Inhomogenous nonlinear transport equation $u_t+uu_x = -Du$

We have the following setup: $$u_t+uu_x = -Du \\ u(x,0)=\sin x.$$ The question is to find the time $T_s$ of a first shock formation. So basically, I need to solve the equation using method of characteristics and go from there.

But when I solve the corresponding quasilinear equation,I am getting the implicit solution $$u(x,t) = \dfrac{\sin(x-tu)}{1+Dt}$$ which does not seem right.

Can anybody provide a solution using the method of characterisics so I can pinpoint where my mistake is?

Solution 1:

First, it should be clearly stated what $D$ is here. Many times it represents a differential operator, but here I assume it is meant to be an arbitrary constant. I'll rewrite the equation with $q = u_x, p = u_t, z = u,$ so that $F(p,q,z) = 0,$ $$p - zq + Dz = 0.$$ Next, apply the method of characteristics, where only the $z, x$ characteristics are needed since its not fully nonlinear.

\begin{align} \frac{dz}{dt} &= q \partial_q F + p \partial_p F = p -zq = -Dz; \;\; z(x_0,0) = sin(x_0)\\ \frac{dx}{dt} &= \partial_q F = -z; \;\; x(t = 0) = x_0. \end{align} Solving these gives us our solution, \begin{align} z(x_0,t) &= \sin(x_0)e^{-Dt}\\ x &= \frac{\sin{(x_0)}(e^{-Dt} - 1) + Dx_0}{D} \end{align} Now if we could solve the latter equation for $x_0 = x_0(x,t),$ we could plug this back into the equation for $z(x_0,t)$ to get $u(x,t)$ explicitly.

If I wanted to show a finite time blow up, I'd formally take the $x$ derivative of $z(x_0,t),$ \begin{align} \frac{dz}{dx} &= \cos{(x_0)}e^{-Dt}\frac{dx_0}{dx} = \cos{(x_0)}e^{-Dt} \frac{1}{\frac{dx}{dx_0}}\\ &= \frac{D\cos{(x_0)}e^{-Dt}}{\cos{(x_0)}(e^{-Dt} - 1) + D}\\ &= \frac{De^{-Dt}}{e^{-Dt} + \frac{D}{\cos{(x_0)}}-1}\\ \end{align} Perhaps I have errored, but this gradient only blows up if $e^{-Dt} = \frac{D}{\cos{x_0}} - 1,$ at some $t.$ Therefore, we need $D \leq 2$ for this to happen at all.

I am a little surprised about this, that the RHS forcing term changes this answer from the normal result, i.e. blow up when $t = \frac{-1}{cos{(x_0)}}.$

Solution 2:

This (dissipative) Burgers' equation with relaxation is a typical example of conditional shock formation. The method of characteristics gives

• $\frac{\text d t}{\text d s} = 1$. Letting $t(0) = 0$, we have $t=s$.
• $\frac{\text d u}{\text d s} = -Du$. Letting $u(0) = \sin x_0$, we have $u = \sin (x_0)\, e^{-Ds}$.
• $\frac{\text d x}{\text d s} = u$. Letting $x(0) = x_0$, we have $x = -\frac{1}{D} \sin(x_0)(e^{-Ds}-1) + x_0$.

This is sometimes written in implicit form as $$u = \sin\left(x - \frac{e^{Dt}-1}{D} u\right) e^{-Dt}\, .$$ Now, let us compute the breaking time. Differentiating $x$ w.r.t. $x_0$ , we have $\frac{\text d x}{\text d x_0} = -\frac{1}{D} \cos(x_0)(e^{-Dt}-1) + 1$. This quantity vanishes at $t = \frac{1}{D}\ln\left(\frac{\cos x_0}{D + \cos x_0}\right)$, where characteristics intersect. The smallest such positive time corresponds to the breaking time $T_s$, i.e. $$ T_s = \frac{1}{D}\ln\left(\frac{1}{1-D}\right) ,\quad \text{if}\quad D < 1 \, . $$ Otherwise, if $D\geqslant 1$, no shock occurs.

Alternatively, one can follow the method leading to Eq. (6.10) p. 36 of (1). Differentiating the PDE with respect to $x$ gives $$ u_{tx} + (u_x)^2 + u u_{xx} = -Du_x \, . $$ Now, we introduce $q = u_x$ and the directional derivative $(\cdot)' = \partial_t + u \partial_x$, so that the previous equations rewrite as $u' = -Du$ and $q' + q^2 = -D q$. These differential equations have the solutions $u = u_0e^{-Dt}$ and $$ q = \frac{q_0 e^{-Dt}}{1 + \frac{q_0}{D} (1-e^{-Dt})} \, , $$ where $q_0 = q(0) = \cos x_0$. Therefore, $q$ blows up at $t = \frac{1}{D}\ln\left(\frac{\cos x_0}{D + \cos x_0}\right)$, and the same expression of $T_s$ is obtained.

(1) P.D. Lax, Hyperbolic systems of conservation laws and the mathematical theory of shock waves. SIAM, 1973. doi:10.1137/1.9781611970562.ch1