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d'Alembert's formula and uniqueness in one dimension

Statement

Let $c > 0$ , $u_{0} \in C^{2} (R)$ and $u_{1} \in C^{1} (R)$ . Then $u (x, t) : = \frac{1}{2} (u_{0} (x - c t) + u_{0} (x + c t)) + \frac{1}{2 c} \int_{x - c t}^{x + c t} u_{1} (y) d y$ is a $C^{2}$ function on $R \times [0, \infty)$ and is the unique classical solution of the homogeneous Cauchy problem $u_{t t} = c^{2} u_{x x}, u (\cdot, 0) = u_{0}, u_{t} (\cdot, 0) = u_{1} .$ Its value depends on $u_{0}$ only through the two endpoints $x \mp c t$ and on $u_{1}$ only through its integral over $[x - c t, x + c t]$ .

Facts & Assumptions

Given: a speed $c > 0$ , data $u_{0} \in C^{2} (R)$ , $u_{1} \in C^{1} (R)$ , and the displayed function $u$ .

[F1]

A continuous function on an interval has the primitive $P (z) = \int_{0}^{z} u_{1} (y) d y$ when the interval is $R$ ; since $u_{1} \in C^{1}$ , this primitive is $C^{2}$ , and $\int_{a}^{b} u_{1} = P (b) - P (a)$ (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ).

[F2]

If $f$ is totally differentiable at $a$ and $g$ at $f (a)$ , then $g \circ f$ is totally differentiable at $a$ with $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ). The required total differentiability follows from continuous coordinate partial derivatives (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[F3]

Sums, constant multiples, products of differentiable functions are differentiable with the usual rules (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[F4]

Every $C^{2}$ solution of $u_{t t} = c^{2} u_{x x}$ on a nonempty open rectangle has the form $F (x - c t) + G (x + c t)$ with $F, G$ of class $C^{2}$ on the projections, and conversely; the pair is unique up to $F \mapsto F + k$ , $G \mapsto G - k$ (General solution of the one-dimensional wave equation).

Proof

1.1F1F2F3algebra

Regularity and the equation. The first summand is $C^{2}$ because $u_{0} \in C^{2}$ , and the integral term is $(P (x + c t) - P (x - c t)) / (2 c)$ with $P \in C^{2}$ by [F1], so it is $C^{2}$ even across $t = 0$ by [F2]: $\partial_{t} u (x, t) = \frac{1}{2} (- c u_{0}^{'} (x - c t) + c u_{0}^{'} (x + c t)) + \frac{1}{2} (u_{1} (x + c t) + u_{1} (x - c t))$ and $\partial_{x} u (x, t) = \frac{1}{2} (u_{0}^{'} (x - c t) + u_{0}^{'} (x + c t)) + \frac{1}{2 c} (u_{1} (x + c t) - u_{1} (x - c t))$ . Differentiating once more by [F2] and [F3], $\partial_{t}^{2} u = \frac{c^{2}}{2} (u_{0}^{''} (x - c t) + u_{0}^{''} (x + c t)) + \frac{c}{2} (u_{1}^{'} (x + c t) - u_{1}^{'} (x - c t))$ and $c^{2} \partial_{x}^{2} u = \frac{c^{2}}{2} (u_{0}^{''} (x - c t) + u_{0}^{''} (x + c t)) + \frac{c}{2} (u_{1}^{'} (x + c t) - u_{1}^{'} (x - c t))$ , so $u_{t t} = c^{2} u_{x x}$ on $R \times [0, \infty)$ .

1.2F1algebra

Both data are attained. At $t = 0$ the displacement terms give $u (x, 0) = \frac{1}{2} (u_{0} (x) + u_{0} (x)) + 0 = u_{0} (x)$ , and the velocity formula of the previous step gives $\partial_{t} u (x, 0) = \frac{1}{2} (- c u_{0}^{'} (x) + c u_{0}^{'} (x)) + \frac{1}{2} (u_{1} (x) + u_{1} (x)) = u_{1} (x)$ .

1.3F2F3F4algebra

Uniqueness. Let $v$ be any classical solution with the same pointwise displacement and velocity limits at zero and put $w = u - v$ . Apply [F4] to the open rectangle $R \times (0, T)$ , where $T > 0$ is arbitrary. Both characteristic projections are all of $R$ , so $w (x, t) = F (x - c t) + G (x + c t)$ with $F, G \in C^{2} (R)$ . At each fixed $x$ , letting $t ↓ 0$ gives $F (x) + G (x) = 0$ and $- c F^{'} (x) + c G^{'} (x) = 0$ . Differentiating the first identity and combining with the second yields $F^{'} = G^{'} = 0$ everywhere; their sum is zero, so $w = 0$ throughout this rectangle. As $T$ is arbitrary, uniqueness holds on the whole time slab.

2.1given∎

The display of $u$ therefore defines the unique classical solution, and its value at $(x, t)$ involves $u_{0}$ only through the endpoint values $u_{0} (x \mp c t)$ and $u_{1}$ only through the integral over $[x - c t, x + c t]$ .

Depends on

Used by

The one-dimensional value depends on the characteristic interval Corollary
Data on one characteristic line do not determine a one-dimensional wave Counterexample
Finite speed of propagation does not imply strong Huygens Counterexample
The strong Huygens principle in the homogeneous Cauchy setting Definition
A compactly supported velocity datum produces an expanding interval Example
Displacement data versus velocity data in one dimension Example
Odd reflection at a Dirichlet endpoint Example
Right- and left-travelling waves Example
The d'Alembert expression attains both initial data Lemma
The forced one-dimensional wave formula over the characteristic triangle Theorem
Wave tails in one and even spatial dimensions: strong Huygens fails Theorem

Dependency tree · two levels

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Sources

Victor Ivrii, Partial Differential Equations (University of Toronto, 2018, CC BY-SA) (standard reference, not scraped)
Jared Speck, MIT 18.152 Introduction to Partial Differential Equations, Class Meeting #10: Introduction to the Wave Equation (Fall 2011) (standard reference, not scraped)
John K. Hunter, Notes on Partial Differential Equations (revised 18 June 2014, UC Davis) (standard reference, not scraped)
Per Kristen Jakobsen, An Introduction to Partial Differential Equations (arXiv:1901.03022) (standard reference, not scraped)