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First-order linear equations and characteristics

The transport equation, characteristic curves, and turning a PDE into a family of ODEs.

The transport equation \( u_t + c\,u_x = 0 \) says that the derivative of \( u \) in the direction \( (1, c) \) of the \( (t, x) \) plane is zero. So \( u \) is constant along every line \( x - ct = \text{const} \), and the solution with initial value \( u(x, 0) = g(x) \) is \( u(x, t) = g(x - ct) \). The example checks this for \( g(x) = x^2 \), \( c = 3 \): \( u_t = -6(x - 3t) \) and \( 3u_x = 6(x - 3t) \).

The same idea handles variable coefficients. For \( a(x,t)\,u_t + b(x,t)\,u_x = 0 \), the curves with \( dx/dt = b/a \) are the characteristics, and along each one \( \frac{d}{dt}u(x(t), t) = u_t + \frac{dx}{dt}u_x = 0 \). For \( u_t + x\,u_x = 0 \) the characteristics are \( x = x_0 e^{t} \), so \( u(x, t) = g(x e^{-t}) \). With a source, \( u_t + c\,u_x = f \), the value changes along the characteristic at rate \( f \), and integrating gives Duhamel's formula \[ u(x,t) = g(x - ct) + \int_0^t f\big(x - c(t - s),\, s\big)\,ds. \]

Picture it: the graph of \( g \) slides to the right at speed \( c \) without changing shape. In the \( (x, t) \) plane, draw the family of parallel lines \( x - ct = x_0 \): each line carries the single value \( g(x_0) \) up from the axis.

Think it: the method converts one PDE into infinitely many ODEs, one per characteristic. It works when the initial curve is non-characteristic, crossing every characteristic exactly once. Data given along a characteristic cannot be arbitrary, because the equation already dictates how \( u \) varies there. In fluids the operator \( \partial_t + u\cdot\nabla \) is the material derivative, transport along particle paths: in two-dimensional ideal flow the vorticity is carried unchanged by exactly this mechanism.

Esempio svolto · does u = (x-3*t)^2 satisfy u_t + 3*u_x = 0

Does u = (x-3*t)^2 satisfy u_t + 3*u_x = 0

u = \left(- 3 t + x\right)^{2},\quad u_{t} + 3 u_{x} = 0

Passo dopo passo

  1. u = \left(- 3 t + x\right)^{2},\qquad u_{t} + 3 u_{x} = 0

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

  2. u_{t} = 18 t - 6 x

    Differentiate with respect to t, holding the other variables constant.

  3. u_{x} = - 6 t + 2 x

    Differentiate with respect to x, holding the other variables constant.

  4. \text{LHS} = 0,\quad \text{RHS} = 0

    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.

Rivela la risposta
\text{Yes: } u = \left(- 3 t + x\right)^{2} \text{ satisfies } u_{t} + 3 u_{x} = 0

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Come: First-order linear equations and characteristics

  1. Write the equation as a directional derivative: a u_t + b u_x = f.
  2. Solve dx/dt = b/a for the characteristic curves through (x_0, 0).
  3. Along each curve, solve the ODE du/dt = f/a with u = g(x_0) at t = 0.
  4. Eliminate x_0 using the characteristic equation to get u(x, t).
  5. Check by substituting back into the PDE.

Domande che la gente chiede

What if two characteristics meet?

For linear equations with smooth coefficients they cannot, since the ODE for x(t) has unique solutions. For nonlinear equations like Burgers they can, and that is how shocks form.

Why does c > 0 move the profile to the right?

Because u(x, t) = g(x - ct) takes at position x the value g had at x - ct, a point to the left: the value has travelled a distance ct rightwards.

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.

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