characteristicshyperbolic-equationsinitial-value-problemspartial differential equations
I want to solve the following PDE initial value problem
$u_t+(u-1)u_x=2$
and
$u (x,0)=\begin{cases}
1 & \text{for } x <0,\\
1-x & \text{for } 0<x <1\\
0 & \text{for } 1 <x
\end{cases}$
However, I find that I have intersecting characteristics between $x=t^2$ and $x=t^2-t+1$.
How would I apply the shockwave method in this case since the PDE is given? Is it possible to solve this PDE as is?
As outlined by OP's self-answer, $$ x(t) = x_0\exp(t\phi(x_0)) \, . $$ Several methods can be used to find the shock times (see e.g. this post). Here we examine the non-ambiguous dependence to the initial data. One observes that $\text d x/\text d x_0$ vanishes at $t=t_S$ such that $$ t_S = \frac{-1}{x_0\phi'(x_0)} = x_0 + \frac{1}{x_0} \, . $$ Thus, for positive times, the classical solution breaks at the breaking time $$t^*_+ = \inf_{x_0>0} t_S = 2\, .$$ Similarly one gets $t^*_- = -2$ for negative times. This may be confirmed by a plot of the characteristics for several $x_0$.
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.
Best Answer
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.