# Physical intuition on the integral contained in D'Alembert's Formula for the wave equation

If $\phi(t,x)$ is a solution to the one dimensional wave equation and if the initial conditions $\phi(0,x)$ and $\phi_t(0,x)$ are given, then D'Alembert's Formula gives

$$\phi(t,x)= \frac 12[ \phi(0,x-ct)+ \phi(0,x+ct) ]+ \frac1{2c} \int_{x-ct}^{x+ct} \phi_t(0,y)dy . \tag{1}$$

Letting $g(x)=\phi(0,x)$ and $h(x)= \phi_t(0,x)$ (with $c=1$, so $ct=t$) this is commonly written

$$\phi(t,x)= \frac 12[ g(x-t)+ g(x+t) ]+ \frac12 \int_{x-t}^{x+t} h(y)dy . \tag{2}$$

My question:
What is the physically intuitive meaning of the integral term?

For example, why does $\phi_t$ show up in the integral whereas $\phi$ shows up in the forward and backward waves? Why does the integral have those specific limits of integration (and region of integration) for $h$ whereas $g$ uses only the end points, $x+ct,x-ct$? Are there examples with specific functions for $\phi$ that would help to understand this?

References:

math.ualberta.ca 337week0405.pdf after equation 180.

stanford univ waveequation3.pdf page 4 Lemma 3 and page 5.

math.nist.gov evolution.pdf page 537

univ. of penn. m425-dalembert-2.pdf first three pages.

univ. of ill. at urbana 286-dalemb.pdf

"Generalized Functions, vol 1", I. M. Gelfand, G. E. Shilov, page 114

"Mathematics for the Physical Sciences", L. Schwartz, pages 253-257

"The Mathematical Theory of Wave Motion", G. R. Baldock, T. Bridgeman, pages 40-45

From these, for example, I know that for a string $\phi_t(0,y)$ represents the velocity at time zero, but why physically (and intuitively) does it end up in the integral. Or I know $(x+ct),(x-ct)$ define the edges of a cone with vertex at $(x,t)$ which form the boundary of the region of the argument of $h$ where $h$ can effect $\phi(t,x)$, but why does the integral have those specific limits of integration for $h$ whereas $g$ uses only the end points, $x+ct,x-ct$?

• What do you mean by "what is the physical meaning of...": it's a differential equation, and being $\phi_t(0,x)$ a derivative of course it has to be integrated back to verify the initial conditions. – gented Jul 13 '16 at 19:40
• @Gennaro; Why physically does the solution depend on the integral of all the values of $\phi_t(0,x)$ lying between x-1 and x+1? I understand that mathematically its required to satisfy the initial conditions. – user45664 Jul 13 '16 at 22:15
• Well, since the variables in the functions are $(x-t)$ and $(x+t)$ they obviously appear as boundary values. If you follow the derivation from the link you posted yourself it is straightforward. – gented Jul 13 '16 at 23:28
• I think the downvoters were thinking "of course there's no meaning, it's just a horrible equation." I wish people didn't do that -- there's a physical explanation for just about everything! – knzhou Jul 17 '16 at 21:35
• Related question by OP physics.stackexchange.com/q/268740 – user45664 Jul 23 '16 at 22:56

The D'Alembert solution has a simple interpretation. Using your notation, it reads $$\phi(x, t) = \frac{g(x-t) + g(x+t)}{2} + \frac12 \int_{x-t}^{x+t} h(x') dx'.$$ where $g(x)$ is the initial position, $h(x)$ is the initial velocity, and $v = 1$.

Mathematically, the first term above solves the wave equation with initial position $g(x)$ but zero initial velocity, while the second term does the same for zero initial position and initial velocity $h(x)$. By the superposition principle, their sum has initial position $g(x)$ and initial velocity $h(x)$, and is hence equal to $\phi(x, t)$ for all times.

We can understand each of these individual terms by physical intuition.

First consider the term for initial position $g(x)$. We know all solutions of the wave equation are a linear combination of functions of the form $f(x\pm t)$, so the only things we can use are $g(x+t)$ and $g(x-t)$, which both have the right initial position. Finally, we notice that averaging them produces zero initial velocity by the chain rule. Intuitively, if you hold a string and let go, you will make equal-sized waves going in both directions.

Next consider the term for initial velocity $h(x)$. To understand it, consider the following simpler question: suppose $h(x)$ is zero everywhere except for a sharp spike at $x = 0$, corresponding to us hitting the string there at time $t = 0$. What is the resulting shape?

Physically, we expect a 'shockwave' to propagate outward from this event. Solving the wave equation using a similar technique to the one in the previous paragraph, we find $$y(x, t) = \frac12\left( \theta(x+t) - \theta(x-t) \right).$$ That is, everything inside the "light cone" of $(x, t) = (0, 0)$ has been raised up by $1/2$.

For a general $h(t)$, then, the solution is to integrate this $1/2$ over all light cones that could have affected the point $(x, t)$. The limits of this light cone are $x-t$ and $x+t$, yielding the term $$\frac12 \int_{x-t}^{x+t} h(x') dx'$$ in agreement with the formula.

• is there not only one light cone? -- the backward one with the vertex at the point (x,t). I am still trying to understand the physical geometry. If there is only one, then is the integral taken over the segment of the x axis lying between (x-1) and (x+1)? :) – user45664 Jul 17 '16 at 20:53
• @user45664 Ah, I should have phrased this better. You can think of a general initial velocity as the sum of individual spikes at every point. Each of those spikes has an associated light cone. So the integral in my answer is over different light cones that contribute to a single point, not over the interval corresponding to a single light cone. – knzhou Jul 17 '16 at 20:58
• Still digesting this (think I understand what you just said). if in the first comment I said (x',t') instead of (x,t), then would there only be one light cone (with vertex at x',t')? I am thinking of the light cone originating at the 'observation point' (x',t') rather than at the points along the x axis which carried the initial t=0 velocity (ie. the support of h). I am looking at a figure in "Mathematics for the Physical Sciences", Laurent Schwartz, p.254 which shows that light cone :) (: – user45664 Jul 17 '16 at 21:26
• @user45664 You're right, the two approaches are equivalent! In either case, though, the physical picture is the same: the integral is over the "causal past" of (x,t). – knzhou Jul 17 '16 at 21:32
• Need to work on this a bit, I assume your spike can be modeled as $\delta(x)\delta(t)$? will be back later :) – user45664 Jul 17 '16 at 21:51