maths.freePartial Differential Equations › Nonlinear first-order equations and shocks

Nonlinear first-order equations and shocks

Burgers' equation, the breaking time, weak solutions, the Rankine-Hugoniot condition and rarefaction fans.

In the inviscid Burgers equation \( u_t + u\,u_x = 0 \) the speed of transport is the solution itself. Characteristics are still straight lines, \( x = x_0 + g(x_0)\,t \), and \( u \) is still constant along each, so the solution is given implicitly by \( u = g(x - u\,t) \). The difference is that faster parts of the profile catch up with slower parts ahead of them.

Differentiate the implicit formula: \( u_x = g'(x_0)/(1 + t\,g'(x_0)) \). Wherever \( g' < 0 \) the denominator reaches zero, so the slope becomes infinite at the breaking time \[ t^* = \frac{-1}{\min_x g'(x)}. \] For \( g(x) = e^{-x^2} \) the example finds \( \min g' = -\sqrt 2\,e^{-1/2} \) at \( x = 1/\sqrt 2 \), so \( t^* = e^{1/2}/\sqrt 2 = \sqrt{e/2} \approx 1.17 \). Smooth data, a perfectly good equation, and a classical solution that ceases to exist in finite time.

After breaking we need weak solutions of the conservation form \( u_t + (u^2/2)_x = 0 \). A jump from \( u_L \) to \( u_R \) moving along \( x = s(t) \) conserves the integral of \( u \) only if its speed obeys the Rankine-Hugoniot condition \( s' = \big(\tfrac12 u_L^2 - \tfrac12 u_R^2\big)/(u_L - u_R) = \tfrac12(u_L + u_R) \). Weak solutions are not unique, so an extra entropy condition selects the physical one: a shock needs \( u_L > u_R \), with characteristics running into it. When \( u_L < u_R \) the right answer is a rarefaction fan \( u = x/t \) filling the gap.

Picture it: a crest of \( e^{-x^2} \) leans forward like a wave approaching a beach, its front face steepening until it is vertical; then the multivalued overhang is replaced by a jump that moves at the average of the states on either side.

Think it: adding viscosity, \( u_t + u\,u_x = \nu\,u_{xx} \), prevents breaking for every \( \nu > 0 \) (the Cole-Hopf transformation turns it into the heat equation), and the entropy solution is the limit as \( \nu \to 0 \). This is the one-dimensional cartoon of the Navier-Stokes question: does dissipation always beat the steepening caused by self-transport? In one dimension it does; in three nobody knows.

工作范例 · critical points of -2*x*exp(-x^2)

Critical points of -2x·e^(-x^2)

- 2 x e^{- x^{2}}

一步

  1. f(x) = - 2 x e^{- x^{2}}

    Critical points are where f′(x) = 0 or is undefined.

  2. f'(x) = 4 x^{2} e^{- x^{2}} - 2 e^{- x^{2}}

    Differentiate.

  3. x = - \frac{\sqrt{2}}{2}, x = \frac{\sqrt{2}}{2}

    Solve f′(x) = 0.

  4. f''(x) = - 8 x^{3} e^{- x^{2}} + 12 x e^{- x^{2}}

    Second-derivative test: f″ < 0 → maximum, f″ > 0 → minimum.

  5. f''(- \frac{\sqrt{2}}{2}) = - \frac{4 \sqrt{2}}{e^{\frac{1}{2}}} \Rightarrow (- \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{e^{\frac{1}{2}}}) \text{ is a local maximum}

  6. f''(\frac{\sqrt{2}}{2}) = \frac{4 \sqrt{2}}{e^{\frac{1}{2}}} \Rightarrow (\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{e^{\frac{1}{2}}}) \text{ is a local minimum}

发送答案
(- \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{e^{\frac{1}{2}}})\ \text{local maximum},\; (\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{e^{\frac{1}{2}}})\ \text{local minimum}

现在轮到你 选择问题, 输入或绘制您自己的 。 每一步, 图片, 答案隐藏到您询问 。

保持你自己的工作

免费账户在每一课上都添加注解, 记录你完成的, 在一个地方解决的, 以及您可以询问的关于此页的导师。 数学本身向每个人开放, 是否签名 。

签名上签名 登录

此处所用符号

输入完整定义、图片和其中每个字母的含义的符号。

如何: Nonlinear first-order equations and shocks

  1. Write the characteristics x = x_0 + g(x_0) t and note u is constant on each.
  2. Compute g'(x) and find its most negative value (a critical point of g').
  3. The breaking time is t* = -1 / min g'; if g' is never negative, no shock forms.
  4. For a jump, use the conservation form and the Rankine-Hugoniot speed s = (u_L + u_R)/2.
  5. Keep a shock only if u_L > u_R; otherwise replace it by the fan u = x/t.

人们对人们提出的问题

Why is the conservation form needed?

u u_x = (u²/2)_x only for smooth u. Across a jump the two forms give different speeds; the conservation form is the one that expresses a physical balance, so it decides the weak solution.

Does every initial profile that decreases somewhere form a shock?

Yes: if g' is negative somewhere, 1 + t g' vanishes at some finite t, and the classical solution breaks at the first such time.

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.

更多 Partial Differential Equations