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I have the following problem:

Be the equation:

$$u(x,t)=\int_{-\infty}^{\infty}c(w)e^{-iwx}e^{-kw^2t}dw$$

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.

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  • $\begingroup$ what are the hypotheses on $c(w)$? $\endgroup$ – V. Moretti May 11 '14 at 19:07
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    $\begingroup$ 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
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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 $|x|\neq 0$, $$|u(x,t)| \leq \frac{K}{|x|}$$

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

i.e.

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

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.

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  • $\begingroup$ 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
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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..

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