Can someone help me to write down a formula for the general solution to the nonlinear partial differential equation $$u_t + u_x + u^2 = 0$$ and how do I show that if the initial data is positive and bounded, $0 ≤ u(0,x) = f(x) ≤ M$, then the solution exists for all $t > 0$, and $u(t,x) → 0$ as $t → ∞$
General solution of $u_t + u_x = -u^2$
characteristicsnonlinear dynamicspartial differential equations
Related Solutions
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:
- $\frac{\text d t}{\text d s} = 1$, letting $t(0)=0$, we know $t=s$.
- $\frac{\text d v}{\text d s} = -cv$, letting $v(0)=h'(x_0)$, we know $v=h'(x_0)e^{-ct}$.
- $\frac{\text d x}{\text d s} = v^2$, letting $x(0)=x_0$, we know $x=\frac{1}{2c}h'(x_0)^2(1-e^{-2ct}) + x_0$.
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
- $\frac{\text d u}{\text d s} = \tfrac23 v^3 - c u$, letting $u(0) = h(x_0)$, we know $u = h(x_0) e^{-ct} + \frac23\! \int_0^t e^{-c(t-s)} v(s)^3 \text d s$.
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.
A graphical representation of the problem in the $x$-$t$ plane may be relevant (see this post). The method of characteristic provides the set of lines $x=t-t_0+x_0$ along which $u$ is constant. Here, the value $u=x_0$ is specified along the unit circle ${x_0}^2+{t_0}^2=1$. Characteristics coming from the region $x_0+t_0< 0$ of the unit circle will cross the circle again in the region $x_0+t_0 >0$ where another boundary-value is specified. Hence, the problem is ill-posed.
To avoid this, we restrict the boundary data to the region $x_0+t_0\leq 0$ of the unit circle. One should note that characteristic curves become parallel to the unit circle at the extremities of this half-circle. To see what happens, we express the characteristic curves as $x-t-x_0=-t_0$ with $t_0=\pm\sqrt{1-{x_0}^2}$. Squaring this identity, we have $$ (x-t)^2 -2x_0(x-t) + {x_0}^2=1-{x_0}^2 $$ so that $$ x_0 = \frac{x-t \pm \sqrt{2-(x-t)^2} }{2} = u(x,t), $$ for $t-\sqrt{2} \leq x\leq t+ \sqrt{2}$. At the extremities of the domain, we have $x-t = \pm\sqrt{2}$. Differentiating the expression of $u$ along these curves would require to differentiate a square root in the vicinity of zero, which is not possible.
Best Answer
You get to solve $$ \frac{dt}1=\frac{dx}1=-\frac{du}{u^2} $$ leading to $$x-t=c_1, ~~u^{-1}-t=c_2$$ along the characteristic curves. As the characteristic curves inside the solution surface form a one-parameter family, there is a function $c_2=\phi(c_1)$ connecting the integration constants, thus $$ u(t,x)=\frac1{t+\phi(x-t)}. $$ Now apply all the other assumptions and try to prove the claims.
As $f(x)=u(0,x)=\frac{1}{\phi(t)}$, the solution can be rewritten as $$ u(t,x)=\frac{f(x-t)}{1+t\,f(x-t)}. $$
If $f(x_*)<0$ at some point, then select $t_*=-\frac1{f(x_*)}$ and consider the point $(t,x)=(t_*, x_*+t_*)$. The denominator will be zero, there will be a singularity at that point.