Unit 4: The 1D & Multi-D Wave Equation: D'Alembert's Formula & Energy Methods

1. Physical Derivation of the 1D Wave Equation

Consider a flexible, perfectly elastic string of constant linear mass density $\rho$ stretched under high uniform tension $T$ along the $x$-axis. Let $u(x, t)$ denote the small transverse displacement at position $x$ and time $t$. Assuming small slopes ($|\partial u/\partial x| \ll 1$), tension variations are negligible, and longitudinal displacements are negligible.

Consider an infinitesimal string element between $x$ and $x + \Delta x$. The vertical component of tension at $x + \Delta x$ is $T \sin\theta_2 \approx T \tan\theta_2 = T u_x(x + \Delta x, t)$. The vertical component of tension at $x$ is $-T \sin\theta_1 \approx -T u_x(x, t)$. By Newton's second law ($F = m a$):

Dividing by $\rho \Delta x$ and taking the limit $\Delta x \to 0$:

Setting the propagation wave speed $c = \sqrt{T/\rho}$ yields the 1D wave equation:

2. Complete Derivation of D'Alembert's Formula

Consider the Cauchy initial value problem on the infinite domain $x \in \mathbb{R}$:

Factor the wave operator into two directional transport derivatives:

Let $v(x, t) = u_t + c u_x$. Then $v$ satisfies the first-order homogeneous advection equation:

whose general solution is $v(x, t) = h(x + ct)$ for an arbitrary function $h$. Now solve the inhomogeneous first-order PDE for $u$:

Using the characteristic coordinates $\xi = x - ct$ and $\eta = x + ct$:

Integrating twice with respect to $\xi$ and $\eta$ yields the general traveling wave solution:

where $\phi(x - ct)$ represents a right-moving wave at speed $c$, and $\psi(x + ct)$ represents a left-moving wave at speed $c$.

Determining $\phi$ and $\psi$ from Initial Data:

At $t = 0$:

1. $u(x, 0) = \phi(x) + \psi(x) = f(x)$
2. $u_t(x, 0) = -c \phi'(x) + c \psi'(x) = g(x)$

Dividing equation (2) by $c$ and integrating from an arbitrary base point $x_0$ to $x$:

We now have a linear system for $\phi(x)$ and $\psi(x)$:

• Adding the two equations:

• Subtracting the two equations:

Substituting $\phi(x - ct)$ and $\psi(x + ct)$ into $u(x, t) = \phi(x - ct) + \psi(x + ct)$:

Combining the integral limits:

Theorem 4.1 (D'Alembert's Formula): The unique solution to the Cauchy problem for the 1D wave equation is:

1. Spacetime Geometry of Hyperbolic Waves

D'Alembert's formula reveals profound relativistic and geometric properties that distinguish hyperbolic PDEs from parabolic and elliptic equations.

Definition 4.1 (Domain of Dependence): For any spacetime point $(x_0, t_0)$ with $t_0 > 0$, the value $u(x_0, t_0)$ depends exclusively on:

1. The values of initial displacement $f$ at the two endpoints: $x_0 - c t_0$ and $x_0 + c t_0$.
2. The values of initial velocity $g$ on the closed spatial interval:

The interval $D(x_0, t_0)$ is called the domain of dependence of the point $(x_0, t_0)$. In the $(x, t)$ spacetime plane, the triangular region with base $D(x_0, t_0)$ and apex $(x_0, t_0)$ bounded by the characteristic lines $x - ct = x_0 - ct_0$ and $x + ct = x_0 + ct_0$ is the past light cone.

2. Range of Influence

Definition 4.2 (Range of Influence): Conversely, an initial disturbance originating at a point $x = \xi$ at $t = 0$ can only affect spacetime points $(x, t)$ satisfying:

The region $I(\xi) = \{ (x, t) \in \mathbb{R} \times [0, \infty) : |x - \xi| \le ct \}$ is the future light cone or range of influence of the point $\xi$.

3. Strict Causality vs Parabolic Infinite Speed

1. Finite Speed of Propagation: Disturbances propagate at the exact finite speed $c$. If initial data $f$ and $g$ are supported on a compact interval $[a, b]$, then at time $t$, $u(x, t) = 0$ everywhere outside $[a - ct, b + ct]$.

2. Contrast with Diffusion: In the heat equation $u_t = \alpha u_{xx}$, an initial point disturbance at $x = 0$ is instantly felt across the entire universe ($u(x, t) > 0$ for all $x \in \mathbb{R}$ for any $t > 0$, no matter how small). Hyperbolic physics strictly respects relativistic causality.

1. The Semi-Infinite String with Fixed End (Dirichlet Boundary)

Consider the semi-infinite string $x \ge 0, t \ge 0$ with a clamped boundary at $x = 0$:

where $f(0) = g(0) = 0$ for compatibility.

For points where $x \ge ct$, the backward characteristic $x - ct \ge 0$ does not touch the boundary $x = 0$, so D'Alembert's formula holds directly. For points where $x < ct$, the characteristic $x - ct < 0$ reflects off the wall $x = 0$.

The Method of Odd Reflection:

To enforce $u(0, t) = 0$ automatically, extend the initial data to the entire real line $\mathbb{R}$ as odd functions:

Applying D'Alembert's formula with these odd extensions: For $x < ct$, since $x - ct < 0$:

Thus, the solution for $0 < x < ct$ is:

Physical Interpretation: The term $-f(ct - x)$ represents an inverted wave packet reflected from the fixed boundary with a phase shift of $\pi$ (inversion).

2. Free End (Neumann Boundary Condition)

If the end at $x = 0$ is free to slide frictionlessly on a vertical rod, the boundary condition is $u_x(0, t) = 0$. By applying even extensions $f_{\text{even}}(-x) = f(x)$, the reflected wave maintains its sign:

No inversion occurs upon reflection from a free boundary.

1. Separation of Variables on a Finite Interval $[0, L]$

Consider the vibrating string of finite length $L$ with clamped endpoints:

Assume a product solution of the form:

Substituting into the wave equation:

where $-\lambda$ is a separation constant.

Spatial Boundary Value Problem (Sturm-Liouville Eigenvalue Problem):

• If $\lambda \le 0$, only trivial solutions $X \equiv 0$ exist.
• For $\lambda > 0$, set $\lambda = k^2$:

$X(0) = 0 \implies A = 0$. $X(L) = B \sin(kL) = 0 \implies k_n L = n\pi \implies k_n = \frac{n\pi}{L}, \quad n = 1, 2, 3, \dots$ Thus, the eigenvalues and eigenfunctions are:

Temporal Equation:

where the natural circular frequencies are:

The general solution for $T_n(t)$ is:

2. General Fourier Superposition

By the principle of linear superposition, the general solution is:

The Fourier coefficients are determined from initial data by orthogonality:

1. Total Mechanical Energy of the Vibrating String

The kinetic energy of the string of linear mass density $\rho$ is:

The potential (elastic strain) energy under tension $T$ is:

Dividing by $\rho$ (recalling $c^2 = T/\rho$), the normalized total energy is defined as:

2. Rigorous Proof of Conservation of Energy

Theorem 4.2 (Conservation of Energy): Let $u(x, t)$ be a $C^2$ solution of the wave equation $u_{tt} = c^2 u_{xx}$ on $[0, L] \times [0, \infty)$ satisfying either Dirichlet boundaries ($u(0, t) = u(L, t) = 0$) or Neumann boundaries ($u_x(0, t) = u_x(L, t) = 0$). Then the total energy is strictly constant in time:

Proof: Differentiate $E(t)$ under the integral sign:

Since $u$ is $C^2$, by Schwarz's theorem $u_{xt} = u_{tx}$. Apply integration by parts to the second term:

Substitute this back into the derivative of energy:

1. In the interior, $u_{tt} - c^2 u_{xx} = 0$ by the wave equation.
2. At the boundaries $x = 0$ and $x = L$:

• For Dirichlet conditions, $u(0, t) = u(L, t) = 0 \implies u_t(0, t) = u_t(L, t) = 0$.
• For Neumann conditions, $u_x(0, t) = u_x(L, t) = 0$.

In either case, the boundary term vanishes identically: $[c^2 u_x u_t]_0^L = 0$. Therefore:

$\blacksquare$

3. Proof of Solution Uniqueness via the Energy Method

Theorem 4.3 (Uniqueness of Solutions to the Wave Equation): The initial-boundary value problem for the 1D wave equation with Dirichlet boundary conditions has at most one $C^2$ solution.

Proof: Suppose $u_1$ and $u_2$ are two $C^2$ solutions with identical initial conditions and boundary conditions. Define the difference function $w(x, t) = u_1(x, t) - u_2(x, t)$. By linearity:

• $w_{tt} - c^2 w_{xx} = 0$ on $(0, L) \times (0, \infty)$
• $w(x, 0) = f(x) - f(x) = 0$
• $w_t(x, 0) = g(x) - g(x) = 0$
• $w(0, t) = 0$ and $w(L, t) = 0$

Now evaluate the energy of the difference function at $t = 0$:

By Theorem 4.2 (Conservation of Energy):

Since the integrand $w_t^2 + c^2 w_x^2 \ge 0$ is non-negative and continuous, $E_w(t) = 0$ forces:

Since all first derivatives vanish identically on a connected domain, $w(x, t)$ is constant:

Using the boundary condition $w(0, t) = 0$, we find $C = 0$. Thus:

The solution is strictly unique. $\blacksquare$