maths.freePartial Differential Equations › The wave equation and d'Alembert's formula

The wave equation and d'Alembert's formula

Factoring the wave operator, travelling waves, d'Alembert's formula and finite speed of propagation.

The wave equation \( u_{tt} = c^2 u_{xx} \) factors: \( (\partial_t - c\,\partial_x)(\partial_t + c\,\partial_x)u = 0 \). It is two transport equations stacked. In the characteristic coordinates \( \xi = x + ct \), \( \eta = x - ct \) the chain rule turns it into \( u_{\xi\eta} = 0 \), whose general solution is \[ u(x,t) = F(x + ct) + G(x - ct): \] a wave moving left plus a wave moving right, each at speed \( c \), each keeping its shape. The example checks \( \sin(x - 2t) \) with \( c = 2 \).

Fitting the initial position \( u(x,0) = f(x) \) and velocity \( u_t(x,0) = g(x) \) gives d'Alembert's formula \[ u(x,t) = \frac{f(x + ct) + f(x - ct)}{2} + \frac1{2c}\int_{x - ct}^{x + ct} g(s)\,ds. \] An initial bump with zero velocity splits into two half-height copies that separate; an initial kick \( g \) spreads out as a widening plateau.

The formula shows at a glance that \( u(x,t) \) depends only on the data in the domain of dependence \( [x - ct, x + ct] \). Conversely, data near \( x_0 \) affect only the cone \( |x - x_0| \le ct \): signals travel at speed at most \( c \). And nothing is smoothed: a corner in \( f \) produces corners in \( u \) that travel along the lines \( x \pm ct = \text{const} \) forever.

Picture it: in the \( (x, t) \) plane, from any point draw two lines back down to the axis with slopes \( \pm 1/c \). Everything the solution knows at that point lies in the triangle between them.

Think it: the heat and wave equations sit at opposite extremes. Heat has infinite speed and infinite smoothing; waves have finite speed and no smoothing, and so the wave equation is time-reversible while the heat equation is not. The Navier-Stokes equations mix both behaviours: the transport term carries features at the fluid velocity, and viscosity smooths them.

Exemple résolu · does u = sin(x - 2*t) satisfy u_tt = 4*u_xx

Does u = sin(x - 2*t) satisfy u_tt = 4*u_xx

u = - \sin{\left(2 t - x \right)},\quad u_{tt} = 4 u_{xx}

Étape par étape

  1. u = - \sin{\left(2 t - x \right)},\qquad u_{tt} = 4 u_{xx}

    To check a solution, compute every derivative the equation uses, substitute, and see whether both sides agree.

  2. u_{tt} = 4 \sin{\left(2 t - x \right)}

    Differentiate 2 times with respect to t.

  3. u_{xx} = \sin{\left(2 t - x \right)}

    Differentiate 2 times with respect to x.

  4. \text{LHS} = 4 \sin{\left(2 t - x \right)},\quad \text{RHS} = 4 \sin{\left(2 t - x \right)}

    Substitute the derivatives into both sides.

  5. \text{LHS} - \text{RHS} = 0

    Subtract and simplify.

  6. \text{LHS} - \text{RHS} = 0\ \checkmark

    The two sides agree for every value of the variables, so it is a solution.

Révèle la réponse
\text{Yes: } u = - \sin{\left(2 t - x \right)} \text{ satisfies } u_{tt} = 4 u_{xx}

À vous Choisissez un problème, ou tapez ou dessinez le vôtre. Chaque étape, une image, la réponse cachée jusqu'à ce que vous le demandiez.

Conservez votre propre travail

Un compte gratuit ajoute des notes sur chaque leçon, un enregistrement de ce que vous avez terminé, vos problèmes résolus en un seul endroit, et un tuteur que vous pouvez demander sur cette page. Les maths eux-mêmes sont ouverts à tout le monde, signé ou non.

Inscription Connexion

Symboles utilisés ici

Appuyez sur n'importe quel symbole pour la définition complète, une image et ce que signifie chaque lettre.

Comment faire: The wave equation and d'Alembert's formula

  1. Identify c, the initial displacement f and the initial velocity g.
  2. Evaluate the average of f at the two feet of the characteristics, x - ct and x + ct.
  3. Integrate g over [x - ct, x + ct] and divide by 2c.
  4. Add the two pieces, and check u(x, 0) = f and u_t(x, 0) = g.

Questions posées par les gens

Is the formula valid for data that are not differentiable?

The formula makes sense for any continuous f and integrable g, and gives the right answer as a weak solution. It is a classical solution only when f is twice and g once continuously differentiable.

Does the wave equation in three dimensions also keep sharp edges?

Yes, and more: in odd dimensions at least three, a pulse leaves no wake behind it (Huygens' principle), which is why sound arrives cleanly. In two dimensions a ripple leaves a tail.

What should I know before starting?

Partial derivatives, the divergence theorem and multiple integrals from multivariable calculus; linear second-order ODEs; eigenvalues from linear algebra; and enough analysis to be comfortable with uniform convergence and integrals over infinite intervals.

Why are there so few formulas for solutions?

Explicit solutions exist for linear equations with constant coefficients on simple domains. Almost everything else (curved domains, variable coefficients, nonlinear terms) has none, so the modern subject proves that a solution exists and estimates its size and smoothness without ever writing it down.

How does this course lead to the Navier-Stokes problem?

The Navier-Stokes equations are a heat equation for the velocity with a transport term and a pressure that is found from a Poisson equation. Reading the Millennium problem needs energy estimates, weak solutions, Sobolev spaces, local existence with a blow-up criterion, and the scaling argument that explains why three dimensions is hard. The last six lessons build those.

What do elliptic, parabolic and hyperbolic mean?

Elliptic equations (Laplace) describe equilibrium and smooth everything; parabolic equations (heat) describe diffusion forward in time; hyperbolic equations (waves) carry signals at finite speed and keep their sharp edges. The type decides which data make a sensible problem.

Plus en Partial Differential Equations