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Suppose that for a given $b\in \mathbb{R}$ \begin{align} 0= \int_{-\infty}^\infty e^{-\frac{(bt+\omega)^2}{2}} f(t+\omega) \frac{1}{i t} dt, \forall \omega \in \mathbb{R} \end{align} where $i =\sqrt{-1}$.

Question: How to find a set of general solutions to this equation? I tried to do the Fourier inversion but things didn't work out.

Few details:

  • the integral above is performed in a sense of Cauchy principal value.
  • Note that the division by the imaginary number is not necessary. However, I keep it so that the final solution is real-valued. (At least I think it guarantees that). One can sertaily remove it.
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  • $\begingroup$ how will you avoid the pole at $t=0$? is this a principal value? you may want to explain where the integral comes from, as it stands it is not well-defined (and since the left-hand-side is 0, what is the meaning of the factor $i\pi$ on the right-hand-side?) $\endgroup$ Mar 7, 2022 at 21:23
  • $\begingroup$ @CarloBeenakker Ok. I have add it. $\endgroup$
    – Boby
    Mar 7, 2022 at 21:35
  • $\begingroup$ @CarloBeenakker. Essentially it originate from Fourier transform of $sign(t)$ $\endgroup$
    – Boby
    Mar 7, 2022 at 21:44
  • $\begingroup$ my mistake, I misread sign for step function; I'll delete this string of comments that confused you, apologies. $\endgroup$ Mar 7, 2022 at 22:02
  • $\begingroup$ @CarloBeenakker No problem. You seem to know these things well. Any idea of how to approach this? $\endgroup$
    – Boby
    Mar 7, 2022 at 22:03

2 Answers 2

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It may be helpful to rewrite this in a way that avoids the principal value: $$0=\int_{-\infty}^\infty e^{-(bt+\omega)^2/2} f(t+\omega) \frac{2}{i t} dt=\int_{-\infty}^\infty dt \int_{-\infty}^{\infty} dx\,e^{-(bt+\omega)^2/2} f(t+\omega) \,\text{sign}(x) e^{-ixt},$$ then define $g_\omega(t)=e^{-(bt+\omega)^2/2}f(t+\omega)$ with Fourier transform $G_\omega(x)=\int_{-\infty}^\infty g_\omega(t)e^{-ixt}\,dt$, and arrive at $$0=\int_{-\infty}^\infty dx\,G_\omega(x)\,\text{sign}(x),\;\;\forall\omega\in\mathbb{R}.$$

For $b=1$ we have the identity $g_\omega(t)=g_0(t+\omega)$, hence $G_\omega(x)=e^{i\omega x}G_0(x)$. Since the Fourier transform only vanishes identically if the function itself vanishes, we must have $G_0(x)\,\text{sign}\,(x)\equiv 0\Rightarrow G_0(x)=\text{constant}\times\delta(x)$, hence $f(t)=\text{constant}\times e^{t^2/2}$. Similarly, for $b=0$ the only solution is $f(t)=\text{constant}$.


For the case of general $b$ I could proceed as follows; substitute $f(t)$ by $$f(t)=e^{\frac{1}{2} (1-b)^2 \omega^2} e^{(1-b) b \omega t} e^{\frac{1}{2} (bt)^2}h(t).$$ Then one has $$g_\omega(t)\equiv e^{-(bt+\omega)^2/2}f(t+\omega)=h(t+\omega).$$ So we are back to case we studied earlier, and we can conclude that $h(t)=\text{constant}$. We thus arrive at the general solution $$f(t)=\text{constant}\times e^{(1-b) b \omega t} e^{\frac{1}{2} (bt)^2}.$$ Note that the earlier special cases $b=0$ and $b=1$ are recovered. Also check that $$\int_{-\infty}^\infty e^{-(bt+\omega)^2/2} f(t+\omega) \frac{2}{i t} dt=\text{constant}\times\int_{-\infty}^\infty dt/t=0,$$ if I regularize the singular integral as $\int_{-\infty}^\infty dt\equiv\lim_{\beta\rightarrow\infty}\lim_{\alpha\rightarrow 0}\left(\int_{-\beta}^{-\alpha}dt+\int_{\alpha}^\beta dt\right)$.

Since this solution is $\omega$-dependent it is not a useful answer. I am inclined to think there is no $\omega$-independent solution for $b\neq 0,1$.

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  • $\begingroup$ Thanks. Any other ideas or places I can look would be greatly appreciated. $\endgroup$
    – Boby
    Mar 8, 2022 at 2:57
  • $\begingroup$ But, if $constant\ne0$ and $b=0$, the integral equals $\infty$ (not $0$) for your general solution. $\endgroup$ Mar 14, 2022 at 16:30
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    $\begingroup$ I don't think so: when $b=0$ and $f(t)=\text{constant}$ we have the principal value integral $\int_{-\infty}^\infty dt/t$ which vanishes --- when interpreted as $\lim_{b\rightarrow\infty}\lim_{a\rightarrow 0}\left(\int_{-b}^{-a}dt/t+\int_{a}^{b}dt/t\right)$ --- my understanding is that this is how the OP wishes to interpret the singular integral $\endgroup$ Mar 14, 2022 at 16:33
  • $\begingroup$ Thanks. Quick question. Shouldn't $f(t)$ be only a function of $t$? Here it is also a function of $\omega$. $\endgroup$
    – Boby
    Mar 14, 2022 at 16:44
  • $\begingroup$ yes, it is also a function of $\omega$, I don't think there is an $\omega$-independent solution for $b\neq 0,1$, but you're right, this is not a useful solution. $\endgroup$ Mar 14, 2022 at 16:47
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$\omega$-dependent solutions are not interesting, since after the substitution $t=x-\omega$ $$0=PV\int_{-\infty}^\infty e^{-\frac{\left(b(x-\omega)+\omega\right)^2}{2}} \frac{f(x)}{i(x-\omega)} \, {\rm d}x \, , \tag{1}$$ you could just set $$f(x) \equiv e^{\frac{\left(b(x-\omega)+\omega\right)^2}{2}} i(x-\omega) h(x)$$ and you would have $$0=PV\int_{-\infty}^\infty h(x) \, {\rm d}x$$ which certainly has many solutions for $h$. Therefore we can assume $f$ to not depend on $\omega$ and define $$f(x) \equiv e^{\frac{b^2x^2}{2}} h(x)$$ after which (1) simplifies to $$0=PV\int_{-\infty}^\infty e^{\omega x b(b-1)} \frac{h(x)}{i(x-\omega)} \, {\rm d}x$$ where the constant, $\omega$-dependent, factor was removed.

Now we assume $h$ to be entire and use contour integration to get $$0=\oint_{-\infty}^\infty e^{\omega z b(b-1)} \frac{h(z)}{i(z-\omega)} \, {\rm d}z \\ = PV\int_{-\infty}^\infty e^{\omega x b(b-1)} \frac{h(x)}{i(x-\omega)}\, {\rm d}x - \pi h(\omega) \, e^{\omega^2 b(b-1)} + \lim_{R\rightarrow\infty} \int_0^\pi e^{\omega Re^{it} b(b-1)} \frac{h\left(Re^{it}\right)}{1-\frac{\omega}{R}e^{-it}} \, {\rm d}t \, .$$ The first line follows by Cauchys theorem and the integral is closed along the semi-circle in the upper half-plane. By the assumption of the vanishing PV-integral it follows $$\pi h(\omega) \, e^{\omega^2 b(b-1)} = \lim_{R\rightarrow\infty} \int_0^\pi e^{\omega Re^{it} b(b-1)} \frac{h\left(Re^{it}\right)}{1-\frac{\omega}{R}e^{-it}} \, {\rm d}t \quad , \quad \forall \omega \in {\mathbb{R}} \, .$$

Now we expand both sides as a power-series in $\omega$. For the LHS we obtain $$\pi h(\omega) e^{\omega^2 b(b-1)} = \pi \sum_{n,m=0}^\infty c_n \omega^n \frac{\left(\omega^2b(b-1)\right)^m}{m!} = \pi \sum_{k=0}^\infty \omega^k \sum_{\substack{n,m=0 \\ n+2m=k}}^\infty \frac{c_n \left(b(b-1)\right)^m}{m!} \, .$$ Similarly the RHS becomes $$\lim_{R\rightarrow \infty} \sum_{n,m,k=0}^\infty c_n R^{n-k+m} \omega^{k+m} \frac{\left(b(b-1)\right)^m}{m!} \int_0^\pi e^{it(n-k+m)} \, {\rm d}t \\ =\lim_{R\rightarrow \infty} \sum_{l=0}^\infty \omega^l \sum_{\substack{n,m,k=0 \\ k+m=l}}^\infty c_n R^{n-k+m} \frac{\left(b(b-1)\right)^m}{m!} \\ \times \left( \pi \left[ n-k+m=0 \right] + 0 \left[n-k+m | \text{even} \right] + \frac{2i}{n-k+m} \left[n-k+m | \text{odd} \right] \right) \\ =\pi \sum_{l=0}^\infty \omega^l \sum_{\substack{n,m=0 \\ n+2m=l}}^\infty \frac{c_n\left(b(b-1)\right)^m}{m!} + 2i \lim_{R\rightarrow \infty} \sum_{l=0}^\infty \omega^l \sum_{\substack{n,m,k,p=0 \\ k+m=l \\ n-k+m=2p+1}}^\infty \frac{c_n R^{n-k+m} \left(b(b-1)\right)^m}{(n-k+m)m!} \, .$$ Because of the limit, $p$ starts at $0$. Now equating the expressions found for the LHS and RHS and canceling common terms, it follows that each $\omega^l$-coefficient has to vanish identically $$0 = \lim_{R\rightarrow \infty} \sum_{\substack{n,m,p=0 \\ n+2m=l+2p+1}}^\infty \frac{c_n R^{2p+1} \left(b(b-1)\right)^m}{(2p+1)m!} \\ = \lim_{R\rightarrow \infty} \sum_{p=0}^\infty \sum_{0\leq 2m \leq l+2p+1} \frac{c_{l+2p+1-2m} R^{2p+1} \left(b(b-1)\right)^m}{(2p+1)m!} \quad , \quad \forall l\in\mathbb{Z}_{\geq 0} \, . \tag{2}$$ If $b=0$ or $b=1$, the $m$-sum disappears except for $m=0$ and so $$\lim_{R\rightarrow \infty} \sum_{p=0}^\infty \frac{c_{l+2p+1} R^{2p+1}}{(2p+1)} = 0 \quad , \quad \forall l \in \mathbb{Z}_{\geq 0} \, . \tag{3}$$

Since the limit must exist for all $l\geq 0$, this is only possible if $$c_k =0 \quad , \quad \forall k \in \mathbb{N}$$

Note however, that in this case, there is no condition on $c_0$. Therefore $h(x)=\text{ const.}$, depending neither on $x$ nor on $\omega$.

In the case $b\neq 0$ and $b\neq 1$, the requirement (2) also forces $c_0=0$.


Some remarks to (3): If I set $$F_l(R)=\sum_{p=0}^\infty \frac{c_{l+2p+1} R^{2p+1}}{(2p+1)} \, ,$$ then (3) implies $$\lim_{R\rightarrow \infty} F_l'(R) = 0 \quad , \quad \forall l\in\mathbb{Z}_{\geq 0} \, .$$ From this a relationship of $F_l'(R)$ to $F_0'(R)$ (for even $l=2m$) or $F_1'(R)$ (for odd $l=2m+1$) can be found $$F_{2m,2m+1}'(R) = \frac{F_{0,1}'(R)}{R^{2m}} - \frac{1}{R^{2m}} \sum_{k=0}^{m-1} c_{2k+1,2k+2} R^{2k} \, .\tag{4}$$ Therefore, if a set of coefficients (not all zero) is found s.t. $F_{0,1}'(R)$ vanishes in the limit (e.g. $c_{2k+1,2k+2}=\frac{(-1)^k}{k!}$), then $F_l'(R)$ vanishes for all $l$. From this and (4) we can then derive a contradiction to (3), i.e. there exists at least one $l>1$ s.t. $$\lim_{R\rightarrow \infty} |F_l(R)|>0 \, .$$

Integrating (4) from $0$ to $\infty$ and assuming $F_{2m,2m+1}(R)$ vanishes in the limit $$0=\lim_{R \rightarrow \infty} F_{2m,2m+1}(R) = \int_{0}^\infty F_{2m,2m+1}'(r) \, {\rm d}r \\ = \int_{0}^\infty {\rm d}r \left( \frac{F_{0,1}'(r)}{r^{2m}} - \frac{1}{r^{2m}} \sum_{k=0}^{m-1} c_{2k+1,2k+2} r^{2k} \right) \, ,$$ we see that the second term under the integral is only the polynomial approximation to $F_{0,1}'(R)$. The latter vanishes asymptotically, while the polynomial approximation does not. Hence, you will generally find $m$ s.t. $$\left| \int_{0}^\infty \left( \frac{F_{0,1}'(r)}{r^{2m}} - \frac{1}{r^{2m}} \sum_{k=0}^{m-1} c_{2k+1,2k+2} r^{2k} \right) {\rm d}r \right| = \text{const.} \neq 0 \, .$$

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