I'm trying to solve the following partial differential equations:
$$
u_t + \frac{1}{3}{u_x}^3 = 0 \tag{a}
$$
$$
u_t + \frac{1}{3}{u_x}^3 = -cu \tag{b}
$$
with the initial value problem
$$
u(x,0)=h(x)=
\left\lbrace
\begin{aligned}
&e^{x}-1 & &\text{for}\quad x<0\\
&e^{-x}-1 & &\text{for}\quad x>0
\end{aligned}
\right.
$$
My idea was to set $v(x,t)=u_x(x,t)$, because then I get
the transport equation in $v$ which I am able to solve:
$v_t + v^2v_x =0$.
But when I do this, my solution for $v$ is
$$
v(x,t)=
\left\lbrace
\begin{aligned}
&\phantom{-}ae^{x-v^2 t} & &\text{for}\quad x<0\\
&{-a}e^{-x+v^2 t} & &\text{for}\quad x>0
\end{aligned}
\right.
$$
Can someone help me with this equation? Is
$u_x = a e^{x-u_x^2 t}$ the correct answer?
Or should I maybe do something very different to solve this equation?
IVP for nonlinear PDE $u_t + \frac{1}{3}{u_x}^3 = -cu$
characteristicshamilton-jacobi-equationhyperbolic-equationspartial differential equations
Related Solutions
Following the comments, we transform this problem by setting $v=u-1$ as $$ v_t + v v_x = 2, \qquad v(x,0) = \left\lbrace \begin{aligned} &0 &&\text{for } x<0\\ &{-x} &&\text{for } 0<x<1\\ &{-1} &&\text{for } 1<x\\ \end{aligned}\right. $$ The breaking time for the Burgers equation above is $t_b = -1/\inf v_x(x,0) = 1$. Hence, there is indeed a shock formation. Before the shock, the solution is given by the method of characteristics. Here, the characteristic curves are $x = v(x_0,0) t + t^2 + x_0$ along which $v = v(x_0,0)+2t$. Hence, for $t<1$, the solution reads $$ v(x,t) = \left\lbrace \begin{aligned} &2t &&\text{for } x<t^2\\ &\tfrac{t^2-x}{1-t} + 2t &&\text{for } t^2<x<t^2-t+1\\ &{-1}+2t &&\text{for } t^2-t+1<x\\ \end{aligned}\right. $$ When characteristic curves intersect, the Rankine-Hugoniot condition gives the shock speed $\dot x_s(t) = \frac{1}{2}(2t - 1 + 2t)$ with $x_s(1) = 1$. Therefore, for $t\geq 1$, the solution reads $$ v(x,t) = \left\lbrace \begin{aligned} &2t &&\text{for } x< \tfrac{1}{2}(4t-1)\\ &{-1}+2t &&\text{for } \tfrac{1}{2}(4t-1)<x\\ \end{aligned}\right. $$ To recover $u$, add $1$ to the values of $v$ above.
If needed, here is a sketch of the characteristic curves in $x$-$t$ plane, which may help to see better what happens:
The solution obtained by the method of characteristics up to the breaking time satisfies $u = \phi (x-ut)$ for $t<1$, i.e. $$ u(x,t) = \left\lbrace\begin{aligned} &1 & &\text{if}\quad x \leq t \\ &\tfrac{1-x}{1-t} & &\text{if}\quad t \leq x\leq 1\\ &0 & &\text{if}\quad x\geq 1 \end{aligned}\right. $$ At the breaking time, the shock speed $s$ is given by the Rankine-Hugoniot condition $s = \tfrac12 (1 + 0)$. Therefore, after the breaking time $t\geq 1$, $$ u(x,t) = \left\lbrace\begin{aligned} &1 & &\text{if}\quad x < 1 +\tfrac12 (t-1) \\ &0 & &\text{if}\quad x > 1 +\tfrac12 (t-1) \end{aligned}\right. $$
To your questions:
- The Rankine-Hugoniot condition $[\![\frac{1}{2}u^2]\!] = s [\![u]\!]$ is a condition satisfied by discontinuous weak solutions of the Burgers' equation. A weak solution which satisfies the Rankine-Hugoniot condition is not necessarily a solution to the proposed initial-value problem $u(x,0) = \phi(x)$. To be a solution, it must be in agreement with the characteristics equation until they cross, i.e. the fact that $u$ is constant along the curves $u = \phi(x - ut)$. Moreover, since weak solutions are not unique, it must satisfy an additional entropy condition. In the case of Burgers' equation, the Lax entropy condition amounts to the following statement: "if characteristics cross, then a shock-wave arises; but if they separate, a rarefaction wave occurs".
- The location of a discontinuity is deduced from the characteristics equation, whereas its speed is deduced from the Rankine-Hugoniot condition. In the present case, a shock occurs at the time $t=1$ (characteristics intersect). Its speed deduced from the Rankine-Hugoniot condition is $s=\frac{1}{2}$.
Best Answer
Both equations (a) and (b) are Hamilton-Jacobi equations. Indeed, they are derived from a canonical transformation involving a type-2 generating function $u(x,t)$ which makes vanish the new Hamiltonian $$ K = H(x,u_x) + (\partial_t + c)\, u \, . $$ Here, $H(q,p)=\frac{1}{3}p^3$ is the original Hamiltonian and $\partial_t + c$ defines the time-differentiation operator. Setting $v=u_x$, we have $$ v_t = u_{tx} = -\left(\tfrac{1}{3}{u_x}^3\right)_x - cu_x = -v^2v_{x} - cv \, . $$ Thus, we consider the first-order quasilinear PDE $v_t + v^2v_{x} = -cv$ with initial data $v(x,0) = h'(x) = \pm e^{\pm x}$ for $\pm x<0$, and we apply the method of characteristics:
Injecting $h'(x_0) = ve^{ct}$ in the equation of characteristics, one obtains the implicit equation $$ v = h'\!\left(x-v^2\frac{e^{2ct}-1}{2c}\right) e^{-ct}\, . $$ Along the same characteristic curves, we have
Thus, we get $$ u = \left(h\!\left(x-v^2\frac{e^{2ct}-1}{2c}\right) + h'\!\left(x-v^2\frac{e^{2ct}-1}{2c}\right)^3 \frac{1-e^{-2ct}}{3c} \right) e^{-ct} \, , $$ where the link between $x_0$ and $v$ along characteristics has been used. For short times, the previous solution is valid. The method of characteristics breaks down when characteristics intersect (breaking time). We use the fact that $\frac{\text d x}{\text d x_0}$ vanishes at the breaking time $$ t_B = \inf_{x_0\in \Bbb R} \frac{-1}{2 c} \ln\left(1 + \frac{c}{h'(x_0)h''(x_0)}\right) . $$ However, it seems pointless to look further for full analytical expressions in the general case.
If $c=0$, the characteristics are straight lines $x=x_0+v^2t$ along which $v=h'(x_0)$ is constant, and along which $u = h(x_0) + \frac23 v^3 t$. A sketch of the $x$-$t$ plane is displayed below:
The breaking time becomes $t_B = \inf_{x_0} -(2h'(x_0)h''(x_0))^{-1} = 1/2$. For short times $t<t_B$, the implicit equation for $v$ reads $v = h'(x-v^2t)$, i.e. $v = \pm e^{\pm (x-v^2t)}$ if $\pm(x-v^2t)<0$. Its analytic solution $$ v(x,t) = \pm\exp\! \left(\pm x- \tfrac{1}{2}W(\pm 2t e^{\pm 2x})\right) \quad\text{for}\quad {\pm (}x-t) < 0 $$ involves the Lambert W function. The expression of $u$ is deduced from $u = h(x-v^2 t) + \frac23 v^3 t$. For larger times $t>t_B$, particular care should be taken when computing weak solutions (shock waves) since the flux $f:v\mapsto \frac{1}{3}v^3$ is nonconvex.