I have the following problem:

Be the equation:


Show that $u\rightarrow 0$ as $x\rightarrow \infty$, even when $e^{-iwx}$ does not falter if $x\rightarrow \infty$.

The problem gives the hint to use integration by parts. I was hoping you explain this problem or help me solve it.

I asked this same question in Prove that $u(x,t)=\int_{-\infty}^{\infty}c(w)e^{-iwx}e^{-kw^2t}dw\rightarrow 0$ if $x\rightarrow \infty$, however, I found no answer. But I remembered that this equation is a solution of many physical systems, so I decided to put it here hoping for a little more luck.

  • $\begingroup$ what are the hypotheses on $c(w)$? $\endgroup$ – V. Moretti May 11 '14 at 19:07
  • 4
    $\begingroup$ This seems to me like a pure math question devoid of a physical context, which our help center specifies is off topic... if so I'll just send it back to Mathematics. Thoughts from anyone else? $\endgroup$ – David Z May 11 '14 at 20:55
  • $\begingroup$ Do it, I agree. I hope our answers will be moved there as well. $\endgroup$ – V. Moretti May 12 '14 at 6:30
  • $\begingroup$ proofwiki.org/wiki/Riemann-Lebesgue_Lemma $\endgroup$ – Kenshin May 12 '14 at 13:29

If $\mathbb R \ni w\mapsto c(w)e^{-kw^2t}$ is $L^1(\mathbb R)$, the result you want is a trivial consequence of Riemann-Lebesgue theorem (also known as Riemann-Lebesgue lemma).

However some direct proof can easily be produced under fair hypotheses on $c$. For instance, if $\mathbb R \ni w \mapsto c(w)$ is $C^1$ with support in, say $[a,b]$, where $a, b$ are finite, you can write $$x u(x,t)=\int_{-\infty}^{+\infty}c(w)x e^{-iwx}e^{-kw^2t}dw = \int_{-\infty}^{\infty}c(w)x \left(i\partial_w e^{-iwx}\right) e^{-kw^2t}dw = -i\int_{a}^{b} e^{-iwx}\partial_w \left(c(w)e^{-kw^2t}\right) dw + \mbox{vanishing boundary terms}.$$ Finally, since $c$ and its derivative are supported in $[a,b]$ you find: $$|x u(x,t)|=\left|\int_{a}^{b} e^{-iwx}\partial_w \left(c(w)e^{-kw^2t}\right) dw\right| \leq \int_a^b (M + |kt| N) dw = K <+\infty \quad (1)$$ where $M,N,K\geq 0$ are finite constants depending on $a,b, k,t$. Since $K$ in the right-hand side of (1) does not depend on $x$, as $|x| \to + \infty$ in the left-hand side, you should also have $|u(x,t)|\to 0$, for every fixed $t$ and $k$. More precisely, for $|x|\neq 0$, $$|u(x,t)| \leq \frac{K}{|x|}$$


This to me looks as if $c(w)$ is the Fourier transformation of some initial condition of a diffusion equation. Then intuitively the claim holds: When the diffused distribution "reaches" infinity, it will be infinitely dilute.

First of all, to substantiate my claim of diffusion, consider the diffusion (heat) equation $$\frac{\partial u}{\partial t} = c\frac{\partial^2 u}{\partial x^2}$$ Take the Fourier transformation of this and solve the resulting first order diff. equation: $$\hat{u}(k, t) = \hat{u}_0(k) \exp(-c k^2 t)$$ Transforming to real space, $$u(x,t) = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{u}_0(k) \exp(-c k^2 t) \exp(i k x)dk$$

Write the initial condition back in real space: $$u(x,t) = \frac{1}{2\pi}\int_{-\infty}^\infty \int_{-\infty}^\infty u_0(y)\exp(-iky)dy \exp(-c k^2 t) \exp(i k x)dk$$

Switch the order of integration $$u(x,t) = \frac{1}{2\pi}\int_{-\infty}^\infty u_0(y) \int_{-\infty}^\infty \exp(-iky -c k^2 t + i k x) dk dy$$

Fill the square, $$u(x,t) = \frac{1}{2\pi}\int_{-\infty}^\infty u_0(y) \int_{-\infty}^\infty \exp\left(-\left(i\sqrt{ct}k - \frac{y-x}{2\sqrt{ct}}\right)^2 - \frac{(y-x)^2}{4ct}\right) dk dy$$


$$u(x,t) = \frac{1}{2\pi}\int_{-\infty}^\infty u_0(y) \exp\left( - \frac{(y-x)^2}{4ct}\right) \int_{-\infty}^\infty \exp\left(-\left(i\sqrt{ct}k - \frac{y-x}{2\sqrt{ct}}\right)^2\right) dk dy$$

I think we can now recognize that the Gaussian should come out independent of $x$ and $y$ as $\sqrt{\pi/ct}$, and thus we have $$u(x,t) = A \int_{-\infty}^\infty u_0(y) \exp\left(- \frac{(y-x)^2}{4ct}\right) dy$$

i.e. convolution of the initial condition with a Gaussian distribution (that it should take this form is not surprising, for the propagator of diffusion is Gaussian). It is then obvious that if $u_0(y)$ has a finite support, that at infinity $u$ tends to zero.

So what I basically did was to Fourier transform $c(w)$ and change the order of some integrals.

This is not mathematically very rigorous even if we suppose I didn't make mistakes (although I guess it can be made so), but this is the physics stackexchange, not the maths one, after all. V. Moretti's answer was very impressive indeed, but I thought I'd write one up with maybe some more physical intuition behind it.

  • $\begingroup$ I agree with your final comment. My answer was purely mathematical but here a more physical approach should be desirable. $\endgroup$ – V. Moretti May 12 '14 at 10:02

I started trying to find a first order differential equation for $$u(x,t),$$ such that for instance..

$$\frac{\partial u}{\partial x} = \int c(w)( -iw )e^{-iwx}e^{-kw^2t}dw = i\int c(w)(- w) e^{-iwx}e^{-kw^2t}dw = i\int c(w) e^{-iwx} (1/2kt) \frac{\partial}{\partial w}(e^{-kw^2t})dw = -i(1/2kt)\int \frac{\partial}{\partial w}(c(w)e^{-iwx}) e^{-kw^2t}dw = -(\frac{x}{2kt})u(x,t) - \frac{i}{2kt}\int \frac{\partial}{\partial w}(c(w)) e^{-iwx} e^{-kw^2t}dw =$$

if c is constant that means $$ u(x,t) = u(x=0,t)e^{-x^2/4kt} $$. Do not know if this can help you..


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.