The Wave Equation and d'Alembert's Solution
~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You must be comfortable with partial differentiation and the concept of a function of two variables. You should also recall the derivation of the wave equation from the continuum limit of coupled oscillators, as covered in the previous lecture.
Why this matters
The wave equation is the fundamental mathematical description of how disturbances propagate through space and time.
While the previous lecture established that a continuous string obeys the wave equation, this lecture provides the general solution to that equation. This solution, known as d'Alembert's solution, reveals that any wave motion can be decomposed into two independent components: a wave travelling to the right and a wave travelling to the left. This insight is crucial not only for understanding mechanical waves but also for later topics such as electromagnetic radiation and quantum field theory, where the same mathematical structure governs the propagation of fields.
The Idea in Plain Language
Imagine a long, taut string. If you pluck a section of the string, the disturbance does not stay in one place; it travels along the string. The wave equation tells us that the acceleration of any point on the string is proportional to the curvature of the string at that point. The remarkable result of d'Alembert is that the most general motion of the string is simply the sum of two "shapes" that move rigidly in opposite directions. One shape moves to the right at speed \(c\), and the other moves to the left at speed \(c\). They do not change shape as they travel; they simply translate. If you have a complex wave, it is just a superposition of these two rigidly moving profiles. This is a profound simplification: instead of solving a complex differential equation for every point, we just need to track two arbitrary functions moving at constant speed.
The General Solution
Consider the one-dimensional wave equation on an infinite domain, where \(x \in \mathbb{R}\): \[ \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2} \] Here, \(u(x,t)\) is the displacement of the string at position \(x\) and time \(t\), and \(c\) is the wave speed. The general solution to this equation is given by d'Alembert's formula. It states that any twice-differentiable function \(u(x,t)\) that satisfies the wave equation can be written as the sum of two arbitrary functions: \[ u(x,t) = f(x - ct) + g(x + ct) \] The term \(f(x - ct)\) represents a wave profile \(f\) that moves to the right with constant velocity \(c\). To see this, consider a specific point on the wave, say where the argument \(x - ct\) is constant. Differentiating \(x - ct = \text{const}\) with respect to time gives \(v = dx/dt = c\). Similarly, the term \(g(x + ct)\) represents a wave profile \(g\) that moves to the left with velocity \(-c\), since \(x + ct = \text{const}\) implies \(dx/dt = -c\).
This form is powerful because \(f\) and \(g\) are completely arbitrary functions (provided they are twice differentiable). Strictly they must be twice differentiable for the equation to hold point by point; physically, kinked profiles (such as a plucked string) are still admitted as generalised (weak) solutions. This means the wave equation does not restrict the shape of the wave, only how it propagates.
Initial Value Problem
In physical problems, we usually do not know the functions \(f\) and \(g\) directly. Instead, we know the initial state of the system: the initial displacement \(\phi(x) = u(x,0)\) and the initial velocity \(\psi(x) = \partial_t u(x,0)\). We can determine \(f\) and \(g\) from these initial conditions.
At \(t=0\), the general solution becomes: \[ u(x,0) = f(x) + g(x) = \phi(x) \] Differentiating the general solution with respect to time: \[ \frac{\partial u}{\partial t} = -c f'(x - ct) + c g'(x + ct) \] At \(t=0\): \[ \frac{\partial u}{\partial t}\bigg|_{t=0} = -c f'(x) + c g'(x) = \psi(x) \] We now have a system of two equations for the unknown functions \(f\) and \(g\). By integrating the second equation and substituting into the first, we can solve for \(f\) and \(g\) explicitly. The result is d'Alembert's explicit formula for the initial value problem: \[ u(x,t) = \frac{1}{2} \left[ \phi(x - ct) + \phi(x + ct) \right] + \frac{1}{2c} \int_{x - ct}^{x + ct} \psi(s) \, ds \] This formula is unique. It tells us that the displacement at any point \((x,t)\) depends only on the initial conditions within the interval \([x-ct, x+ct]\). This region is known as the domain of dependence.
Physical Interpretation
The explicit formula has a clear physical interpretation. The first term, \(\frac{1}{2} [\phi(x - ct) + \phi(x + ct)]\), represents the initial shape \(\phi\) splitting into two identical halves, each travelling in opposite directions at speed \(c\). The second term, \(\frac{1}{2c} \int_{x - ct}^{x + ct} \psi(s) \, ds\), accounts for the initial velocity. It is the average of the initial velocity over the interval \([x-ct, x+ct]\), scaled by time. This term vanishes if the initial velocity is zero everywhere.
The speed \(c\) is determined by the physical properties of the medium. For a string with tension \(T\) and linear mass density \(\mu\), the wave speed is \(c = \sqrt{T/\mu}\). This means that heavier strings (larger \(\mu\)) or looser strings (smaller \(T\)) support slower waves. The wave equation itself does not specify \(c\); it is a parameter that must be determined by the physics of the medium.
The derivation
d'Alembert's General Solution
On the whole line \(x\in\mathbb{R}\), every twice-differentiable solution of the one-dimensional wave equation \(\partial_{tt}u=c^{2}\,\partial_{xx}u\) can be written as \(u(x,t)=f(x-ct)+g(x+ct)\), a rigidly translating right-moving profile \(f\) superposed on a rigidly translating left-moving profile \(g\); conversely any such \(C^2\) sum solves the equation. When the initial displacement \(u(x,0)=\phi(x)\) and velocity \(\partial_t u(x,0)=\psi(x)\) are prescribed, the solution is unique and given explicitly by \(u(x,t)=\tfrac12\big[\phi(x-ct)+\phi(x+ct)\big]+\tfrac{1}{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds\).
A common misconception is that the wave equation implies energy is created or destroyed as the wave travels. In reality, the wave equation is linear and conservative (in the absence of damping). The energy density is proportional to the sum of the kinetic and potential energy densities, and the total energy is conserved. However, the *shape* of the wave can change if the medium is dispersive, but in the ideal string case, the shape remains rigid. Another subtle point is that the solution \(u(x,t) = f(x-ct) + g(x+ct)\) is valid only for the infinite line. For finite strings with boundaries, reflections occur, and the solution must be constructed using the method of images or Fourier series, which will be covered in the next lecture. C
Worked example
Problem. A string of infinite length has an initial displacement \(\phi(x) = e^{-x^2}\) and zero initial velocity \(\psi(x) = 0\). The wave speed is \(c = 2\ \mathrm{m/s}\). Find the displacement \(u(x,t)\) at \(t = 1\ \mathrm{s}\) and \(x = 0\).
- Identify the given parameters: \(\phi(x) = e^{-x^2}\), \(\psi(x) = 0\), \(c = 2\ \mathrm{m/s}\), \(t = 1\ \mathrm{s}\), \(x = 0\).
- Use d'Alembert's formula: \(u(x,t) = \frac{1}{2} [\phi(x - ct) + \phi(x + ct)] + \frac{1}{2c} \int_{x - ct}^{x + ct} \psi(s) \, ds\).
- Since \(\psi(x) = 0\), the integral term vanishes.
- Calculate the arguments: \(x - ct = 0 - 2(1) = -2\) and \(x + ct = 0 + 2(1) = 2\).
- Evaluate \(\phi(-2) = e^{-(-2)^2} = e^{-4}\) and \(\phi(2) = e^{-(2)^2} = e^{-4}\).
- Substitute into the formula: \(u(0,1) = \frac{1}{2} [e^{-4} + e^{-4}] = e^{-4}\).
- Calculate the numerical value: \(e^{-4} \approx 0.0183\).
Units check. The displacement \(u\) is in metres (m). The function \(\phi(x)\) is in metres. The arguments \(x \pm ct\) are in metres, so the exponent is dimensionless. The result is consistent with SI units.
Three quick tests
- What is the physical meaning of the term \(f(x - ct)\) in d'Alembert's solution?
Answer
It represents a wave profile moving to the right with constant velocity \(c\). - If the initial velocity \(\psi(x)\) is non-zero, how does it affect the solution?
Answer
It adds a term \(\frac{1}{2c} \int_{x - ct}^{x + ct} \psi(s) \, ds\) to the displacement, which accounts for the initial motion of the string. - What is the domain of dependence for the point \((x,t)\)?
Answer
The interval \([x - ct, x + ct]\) on the initial line \(t=0\). The value of \(u(x,t)\) depends only on the initial conditions within this interval.
Problem set for PU-103 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.