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Consider an inviscid Burgers equation:

$u_t + u u_x = 0.$

With the initial data:

$ u(x,0) = \left\{ \begin{array}{ll} 0 \quad & \text{if} \quad x < 0, \\[0.5em] 1 & \text{if} \quad x > 0.\end{array} \right. $

Possible weak solutions to this problem include, for example:

$ u_1(x,t) = \left\{ \begin{array}{ll} 0 \quad & \text{if} \quad x < t/2, \\[0.5em] 1 & \text{if} \quad x > t/2.\end{array} \right. \hspace{10mm} $ $ u_2(x,t) = \left\{ \begin{array}{lll} 0 \quad & \text{if} \quad x < 0, \\[0.5em] x/t & 0 < x < t,\\[0.5em] 1 & \text{if} \quad x > t.\end{array} \right. $

Every material that provides such examples states that it is easy to check that these are both solutions of our problem. I do not see how that is done.

I know how to obtain the equation:

$ \int_{0}^\infty \int_{-\infty}^\infty [ q \phi_t + f(q) \phi_x ] dx dt + \int_{-\infty}^\infty q(x,0) \phi(x,0) dx = 0, $

But I do not see how to "test" if the weak solutions provided above satisfy this. I tried writing the solutions as Heavside functions and integrating them, but I am not sure how to handle an arbitrary test function.

I assume there are some solution I could construct that would not be the weak solution to our initial data, how to check that?

nevermind
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    The wikipedia page on Weak Solutions says the following. I cannot really follow it but maybe it is helpful to you. if you manage to understand, I would appreciate an answer to your own question.

    "That u defined in this way satisfies equation (2) is easy enough to check, one needs to integrate separately on the regions above and below the line x = t and use integration by parts.) A solution u of equation (2) is called a weak solution of equation (1)."

     – lucidbrot Jul 06 '17 at 14:10
  • One key point seems to be that this must hold for all test functions phi, so it must especially hold for a testfunction where phi is 0 for t=0. Then you have only the left integral and must show that it equals to 0. – lucidbrot Jul 07 '17 at 06:18
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    @lucidbrot By definition of weak solutions, the proof must be carried out for all test funtions $\phi$ in $C_0^1(\mathbb{R}\times\mathbb{R}^+)$. Luckily, it turns out to be not such a hard task, since integration is performed over all of $\mathbb{R}\times\mathbb{R}^+$, which is a fixed domain. See answer for details. – EditPiAf Sep 21 '17 at 09:24

1 Answers1

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To show that $u$ is a weak solution of this initial-value problem, we show that $$ \int_{0}^{\infty}\int_{-\infty}^{\infty} \left[ \phi_t u + \phi_x f(u)\right] \mathrm{d}x\, \mathrm{d}t = -\int_{-\infty}^{\infty} \phi(x,0)\, u (x,0)\,\mathrm{d}x $$ is satisfied for all $\phi$ in $C_0^1(\mathbb{R}\times \mathbb{R}^+)$. Let us prove this identity in the case $$ u(x,t) = u_1(x,t) = \left\lbrace\begin{aligned}&0 &&\text{if } x < t/2\, ,\\ &1 &&\text{if } x > t/2\, ,\end{aligned}\right. $$ with the flux function of Burgers' equation $f(u) = \frac{1}{2}u^2$. To do so, we split the integral in two parts, and switch the integrals according to the Fubini theorem: \begin{aligned} \int_{0}^{\infty}\!\int_{-\infty}^{\infty} \left[ \phi_t u + \phi_x f(u)\right] \mathrm{d}x\, \mathrm{d}t &= \int_{-\infty}^{\infty}\int_{0}^{\infty} \phi_t u\, \mathrm{d}t\, \mathrm{d}x + \int_{0}^{\infty}\!\int_{-\infty}^{\infty} \phi_x f(u)\, \mathrm{d}x\, \mathrm{d}t \\ &= \int_{0}^{\infty}\int_{0}^{2x} \phi_t \, \mathrm{d}t\, \mathrm{d}x + \frac{1}{2}\int_{0}^{\infty}\!\int_{t/2}^{\infty} \phi_x \, \mathrm{d}x\, \mathrm{d}t \\ &= \int_{0}^{\infty}\left[\phi(x,2x) - \phi(x,0)\right]\mathrm{d}x - \frac{1}{2}\int_{0}^{\infty} \phi(t/2,t)\, \mathrm{d}t \\ &= -\int_{0}^{\infty}\phi(x,0)\,\mathrm{d}x \\ &= -\int_{-\infty}^{\infty}\phi(x,0)\,u(x,0)\,\mathrm{d}x \, . \end{aligned} In the case where $u(x,t) = u_2(x,t)$, the proof is quite similar. Note however that now $u$ is continuous, hence its first derivatives include jumps, but no higher-order singularity such as Dirac deltas. Also, one observes that $u_t$, $u_x$ equal zero outside the domain $\lbrace 0\leq t, 0\leq x\leq t\rbrace$. Furthermore, \begin{aligned} \int_{0}^{\infty}\int_{-\infty}^{\infty} u_t \phi\, \text d x\,\text d t &= -\int_{0}^\infty\int_{0}^{t} \frac{x}{t^2} \phi\, \text d x\,\text d t , \end{aligned} \begin{aligned} \int_{0}^{\infty}\int_{-\infty}^{\infty} \tfrac12(u^2)_x \phi\, \text d x\,\text d t &= \int_{0}^\infty\int_{0}^{t} \frac{x}{t^2} \phi\, \text d x\,\text d t . \end{aligned} The sum of these two integrals vanishes, thus ending the proof (via IBP).

This is a particular case of exercise 3.4 p 29 of the book [1].

[1] R.J. LeVeque: Numerical Methods for Conservation Laws, 2nd ed., Birkhäuser, 1992.

EditPiAf
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