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d’Alembert’s solution of one-dimensional wave
equation
A. Eremenko
January 26, 2021
- Solution on the line
Problem. To find the general solution of the one-dimensional wave equa- tion on the whole line, utt = c^2 uxx. (1)
Solution (due to d’Alembert). Let us introduce new independent variables:
ξ = x + ct, η = x − ct,
so that x = (ξ + η)/ 2 , t = (ξ − η)/(2c).
Then the new function is
w(ξ, η) = u(x, y) = u((ξ + η)/ 2 , (ξ − η)/(2c)).
To obtain a differential equation for w, we differentiate with respect to ξ:
wξ =
ux
( ξ + η 2
ξ − η 2 c
)
2 c
ut
( ξ + η 2
ξ − η 2 c
) ,
and then differentiate with respect to η:
wξη =
uxx −
4 c
uxt +
4 c
utx −
4 c^2
utt.
Since uxt = utx the two middle terms cancel, and we obtain from (1) that
wξη = 0. (2)
This is easy to solve: wξ must be independent of η, so wξ = f 1 (ξ) for some function f 1 , and integrating this with respect to ξ we obtain that w(ξ, η) = f (ξ) + g(η), for some function f , such that f ′^ = f 1 , and some function g, the “constant of integration”. Returning to our original variables we obtain
u(x, t) = f (x + ct) + g(x − ct), (3)
Now it is easy to check directly that such a function u with arbitrary differ- entiable functions f and g satisfies equation (1). So we obtained a general solution which depends on two arbitrary functions.
Equation (1) describes oscillations of an infinite string, or a wave in 1- dimensional medium. To single out a unique solution we impose initial con- ditions at t = 0: u(x, 0) = φ(x) ut(x, 0) = ψ(x), (4)
that is we specify the initial position and initial velocity of the string. To simplify our computation, we can use the Superposition Principle: first find a solution with arbitrary given φ and ψ = 0, then find a solution with φ = 0 and arbitrary ψ, and then take the sum of these two solutions. For the first solution we plug t = 0 into (3) and obtain
f + g = φ, f − g = 0,
so f = g = φ/2, and the first solution, with zero initial velocity, is
u 1 (x, t) =
(φ(x + ct) + ψ(x − ct)). (5)
For the second solution we differentiate (3) with respect to t and plug t = 0. We obtain f + g = 0, c(f ′^ − g′) = ψ.
Solving this we obtain the second solution
u 2 (x, t) =
2 c
∫ (^) x+ct
x−ct
ψ(y)dy, (6)
corresponding to zero initial position. Thus the complete solution u of the initial value problem (1), (4) is given by
u(x, t) = u 1 (x, t) + u 2 (x, t) =
(φ(x + ct) + φ(x − ct)) +
2 c
∫ (^) x+ct
x−ct
ψ(y)dy.
Why this extension is works? The even extension of a smooth function with φ(0) = 0 will in general be non smooth at 0. But the main point is that every odd 2 L-periodic function is zero at the points 0 and L! So if u(x, t) remains odd and 2L-periodic with respect to x at all times t, then the boundary condition will be automatically satisfied. Check that if φ is continuous and differentiable on [0, L], and satisfies (9), then the proposed odd 2L periodic extension will be also continuous and differentiable. Suppose now that φ is an odd 2L periodic function on the whole real line, and consider the function u defined by (5). It is also an odd 2L-periodic function of x for each fixed t That it is 2L periodic is evident. To check that it is odd we write
u(−x, t) = (1/2) (φ(−x + ct) + φ(−x − ct)) = −(1/2) (φ(x + ct) + φ(x − ct)) = −u(x, t),
where we used that φ is odd. Now it remains to notice that any odd 2L-periodic function f satisfies f (0) = 0 and f (L) = 0.
Thus we obtain a unique solution of (1) on [0, L] with boundary condition (7) and initial condition (9): it is given by formula (5) where φ is the odd 2 L-periodic extension of the initial shape.
It is very instructive to look what really happens in a series of pictures showing the shape of the string at various moments. Let us take some very simple initial shape φ, say constant on a small interval near L/4 and zero outside a slightly bigger interval. I urge you to make few pictures!
Initially, for small times t, the original shape splits into two equal halves, and they start moving at speed c into opposite directions. But look carefully what happens when the bump reaches the left end x = 0. Since we made an odd 2L-periodic extension, a similar negative bump comes to 0 from the negative side. In the next moments, these two bumps interfere and eventually may cancel each other, so at some moment the solution is zero near the end x = 0. And in the next moments the negative bump re-appears at the left end
of the string and moves to the right. This is called “reflection of the wave” at the end. Same thing happens a bit later on the right end. Eventually the two negative reflected bumps will collide near the place where we had a positive bump at t = 0. Then they pass each other and this motion continues indefinitely.
Example. Suppose that c = 1 in (1), and we are solving (1) on the interval [− 1 , 1] with the boundary conditions
u(− 1 , t) = u(1, t) = 0
and the initial conditions are
u(x, 0) = φ :=
{ 1 − |x|, |x| < 1 , 0 , |x| ≥ 1 ,
and ut(x, t) = 0. Find u(0, 2021) and u(0, 2020).
Solution. By d’Alembert’s formula, u(x, t) = ( φ˜(x + t) + φ˜(x − t))/2, where φ˜ is the 2-periodic extension of φ. So
u(0, 2021) = ( φ˜(−2021) + φ˜(2021))/2 = (φ(1) + φ(−1))/2 = 0.
and similarly u(0, 2020) = 1.
Remarks. d’Alembert’s formula explains why we can communicate using light or sound. Suppose that at the time t = 0 at the place x = 0 someone makes a click. We describe this click by some function φ(x) which is localized near x = 0: φ(x) = 0 for |x| > , where is a small number, like in the previous example. If you are sitting at some point x = A > 0, first you hear nothing. But at the time t = A/c the wave φ(x − ct) moving with speed c to the right reaches you and passes. So you hear a click. It is important that the click arrives to the point A at time A/c undistorted. It is just 2-ce smaller that the original click. So you can hear the original click, only with a time delay. Similarly if we are talking about electromagnetic waves, you can see a flash as a flash. A localized signal (a flash) arrives to your place with the speed of light undistorted. All this applies to one space dimension while we are living in three di- mensions. It turns out that in dimension 3 there is a formula for the solution
