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I am stuck on the following problem: given the following Volterra integral inhomogeneous equation: $$\phi(x)=\exp(-x)+\lambda\int_0^x\frac{1}{x^2+t^2}\phi(t)dt$$ is it possible to solve it given the constraint: $$\lim_{x\to\infty}\phi(x)=\alpha$$ where $\alpha$ is a constant ($\alpha\in R$) and $\lambda$ a unknown parameter? I'm looking for closed form expressions of $\phi(x)$. Thanks.

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Is this a textbook problem? – Norbert Jul 26 '12 at 11:34
@Norbert: no it doesn't. It's a pquestion arising from a radar detection problem. – Riccardo.Alestra Jul 26 '12 at 11:42
Just guessing: Would it help to take the derivative w.r.t. $x$ and treat it as differential equation? – draks ... Jul 26 '12 at 12:16
@draks: it could be a good suggestion. If you expand your hint, and give an answer I will accept it. – Riccardo.Alestra Jul 26 '12 at 12:28
@Riccardo.Alestra puh, a look at W|A makes me think, that it just might complicate things... but let's see... – draks ... Jul 26 '12 at 12:43
up vote 1 down vote accepted

This equation can be cast into a Fredholm-like integral equation. This can be seen as follows $$ \phi(x)=e^{-x}+\frac{\lambda}{x^2}\int_0^x\frac{1}{1+\frac{t^2}{x^2}}\phi(t)dt $$ and change the integration variable to $y=\frac{t}{x}$. We are assuming that exists and is finite the limit $\lim_{x\rightarrow 0}\phi(x)$ after this variable change. Then $$ \phi(x)=e^{-x}+\frac{\lambda}{x}\int_0^1\frac{1}{1+y^2}\phi(xy)dy. $$ This yields the identical conclusion as in the other answers that $\lim_{x\rightarrow\infty}\phi(x)=0$.

A possible approach to solve this equation is by integration by parts and truncating at a given iteration. So, a first integration would give $$ \phi(x)=e^{-x}+\frac{\lambda}{x}\left[\left(\frac{\pi}{4}+k\pi\right)\phi(x)-k\pi\phi(0)\right]+R(x) $$ being the remainder $$ R(x)=-\lambda\int_0^1 dy\arctan y\phi'(xy) $$ to be evaluated after solving the above algebraic equation. The improved correction is given through a first order ODE and so on.

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Any solution $\phi$ converges to zero at infinity.

To see this, note that $\displaystyle\int_0^x\frac{\mathrm dt}{x^2+t^2}\leqslant x^{-1}$ for every $x\gt0$ hence, for every solution $\phi$ and every $x\gt0$, $|\phi(x)|\leqslant\mathrm e^{-x}+\lambda x^{-1}\|\phi\,\mathbf 1_{[0,x]}\|_\infty$. Since $\mathrm e^{-x}\to0$ and $\lambda x^{-1}\lt1$ for every $x$ large enough, this shows that $|\phi(x)|\leqslant\frac12\|\phi\,\mathbf 1_{[0,x]}\|_\infty$, say, for every $x$ large enough. Hence, $\|\phi\,\mathbf 1_{[0,x]}\|_\infty$ stays constant after a while, that is, $\|\phi\|_\infty$ is finite.

Now, $|\phi(x)|\leqslant\mathrm e^{-x}+\lambda x^{-1}\|\phi\|_\infty\to0$ when $x\to+\infty$, hence $\phi(x)\to0$ when $x\to+\infty$. (And more is true, for example, $x\mapsto x\phi(x)$ is bounded.)

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If $\varphi$ has a limit in $+\infty$, then I think this limit is $0$. To prove it, let me consider $I(x)=\displaystyle\int_0^x\frac{1}{x^2+t^2}\varphi(t)dt$, and make the change of variable $u = t/x$:

$I(x) = \displaystyle\int_0^x \frac{1}{x^2}\frac{1}{1+(\frac{t}{x})^2}\varphi(t) dt = \frac{1}{x}\int_0^1\frac{1}{1+u^2}\varphi(ux)du$.

Now assume that $\displaystyle\lim_{x\rightarrow +\infty}\varphi(x) = \alpha$.

By the theorem of the dominated convergence, we have $\displaystyle\lim_{x\rightarrow +\infty} I(x)= 0$. (because $\int_0^1\frac{1}{1+u^2}\varphi(ux)du \rightarrow \alpha\arctan(1)$)

So we have $\alpha = \displaystyle\lim_{x\rightarrow +\infty}\varphi(x) = 0+\lambda\times 0 = 0$.

This doesn't provide a closed formula for $\varphi$, but proves that if it has a limit, then this limit is $0$.

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@Sasha Well, this is due to the dominated convergence theorem explained here: . Because $\varphi$ has a limit in $+\infty$, I verify all the conditions of the theorem. And anyway it's easy to prove that this quantity is bounded, and thus $I(x)$ goes to $0$. – S4M Jul 26 '12 at 12:36

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