0

I've been going through 'Detection, Estimation, and Modulation Theory 2e' by Van Trees et al and am having trouble with Problem 6.4.1:

Consider the integral equation $$\int_{-T}^{+T}\exp\left(-\alpha|t-u|\right)\phi(u)\mathrm{d}u=\lambda\phi(u).$$ a) Prove that $\lambda=0$ and $\lambda=2/\alpha$ are not eigenvalues.

I can't see the best way to do this. If we do a change of coordinates $\alpha u=u'$, we can re-write the left-hand side as $$\frac{1}{\alpha}\left[\int_{-\alpha T}^t\exp\left(-(t'-u')\right)\phi(u')\mathrm{d}u'+\int_t^{\alpha T}\exp\left(-(u'-t')\right)\phi(u')\mathrm{d}u'\right],$$ and it reduces to showing that the quantity in square brackets may not have eigenvalues $0$ or $2$, but I don't know how much that helps.

There is also a part b:

b) Prove that all values of $\lambda>2/\alpha$ cannot be eigenvalues of the integral equation.

but I am hoping that once I understand how to approach a), it will be clearer how to approach this.

Ruvi Lecamwasam
  • 2,002

1 Answers1

1

I will assume that $\alpha >0$ and we are looking for continuous solutions.

Let's define $f$ as $$f(x)=\exp(-\alpha|x|)$$ In this case, the equation can be written as $$(f*\phi [-T,T])(x)=\lambda \phi(x)[-T,T](x)$$ With arbitrary extension outside $[-T,T]$. If we make it $0$ there, we can abandon the indicator function, i.e.: $$(f*\phi)(x)=\lambda \phi(x)$$ If we are looking for continuous solutions (we can loosen this assumption), we can take the Fourier-transform of both sides: $$\hat{f}(\omega)\hat{\phi}(\omega)=\frac{2\alpha}{\alpha^2+\omega^2}\hat{\phi}(\omega)=\lambda \hat{\phi}(\omega)$$ I.e. $$\hat{\phi}(\omega)\left(\frac{2\alpha}{\alpha^2+\omega^2}-\lambda\right)=0$$ One of the terms need to be zero for all $\omega$. So let's look for the zeroes of the second expression: $$\frac{2\alpha}{\lambda}-\alpha^2=\omega^2$$ i.e. $$\omega = \pm \sqrt{\frac{2\alpha}{\lambda}-\alpha^2}$$ Which has a solution if $$\frac{2\alpha}{\lambda}-\alpha^2\geqslant 0$$ i.e. $$\lambda \leqslant \frac{2}{\alpha}$$ Buf if $\lambda =\frac{2}{\alpha}$, we only have one solution: $\omega=0$, so only $\hat{\phi}(0)$ can be non-zero.

To rule out the $\lambda=0$ possibility, we could use an orthonormal series in $L^2$ and derive that all of the coefficients of $\phi$ must be zero, but I wouldn't do it, because I don't like the absolute value in the integrals. But if we substitute $\lambda=0$ into the equation in the Fourier space, we get that $$\frac{2\alpha}{\alpha^2+\omega^2}=0$$ Which does not have a solution unless $\alpha=0$.

Edit: Clearly, there are some flaws in the answer, but I think the idea is not bad: the Fourier transform gives us an algebraic equation ehich reveals the possible eigenvalues. It also reveals that the solution (with the extension and cutoff) cannot be continuous, because the Fourier-equation gives us $2$ possible non-zero value, which is a measure-zero set (and the FT of a continuous function is continuous). The solution might need distributions and/or a better cutoff.

Botond
  • 11,938
  • Thanks very much, I think the most important point is there, that since there is a convolution we need to move to the Fourier basis, and the algebraic constraint you derive places the appropriate bounds on the eigenvalue. We do have this strange behaviour where $\lambda$ depends on $\omega$, but I think that's because $\hat{\phi}(\omega)$ is in fact a sum of $\delta$-functions ($\phi(t)$ is a cosine function), so $\lambda$ will end up being a constant. – Ruvi Lecamwasam Aug 06 '20 at 01:52
  • @RuviLecamwasam I think it's normal that $\lambda$ depends on $\omega$. In the case of the negative differential operator squared, the eigenvalue is the square root of the frequency of the solution (i.e. $u(t)=A\cos(\omega t)+B\sin(\omega t)$ satisfies $-D^2u=\omega^2 u$). – Botond Aug 06 '20 at 15:37