d'Alembert's General Solution

Derivation

d'Alembert's General Solution

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Statement

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 \(u(x,t)=\tfrac12\big[\phi(x-ct)+\phi(x+ct)\big]+\tfrac{1}{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds\).

Why it matters

d'Alembert's formula is the exact, closed-form general solution of the archetypal hyperbolic PDE. It shows that disturbances on a string, in a column of air, or along a lossless transmission line do not smear out but propagate at a fixed speed \(c\) with their shape intact, splitting cleanly into two counter-propagating pieces. This is the mechanical origin of the very idea of a travelling wave: understanding one arbitrary pulse suffices to understand all 1D wave motion.

It also introduces two ideas that pervade all of wave physics: characteristics — the lines \(x\mp ct=\text{const}\) along which information travels — and the domain of dependence, the finite interval of initial data that can influence a given event. Together these fix causality and the finite speed of signalling for every hyperbolic problem.

Assumptions

The medium obeys \(\partial_{tt}u=c^{2}\partial_{xx}u\) with constant \(c>0\).If \(c=c(x)\) the characteristics curve and the profiles distort; the clean \(f,g\) split is replaced by scattering and partial reflection at gradients.

The equation is linear, undamped and source-free.Drop linearity and \(f+g\) superposition fails (nonlinear terms couple the modes, producing shocks); add damping \(\gamma\,\partial_t u\) and each profile decays and disperses rather than translating rigidly.

The domain is the full line with no boundaries.On a finite string, boundary conditions force reflections; the solution becomes a sum of images and the raw \(f(x-ct)+g(x+ct)\) form holds only until a signal reaches an end.

The solution is \(C^{2}\) (twice continuously differentiable).Kinks or steps in \(\phi,\psi\) violate classical differentiability; the formula still returns the physically correct weak solution, but "\(u\) solves the PDE pointwise" must be reinterpreted distributionally.

Derivation

\[ \frac{\partial^{2}u}{\partial t^{2}}=c^{2}\frac{\partial^{2}u}{\partial x^{2}} \]

Starting point: the 1D wave equation, taken as established from the loaded string. A

\[ \left(\frac{\partial}{\partial t}-c\frac{\partial}{\partial x}\right)\left(\frac{\partial}{\partial t}+c\frac{\partial}{\partial x}\right)u=0 \]

Factor the linear operator. Because \(c\) is constant the two first-order operators commute and their product reproduces \(\partial_{tt}-c^{2}\partial_{xx}\); the factoring is exact, not approximate. B

\[ \xi=x-ct,\qquad \eta=x+ct \]

Introduce characteristic coordinates — the two families of lines along which the first-order factors act as directional derivatives. This change of variables diagonalises the operator. B

\[ \frac{\partial}{\partial x}=\frac{\partial}{\partial\xi}+\frac{\partial}{\partial\eta},\qquad \frac{\partial}{\partial t}=-c\frac{\partial}{\partial\xi}+c\frac{\partial}{\partial\eta} \]

Chain rule, using \(\partial\xi/\partial x=\partial\eta/\partial x=1\), \(\partial\xi/\partial t=-c\), \(\partial\eta/\partial t=+c\). Symbolic transformation of the derivatives. A

\[ \frac{\partial}{\partial t}+c\frac{\partial}{\partial x}=2c\frac{\partial}{\partial\eta},\qquad \frac{\partial}{\partial t}-c\frac{\partial}{\partial x}=-2c\frac{\partial}{\partial\xi} \]

Substitute Step 4 into the two factors; the cross terms combine so each factor collapses to a single characteristic derivative. Pure operator algebra. B

\[ (-2c)(2c)\,\frac{\partial^{2}u}{\partial\xi\,\partial\eta}=0 \;\Longrightarrow\; \frac{\partial^{2}u}{\partial\xi\,\partial\eta}=0 \]

Insert Step 5 into Step 2. Since \(c\neq0\) the prefactor \(-4c^{2}\) is nonzero and divides out, leaving the canonical hyperbolic form. B

\[ \frac{\partial}{\partial\xi}\!\left(\frac{\partial u}{\partial\eta}\right)=0 \;\Longrightarrow\; \frac{\partial u}{\partial\eta}=G'(\eta) \]

Integrate once in \(\xi\): a quantity whose \(\xi\)-derivative vanishes is independent of \(\xi\), hence a function of \(\eta\) alone (written \(G'(\eta)\) for later convenience). B

\[ u=\int G'(\eta)\,d\eta + f(\xi)=g(\eta)+f(\xi) \]

Integrate once in \(\eta\): the \(\eta\)-antiderivative \(g\) plus an "\(\xi\)-constant" of integration \(f(\xi)\). Here \(f,g\) are arbitrary \(C^{2}\) functions. B

\[ u(x,t)=f(x-ct)+g(x+ct) \]

Undo the substitution \(\xi=x-ct,\ \eta=x+ct\). This is the general solution: every solution has this form and any \(C^2\) choice of \(f,g\) yields a solution. A

\[ u(x,0)=f(x)+g(x)=\phi(x),\qquad u_t(x,0)=-c\,f'(x)+c\,g'(x)=\psi(x) \]

Impose the initial data on the general solution to pin down \(f,g\). Two equations for the two unknown profiles. B

\[ -f'(x)+g'(x)=\frac{1}{c}\psi(x)\;\Longrightarrow\; -f(x)+g(x)=\frac{1}{c}\int_{0}^{x}\psi(s)\,ds + K \]

Divide the velocity condition by \(c\) and integrate from a reference point \(0\) to \(x\); \(K\) is a constant of integration. B

\[ g(x)=\tfrac12\phi(x)+\tfrac{1}{2c}\int_{0}^{x}\psi + \tfrac{K}{2},\qquad f(x)=\tfrac12\phi(x)-\tfrac{1}{2c}\int_{0}^{x}\psi - \tfrac{K}{2} \]

Solve the linear pair (displacement condition of Step 10 and Step 11) for \(f\) and \(g\) by addition and subtraction. B

\[ u(x,t)=\tfrac12\big[\phi(x-ct)+\phi(x+ct)\big]+\tfrac{1}{2c}\left[\int_{0}^{x+ct}\psi-\int_{0}^{x-ct}\psi\right] \]

Evaluate \(f(x-ct)+g(x+ct)\); the \(\pm K/2\) cancel, and the two lower-limit integrals combine. B

\[ u(x,t)=\frac{\phi(x-ct)+\phi(x+ct)}{2}+\frac{1}{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds \]

Combine via \(\int_0^{x+ct}-\int_0^{x-ct}=\int_{x-ct}^{x+ct}\). This is d'Alembert's formula; uniqueness follows because the construction of \(f,g\) was forced at every step. A

Result

\[ u(x,t)=f(x-ct)+g(x+ct)=\frac{\phi(x-ct)+\phi(x+ct)}{2}+\frac{1}{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds \]

Reading. Any 1D wave is exactly two independent, shape-preserving pulses: one sliding right at speed \(c\), one sliding left at speed \(c\). Given a snapshot of shape \(\phi\) and velocity \(\psi\), the displacement at \((x,t)\) is the average of the initial displacement carried in along the two characteristics through \((x,t)\), plus the mean initial velocity over the interval \([x-ct,\,x+ct]\) they enclose — that interval is the domain of dependence.

Units check. \(\phi\) has units of displacement (m), so the averaged term is in m. In the second term \(\psi\) is m·s\(^{-1}\) and \(ds\) is m, so \(\int\psi\,ds\) is m\(^2\)·s\(^{-1}\); dividing by \(2c\) (m·s\(^{-1}\)) gives m. Both terms are metres, matching \(u\). The arguments \(x\pm ct\) are consistent since \(ct\) is (m·s\(^{-1}\))·s = m.

Limiting cases

Pure right-mover: choose \(\psi=-c\,\phi'\). Then \(g\equiv\text{const}\) and \(u=\phi(x-ct)\) translates rigidly to the right with no left-going piece.
Released from rest, \(\psi=0\): \(u=\tfrac12[\phi(x-ct)+\phi(x+ct)]\). The initial hump splits into two half-amplitude copies moving oppositely (the plucked string releasing two half-pulses).
Struck at rest, \(\phi=0\) (a hammer blow): \(u=\tfrac{1}{2c}\int_{x-ct}^{x+ct}\psi\), a spreading plateau bounded by the outgoing characteristics.
\(c\to\infty\): the domain of dependence widens without bound and the finite-speed causality structure degenerates — the characteristic cone opens flat.
\(t=0\): the formula collapses to \(u=\phi(x)\), and its time derivative to \(\psi(x)\), recovering the initial data.

Breaks when

Dispersive media. If the dispersion relation is not \(\omega=ck\) (a stiff bar with \(\omega\propto k^{2}\), or a waveguide with cutoff), Fourier components travel at different speeds. No single \(c\) exists, the profile spreads, and \(f(x-ct)+g(x+ct)\) is not a solution.
Nonlinearity. For large amplitude or amplitude-dependent speed \(c(u)\) (shallow water, gas dynamics), characteristics cross, the two families couple, and smooth \(f,g\) steepen into shocks — superposition fails entirely.
Damping or driving. Adding \(\gamma\,\partial_t u\) or a source \(F(x,t)\) makes \(\partial_\xi\partial_\eta u\neq0\); the telegrapher-type equation no longer factors into two commuting transport operators.
Higher dimensions. In 2D and 3D there is no exact left/right split; Huygens' principle differs (a sharp 3D pulse stays sharp, a 2D pulse leaves a wake), so the clean d'Alembert structure is special to one spatial dimension.

Failure modes

Sign of the moving argument. Writing \(f(x+ct)\) for a right-mover. The right-moving profile keeps \(x-ct\) constant, so it must be \(f(x-ct)\); the two are routinely swapped.
Missing the factor of \(\tfrac12\). For rest initial data the hump splits into two half-amplitude pulses; forgetting the \(\tfrac12\) doubles the energy.
Dropping the velocity integral. Using \(u=\tfrac12[\phi(x-ct)+\phi(x+ct)]\) even when \(\psi\neq0\), thereby ignoring the entire second term.
Reversed integration limits. Writing \(\int_{x+ct}^{x-ct}\), flipping the sign of the velocity contribution.
Treating \(f,g\) as the same function. They are independent; only initial data relate them. Assuming \(f=g\) forces an even solution and loses half the general solution.
Confusing phase speed with particle speed. \(c\) is how fast the pattern moves, not how fast a string element moves (\(\partial_t u\)); these are distinct quantities.

Discussion

The heart of the derivation is that a constant-coefficient hyperbolic operator factors into two commuting transport operators, and each transport operator \(\partial_t\pm c\,\partial_x\) is a directional (convective) derivative along a characteristic. Setting a directional derivative to zero says the solution is constant along that direction — hence a rigidly translating profile. The entire content of "waves travel" is therefore the geometry of the characteristics \(x\mp ct=\text{const}\).

The characteristics also encode causality. The value at \((x,t)\) depends only on \(\phi\) at the two points \(x\pm ct\) and on \(\psi\) over the interval between them: the domain of dependence. Reciprocally, initial data at a point \(x_0\) influence only the forward cone \(|x-x_0|\le ct\): the range of influence. Signals cannot outrun \(c\). This finite propagation speed is precisely what distinguishes the wave equation from the heat equation, whose Gaussian kernel is instantaneously nonzero everywhere.

Energetically the two profiles are non-interacting: the total energy splits into a right-going part built from \(f\) and a left-going part built from \(g\), each separately conserved on the infinite line, because the cross term \(\int f'(x-ct)\,g'(x+ct)\,dx\) reduces to a boundary quantity that vanishes for localised data. The energy flux \(-T\,\partial_x u\,\partial_t u\) for a pure right-mover equals \(+c\) times the local energy density — energy is transported at exactly the phase speed. This clean separation is the mechanical prototype of the decoupled forward/backward modes later seen in transmission lines and in the light-cone structure of relativity, where \(x\pm ct\) become null coordinates.

Common misconceptions. d'Alembert's result is often misread as "waves are sinusoids". It says nothing of the sort: \(f\) and \(g\) are arbitrary shapes. Sinusoids enter only as a convenient Fourier basis when boundaries or dispersion make normal modes natural. A second confusion is that the superposition here is between the two directions of travel, not among many modes — that broader superposition is a separate consequence of linearity.

Worked examples

Example 1 — Released Gaussian hump (rest initial data).

\[ \phi(x)=A\,e^{-x^{2}/a^{2}},\qquad \psi(x)=0 \]

Gaussian displacement, string released from rest. A

\[ u(x,t)=\tfrac12\phi(x-ct)+\tfrac12\phi(x+ct)=\frac{A}{2}\Big[e^{-(x-ct)^2/a^2}+e^{-(x+ct)^2/a^2}\Big] \]

Apply d'Alembert with \(\psi=0\): the velocity integral drops. A

\[ A=2.0\times10^{-3}\,\text{m},\quad a=0.10\,\text{m},\quad c=50\,\text{m·s}^{-1},\quad t=4.0\times10^{-3}\,\text{s} \]

Insert numbers; \(ct=50\times4.0\times10^{-3}=0.20\,\text{m}\). A

\[ u(0,t)=\frac{2.0\times10^{-3}}{2}\Big[e^{-(0.20/0.10)^2}+e^{-(0.20/0.10)^2}\Big]=2.0\times10^{-3}\,e^{-4} \]

Evaluate at the origin; both Gaussians are centred at \(\mp0.20\,\text{m}\), each contributing with argument \((0.2/0.1)^2=4\). A

\[ u(0,\,4\,\text{ms})=2.0\times10^{-3}\,e^{-4}\approx3.7\times10^{-5}\,\text{m}\approx37\,\mu\text{m} \]

Reading. The single hump has split into two half-amplitude Gaussians, each moved \(0.20\,\text{m}\) outward; at the centre only their overlapping tails remain, suppressed by \(e^{-4}\).

Units check. \(A\) in m, exponent dimensionless (m/m, squared), so \(u\) in m.

Example 2 — Hammer blow (rest displacement, localized velocity).

\[ \phi(x)=0,\qquad \psi(x)=\begin{cases}v_0,&|x|\le b\\[2pt]0,&|x|>b\end{cases} \]

String flat at \(t=0\), struck over \([-b,b]\) with uniform transverse velocity \(v_0\). A

\[ u(x,t)=\frac{1}{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds=\frac{v_0}{2c}\,\big|[x-ct,\,x+ct]\cap[-b,\,b]\big| \]

d'Alembert with \(\phi=0\); the integral is \(v_0\) times the length of overlap between the dependence interval and the struck region. B

\[ b=0.020\,\text{m},\quad v_0=0.30\,\text{m·s}^{-1},\quad c=120\,\text{m·s}^{-1},\quad x=0 \]

Insert numbers; at \(x=0\) the dependence interval is \([-ct,\,ct]\), symmetric about the struck region. A

\[ ct=120\,t;\quad \text{for } ct\ge b\ (t\ge1.67\times10^{-4}\,\text{s}):\ \text{overlap}=2b \]

Once the characteristics sweep past the whole strike zone the overlap saturates at the full width \(2b\). Take \(t=1.0\times10^{-3}\,\text{s}\Rightarrow ct=0.12\,\text{m}>b\). A

\[ u(0,t)=\frac{v_0}{2c}(2b)=\frac{v_0 b}{c}=\frac{0.30\times0.020}{120} \]

Substitute the saturated overlap. A

\[ u(0,\,1\,\text{ms})=\frac{0.30\times0.020}{120}=5.0\times10^{-5}\,\text{m}=50\,\mu\text{m} \]

Reading. The struck patch launches two step-fronts; between them the centre has risen to a constant plateau \(v_0 b/c\) and stays there until reflections arrive. The displacement is set by the impulse per unit length, not by the strike duration.

Units check. \(v_0 b/c=(\text{m·s}^{-1}\cdot\text{m})/(\text{m·s}^{-1})=\text{m}\).

Problems

Verify by direct differentiation that \(u(x,t)=f(x-ct)+g(x+ct)\) satisfies \(u_{tt}=c^{2}u_{xx}\) for arbitrary twice-differentiable \(f,g\).
Solution
With \(\xi=x-ct,\ \eta=x+ct\): \(u_x=f'+g'\), so \(u_{xx}=f''+g''\). And \(u_t=-cf'+cg'\), so \(u_{tt}=c^2 f''+c^2 g''=c^2(f''+g'')\). Hence \(u_{tt}=c^2 u_{xx}\) identically, for any \(C^2\) \(f,g\).
A string has \(c=200\,\text{m·s}^{-1}\). Its initial shape is a triangular pulse of peak height \(4.0\,\text{mm}\) and half-width \(0.15\,\text{m}\) centred at the origin, released from rest. Find the peak height of each outgoing pulse and the locations of their peaks at \(t=2.0\,\text{ms}\).
Solution
Rest data give \(u=\tfrac12[\phi(x-ct)+\phi(x+ct)]\): the pulse splits into two copies of half the amplitude, so peak height \(=\tfrac12(4.0\,\text{mm})=2.0\,\text{mm}\). The shift is \(ct=200\times2.0\times10^{-3}=0.40\,\text{m}\). Peaks lie at \(x=+0.40\,\text{m}\) (right-mover) and \(x=-0.40\,\text{m}\) (left-mover). The half-width \(0.15\,\text{m}<0.40\,\text{m}\), so the two triangles no longer overlap.
Show that initial data \(\phi(x),\psi(x)\) produce a purely right-moving wave (no left-mover) if and only if \(\psi(x)=-c\,\phi'(x)\).
Solution
Purely right-moving means \(g'\equiv0\). From \(f+g=\phi\) differentiate: \(f'+g'=\phi'\). The velocity condition is \(-cf'+cg'=\psi\). If \(g'=0\) then \(f'=\phi'\) and \(\psi=-cf'=-c\phi'\). Conversely if \(\psi=-c\phi'\) then \(-cf'+cg'=-c(f'+g')\Rightarrow 2cg'=0\Rightarrow g'=0\), so \(g\) is constant and \(u=\phi(x-ct)+\text{const}\).
Initial data are \(\phi(x)=0\) and \(\psi(x)=U\cos(kx)\) on the whole line, with \(U=0.50\,\text{m·s}^{-1}\), \(k=3.0\,\text{m}^{-1}\), \(c=100\,\text{m·s}^{-1}\). Find \(u(x,t)\) in closed form and its amplitude.
Solution
\(u=\tfrac{1}{2c}\int_{x-ct}^{x+ct}U\cos(ks)\,ds=\tfrac{U}{2ck}\big[\sin(k(x+ct))-\sin(k(x-ct))\big]\). Using \(\sin A-\sin B=2\cos\tfrac{A+B}{2}\sin\tfrac{A-B}{2}\): \(u=\tfrac{U}{ck}\cos(kx)\sin(kct)\), a standing wave. Amplitude \(=\dfrac{U}{ck}=\dfrac{0.50}{100\times3.0}=1.7\times10^{-3}\,\text{m}\approx1.7\,\text{mm}\).
At \(t=0\), two identical rectangular pulses of height \(h=1.0\,\text{mm}\) and width \(w=0.10\,\text{m}\) sit on a rest string (\(\psi=0\)), one centred at \(x=-1.0\,\text{m}\), one at \(x=+1.0\,\text{m}\), with \(c=250\,\text{m·s}^{-1}\). At what time do a right-mover from the left pulse and a left-mover from the right pulse first fully overlap at the origin, and what is the displacement there at that instant?
Solution
Each pulse splits into two half-height (\(0.50\,\text{mm}\)) copies. The right-mover from the left pulse is centred at \(x=-1.0+ct\); the left-mover from the right pulse at \(x=+1.0-ct\). Both centre on \(x=0\) when \(ct=1.0\,\text{m}\), i.e. \(t=1.0/250=4.0\times10^{-3}\,\text{s}=4.0\,\text{ms}\). Being rectangular of equal width they overlap completely, so the displacement is \(0.50+0.50=1.0\,\text{mm}\) — momentarily the full original height is reconstructed at the origin.