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I'm having troubles with understanding the physical meaning of a certain transform.

If $u$ is a solution to the wave equation $$\partial_t^2u-\Delta u=0\ \mathrm{in}\ \mathbb{R}^n\times(0,\infty)\\u=g,\ \partial_tu=0\ \mathrm{on}\ \mathbb{R}^n\times\{0\}$$ then $$v(x,t)=\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^\infty{\exp\left(-\frac{s^2}{4t}\right)u(x,s)\,\mathrm{d}s}$$ solves the heat equation $$\partial_tv-\Delta v=0\ \mathrm{in}\ \mathbb{R}^n\times(0,\infty)\\v=g\ \mathrm{on}\ \mathbb{R}^n\times\{0\}$$

What is the physical interpretation of this behaviour?

And why is that so?

Andrews
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Maximilian M.
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  • That Gauss factor in $v$ is almost a fundamental solution of the heat equation. But I can not use that to help here, either I get it wrong or there are mistakes in the above. No I fear you have to calculate $v_t$ and $\Delta v$ which leads to some expressions with $u_t$ and $\Delta u$ and then use $\Delta u = u_{tt}$. Etc. – mvw Jun 30 '15 at 19:24
  • Yes, you can use this to show that the above is actually true. I got it from the book "Partial Differential Equations" from L.C.Evans in the chapter on Laplace transform, where he uses this approach to find Kirchhoff's formula. However he gives no motivation to this Ansatz at all. – Maximilian M. Jun 30 '15 at 20:23
  • This looks like an interesting twist on the angular spectrum representation of solutions to the wave equation. At least within angular spectrum, though, the physical interpretation is that any propagating wave is simply a sum over plane waves at all angles (including evanescent waves). In this case, though, what is $s$? An angle? Something else? – Ron Gordon Jan 28 '16 at 16:17

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Remark

This is actually more general than it may seem. Consider an equation $$Tf-\Delta f=0\tag1$$ where $T$ is an operator acting on the time variable. (For instance if $T=\partial_t^2+2\partial_t$, the equation is called the telegrapher's equation). Let $h$ be the Green's function of the equation (1) in one dimension, where we call $s$ the "space" coordinate: $$ Th(s,\,t)=\partial^2_sh(s,t),\qquad h(s,0)=\delta(s). $$ We can define $$v(x,t)=\int h(s,t)u(x,s)\mathrm ds$$ where $u$ is a solution of the wave equation, $\partial_t^2u-\Delta u=0$, with initial condition $g(x)$. Note that $v(x,0)=u(x,0)=g(x)$ and $$T v(x,t)=\int T h(s,t)u(x,s)\mathrm ds=\int \partial^2_s h(s,t)u(x,s)\mathrm ds.$$ One can now integrate twice by parts, with respect to the variable $s$. If $h$ and $\partial_xh$ goes to $0$ at infinity (this is required by causality in physics), we obtain $$Tv(x,t)=\int h(s,t)\partial_s^2 u(x,s)\mathrm ds=\int h(s,t)\Delta_x u(x,t)=\Delta_x v(x,t).$$ Consequently, it follows that $v$ satisfies the equation (1) and was constructed using a solution a the wave equation with the same initial condition. You should note that one can replace $\Delta$ by any polynomial in $\Delta$, this would still work.

Attempt for a physical interpretation

As you observed, provided that an equation possesses a Green's function, the same construction can be done. The fundamental property that makes this work is the fact that the wave equation is no more that a change of reference frame.

The solution to the wave equation describes, in our reference frame, how some disturbance propagates in a medium. But if we move together with this disturbance, at the same speed, we will observe it as static. Physical equations describing another phenomenon, like in your case heat transfer or diffusion, should not be changed in another referential. This principle is called the relativity principle in Galileo's sense. The relativity principle states that equations of physics must satisfy the same equations in reference frames under uniform translation with respect to each other. Equation (1) describes therefore acceptable equations for physical phenomena. They all have the same property as the one you discovered for the heat equation.

All of this would be nice if there were no limit for the speed. As Einstein first understood, this is not the case and physical equations must be preserved under Galileo's transform only for speed small compared to $c$, the speed of light in vacuum. But this is another story.

Tom-Tom
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    Nice. It's really great to see this in a more general setting. Your interpretation also gives a nice explanation as to why the dependencies are $h(s,t)\cdot u(x,s)$ when we see ourselves moving with the wave. And (maybe too figuratively spoken) by integration we just travel through the whole system like this. – Maximilian M. Feb 03 '16 at 23:29
  • I'm a little confused by the dimensions here, since $x \in \mathbb{R}^n$ but $s \in \mathbb{R}$. When you say “Let $h$ be the Green's function of this equation”, do you mean the equation $Tf-f_{xx}=0$ with $x \in \mathbb{R}$ (rather than $Tf-\Delta x=0$ with $x \in \mathbb{R}^n$)? And in the formula where you write $\Delta h(s,t)$, do you mean $h_{ss}(s,t)$? – Hans Lundmark Apr 02 '23 at 17:16
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    @HansLundmark Thanks for your comment. Yes, $\Delta h$ is indeed $\partial_{ss}h$. I agree that the definition of $h$ was unclear. I have improved it and also clarified the notations. This will hopefully clarify this point. – Tom-Tom Apr 06 '23 at 17:02