as a power-series in $f_0$ and $g_0$ respectively..)

Similarly when sources are all turned off, one can eliminate $g$ between the two equations to get a 2nd order ODE for f as, $f'' + (a b'/b)f'+(a'-(ab')/b-b)f=0$ and then let $G_f$ be the Green's function for this.

• Given so much of data $G_g$, $G_f$, $f_{13}$ and $g_{13}$ can one build the full solutuions when all the 3 sources are on?

• Will it be an approximation (how good?) if one says that the full $g$ is given by,

$g (x) = g_{13}(x) + \int _{0}^x G_g(x,x')S_2 (f_{13},g_{13}) dx'$

and similarly can one write (approximate?) that,

$f(x) = f_{13}(x) + \int _0 ^x G_f(x,x')S_2(f_{13},g_{13})dx'$

(..I guess that in the above the "=" should be replaced by "~" and one can pertrubatively develop a solution by taking the $g$ or $f$ obtained in the first pass and then using that again as the new $f_{13}$ and $g_{13}$ respectively..though such an iteration is a bit hard to make sense of given that $f_{13}$ and $g_{13}$ themselves are given as an infinite power-series - so it seems like for any truncation of that power-series one would have to still do infinite updates here!..)

• Rewriting in vector notation your equation is $F' = A(x)F + S_{13}(F) + S_2(F)$. You want to update [F(x+\delta) \approx F(x) + \int_x^{x+\delta} G_F(x,x')\left[S_{13}(F(x')) + S_2(F(x'))\right]dx'] and iterate. Its tempting to write the first two terms as $F_{13}(x+\delta)$ but this of course depends on the initial condition $F_{13}(x)$. The $S_2$ term kicks the $F_{13}$ term away from a true solution. Upon iteration error will accrue. – Aaron Hoffman Apr 18 '13 at 15:40
• @Aaron Hoffman Can you state what is the exact correct answer for $f$ and $g$ in terms of $G_f$, $G_g$, $f_{13}$ and $g_{13}$? (..i am hoping an integral representation..) – Anirbit Apr 18 '13 at 16:36
• The standard Duhamel integral representation is [ F(x) = G_F(x,0)F(0) + \int_0^x G_F(x,x')[S_{13}[F(x') + S_2(F(x')]dx' ] Of course this is not a closed form as the unknown F appears on the right hand side. My claim (not at all carefully checked) is that in the generality that you've stated the question the terms $f_{13}$ and $g_{13}$ don't help you. – Aaron Hoffman Apr 18 '13 at 17:53
• @Aaron Hoffman Are there a few "]" missing in your expression? So perturbatively if one is solving this expression then all of the $F(x')$ on your RHS be replaced by $f_{13}(x)$? – Anirbit Apr 18 '13 at 18:49
• And what do you denote as $S_{13}$? I am not getting your notation. And similarly for $g$? (..here the confusion stems from the fact that once there are these sources I can't anymore think of a separate equation for $f$ and $g$ with sources..) – Anirbit Apr 18 '13 at 18:51