Unbounded entropy solution to Burgers' equation

Question

I need to solve Problem 3.5 - 11 p. 164 of the book Partial Differential Equations by Lawrence C. Evans (2nd ed., AMS, 2010):

11. Show that $$ u(x,t) = \begin{cases} -\dfrac{2}{3}\left(t+\sqrt{3x+t^2}\right); & \text{if } 4x + t^2 >0\\ 0; & \text{if } 4x + t^2<0 \end{cases} $$ is an (unbounded) entropy solution of $u_t + \left(\dfrac{u^2}{2}\right)_x=0$.

Clearly it is easy to see that this is both unbounded and a solution to the given PDE, however I am not sure how to gather any information to say that it satisfies the entropy condition from the solution alone. Any hints would be welcomed.

Edit. I know the definition (Evans, §3.4.3.b p. 150). A weak solution $u\in L^\infty(\Bbb R\times (0,\infty))$ of the initial value problem $u_t + \left(\dfrac{u^2}{2}\right)_x = 0$ with data $u|_{t=0} = g$ is an entropy solution if

$$ u(x+z,t) - u(x,t) \leq C \left(1 + \frac{1}{t}\right) z \tag{ii} $$ for some constant $C>0$ and a.e. $x$, $z \in \Bbb R$, $t>0$, with $z>0$.

Is it the one needed to solve the problem? How to use it?

Answer 1

$\DeclareMathOperator{\supp}{supp}\def\d{\mathrm{d}}\def\peq{\mathrm{\phantom{=}}{}}$Note that $u_t + u u_x = u_t + \left( \dfrac{u^2}{2} \right)_x = 0$ holds for $x > -\dfrac{t^2}{4}$, and $u = 0$ for $x < -\dfrac{t^2}{4}$. Denote$$ g(x) = u(x, 0) = \begin{cases} -\dfrac{2\sqrt{x}}{\sqrt{3}}; & x > 0\\ 0; & x < 0 \end{cases}. $$

For any test function $v$, suppose $\supp(v) \subseteq \left( -\dfrac{T^2}{4}, \dfrac{T^2}{4} \right) × [0, T)$, then\begin{align*} &\peq \int_0^{+∞} \int_{-∞}^{+∞} uv_t \,\d x\d t = \int_0^T \int_{-\tfrac{T^2}{4}}^{\tfrac{T^2}{4}} uv_t \,\d x\d t = \int_{-\tfrac{T^2}{4}}^{\tfrac{T^2}{4}} \int_0^T uv_t \,\d t\d x\\ &= \int_{-\tfrac{T^2}{4}}^0 \int_{\sqrt{-4x}}^T uv_t \,\d t\d x + \int_0^{\tfrac{T^2}{4}} \int_0^T uv_t \,\d t\d x\\ &= \int_{-\tfrac{T^2}{4}}^0 \left( uv \Biggr|_{t = \sqrt{-4x}}^{t = T} uv_t - \int_{\sqrt{-4x}}^T u_tv \,\d t \right)\d x + \int_0^{\tfrac{T^2}{4}} \left( uv \Biggr|_{t = 0}^{t = T} - \int_0^T u_tv \,\d t \right)\d x\\ &= - \left( \int_{-\tfrac{T^2}{4}}^0 \int_{\sqrt{-4x}}^T u_tv \,\d t\d x + \int_0^{\tfrac{T^2}{4}} \int_0^T u_tv \,\d t\d x \right)\\ &\peq + \int_{-\tfrac{T^2}{4}}^0 uv \Biggr|_{t = \sqrt{-4x}}^{t = T} \,\d x + \int_0^{\tfrac{T^2}{4}} uv \Biggr|_{t = 0}^{t = T} \,\d x\\ &= -\int_{-\tfrac{T^2}{4}}^{\tfrac{T^2}{4}} \int_0^T u_tv \,\d t\d x - \int_{-\tfrac{T^2}{4}}^0 uv \Biggr|_{t = \sqrt{-4x}} \,\d x - \int_0^{\tfrac{T^2}{4}} uv \Biggr|_{t = 0} \,\d x\\ &= -\int_0^T \int_{-\tfrac{T^2}{4}}^{\tfrac{T^2}{4}} u_tv \,\d x\d t + \int_{-\tfrac{T^2}{4}}^0 2\sqrt{-x} · v(x, \sqrt{-4x}) \,\d x - \int_0^{\tfrac{T^2}{4}} g(x) v(x, 0) \,\d x\\ &= -\int_0^T \int_{-\tfrac{T^2}{4}}^{\tfrac{T^2}{4}} u_tv \,\d x\d t + \int_0^T \frac{t^2}{2} · v\left( -\frac{t^2}{4}, t \right) \,\d t - \int_0^{\tfrac{T^2}{4}} g(x) v(x, 0) \,\d x, \end{align*}\begin{align*} &\peq \int_0^{+∞} \int_{-∞}^{+∞} \frac{u^2}{2} · v_x \,\d x\d t = \int_0^T \int_{-\tfrac{T^2}{4}}^{\tfrac{T^2}{4}} \frac{u^2}{2} · v_x \,\d x\d t = \int_0^T \int_{-\tfrac{t^2}{4}}^{\tfrac{T^2}{4}} \frac{u^2}{2} · v_x \,\d x\d t\\ &= \int_0^T \left( \frac{u^2}{2} · v\Biggr|_{x = -\tfrac{t^2}{4}}^{x = \tfrac{T^2}{4}} - \int_{-\tfrac{t^2}{4}}^{\tfrac{T^2}{4}} \left( \frac{u^2}{2} \right)_x · v \,\d x \right) \d t\\ &= -\int_0^T \frac{t^2}{2} · v\left( -\frac{t^2}{4}, t \right) \,\d t - \int_0^T \int_{-\tfrac{T^2}{4}}^{\tfrac{T^2}{4}} uu_xv \,\d x\d t, \end{align*}$$ \int_{-∞}^{+∞} g(x) v(x, 0) \,\d x = \int_0^{\tfrac{T^2}{4}} g(x) v(x, 0) \,\d x, $$ thus\begin{align*} &\peq \int_0^{+∞} \int_{-∞}^{+∞} \left( uv_t + \frac{u^2}{2} · v_x \right) \,\d x\d t + \int_{-∞}^{+∞} g(x) v(x, 0) \,\d x\\ &= -\int_0^T \int_{-\tfrac{T^2}{4}}^{\tfrac{T^2}{4}} u_tv \,\d x\d t - \int_0^T \int_{-\tfrac{T^2}{4}}^{\tfrac{T^2}{4}} uu_xv \,\d x\d t\\ &= -\int_0^T \int_{-\tfrac{T^2}{4}}^{\tfrac{T^2}{4}} (u_t + uu_x) v \,\d x\d t = 0. \end{align*} Therefore, $u(x, t)$ is a weak solution.

Finally, since $u(x, t)$ is decreasing with respect to $x$, then$$ u(x + z, t) - u(x, t) \leqslant 0 \leqslant \left( 1 + \frac{1}{t} \right) z. \quad \forall x \in \mathbb{R},\ z > 0,\ t > 0 $$

Answer 2

As suggested by @Saad, setting $t=0$ in $u(x,t)$ gives the initial data $$ u(x,0) = g(x) = \left\lbrace \begin{aligned} &{-{2}}\sqrt{{x}/{3}} & &\text{if}\quad x>0, \\ &0 & &\text{if}\quad x<0 . \end{aligned} \right. $$ Instead of proving directly that $u$ is an entropy-condition-satisfying weak solution, let us derive the entropy solution to this initial-value problem and compare with the proposed $u$. We first derive a classical solution by applying the method of characteristics for short times. As long as the method of characteristics is valid, we have $u=g(x-u t)$. According to the expression of $g$, two cases must be considered:

• for $x$ sufficiently small, $g(x)=0$, so that $u(x,t)=0$;
• for $x$ sufficiently large, $g(x) = -2\sqrt{x/3}$, so that $u(x,t)={-\frac{2}{3}}\big(t+\sqrt{3x+t^2}\big)$.

Below is a plot of the characteristic curves in the $x$-$t$ plane deduced from the initial data:

An intersection of the characteristic lines is observed around the origin, and the method of characteristics fails there. According to the Lax entropy condition, a shock wave is generated. Its position $x_s(t)$ must satisfy the Rankine-Hugoniot condition to ensure that such a wave is a weak solution. Therefore, the shock speed is given by $$ x'_s(t) = \frac{1}{2}\left( 0 - \frac{2}{3}\big(t+\sqrt{3 x_s(t)+t^2}\big) \right) $$ with initial position $x_s(0)=0$, i.e., $x_s(t) = -t^2/4$. By construction, the (unique) entropy solution is therefore $$ u(x,t) = \left\lbrace \begin{aligned} &{-\frac{2}{3}}\big(t+\sqrt{3x+t^2}\big) & &\text{if}\quad x>-t^2/4, \\ &0 & &\text{if}\quad x<-t^2/4 , \end{aligned} \right. $$ which ends the proof.