# Coupled differential-integral equation

This is coming from a physics paper I'm reading. It's been a while since I've done much differential equation solving and the system here is a bit unorthodox in that I'm actually searching for the initial conditions.

I would like to show that

$f(0) = g(0) (1 - e^{-a \tau I}) + I$ where $I = \int_0^{\infty}\frac{f(t)}{\tau} dt$

given the differential equations:

$\frac{df}{dt} = - a f(t) g(t) - f(t)/\tau$

$\frac{dg}{dt} = - a f(t) g(t)$.

It might be important to state that f(t) and g(t) $\rightarrow 0$ as $t\rightarrow \infty$.

Thanks!

EDIT: $a$ and $\tau$ are constants greater than zero and everything's real.

EDIT: Actually $g(\infty)$ cannot be 0 [see comments below].

• What is $\tau$? Jun 14 '16 at 15:43
• If $\tau$ is a constant this is pretty simple, if not it may be more difficult Jun 14 '16 at 15:45
• @M10687 sorry, $\tau$ is constant, as is $a$. Jun 14 '16 at 15:45
• @M10687 this is not all that simple even with constant $\tau$. Scattering theory can contain some tough math issues. BTW, the asymptotic behavior of $f(t)$ and $g(t)$ is an essential part of this problem; the differential equations are well-posed for any $f(0)$ and $g(0)$ but for a given value of $g(0)$ not every $f(0)$ leads to $f$ and $g$ going to zero at $+\infty$. Jun 14 '16 at 17:10
• The exercise seems absurd: note that $g'(t)=-af(t)g(t)$ hence $g(t)=g(0)e^{-aF(t)}$ with $F(t)=\int_0^tf(s)ds$. Now, one assumes that $g(\infty)=0$ hence either $g(0)=0$ or $F(\infty)=\infty$. If $g(0)=0$ then $g(t)=0$ for every $t$ hence $f'(t)=-f(t)/\tau$, which, together with $f(\infty)=0$, implies $f(0)=F(\infty)/\tau=I$ hence indeed, $f(0)=g(0) (1 - e^{-a \tau I}) + I$... Otherwise, $F(\infty)=\infty$, that is, $I=\infty$ and $f(0)=g(0) (1 - e^{-a \tau I}) + I$ only holds if $f(0)=\infty$... Sure about the text of the exercise?
– Did
Jun 15 '16 at 16:51

First we find $g(t)$ and as $g'(t)/g(t)=-af$ then we have

$$g(t)= g(0)e^{-a\int_0^t f(s)ds}$$

Now integrate the first equation between 0 and $t$ to give

$$\int_0^t f'(s) ds= -\int_0^t a f(s)g(s)ds - \int_0^t \frac{f(s)}{\tau} ds$$

Take the limit $t \to \infty$ then

$$-f(0)= -\int_0^{\infty} af(z)g(0) e^{-a\int_0^zf(s) ds}dz -\int_0^\infty \frac{f(s)}{\tau} ds$$ So the second term is intergated to give

$$g(0)e^{-a \int_0^z f(s) ds} |_{z=0}^{\infty}= -g(0)[1-e^{-a\int_0^ \infty f(s) ds}]$$

which gives the result

• in the third equation down, the $t$ in the exponentiated integral should be ∞? And the $t$ in$f(t)dt$ is not the same tt that was taken to ∞, right? Lastly, in the last equation how does LHF = RHS? Jun 15 '16 at 17:08
• Sorry I had a few typing mistakes. I have replaced $t$ with the dummy variable $z$, The exponential inside the integral should have upper limit z as this the function g(z) we are integating. The final step I had the limits missing. Put the limits in and lhs=rhs
– aw80
Jun 15 '16 at 17:47
• your result seems to entail that $\int_0^{\infty} a f(z)e^{-a\int_0^zf(s) ds}dz = e^{-a\int_0^zf(s) ds}|^{\infty}_0$. Is that right? What happened to the $f(z)$? Jun 15 '16 at 18:03
• differentiate rhs with respect to z
– aw80
Jun 15 '16 at 18:11
• that's awesome. I wish I knew math, lol. thnx Jun 15 '16 at 18:19