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Consider the viscous conservation law \begin{equation} u_t+F(u)_x-au_{xx}=0,\quad\text{in }\mathbb{R}\times (0,\infty),\quad(1) \label{3-1} \end{equation} where $a>0$ and $F$ is uniformly convex.

(a). Show $u$ solves above PDE if $u(x,t)=v(x-\sigma t)$ and $v$ is defined implicitly by the formula $$ s=\int_{c}^{v(s)} \frac{a}{F(z)-\sigma z+b} d z \quad(s \in \mathbb{R}), $$ where $b$ and $c$ are constants.

(b). Demonstrate that we can find a traveling wave satisfying $$ \lim\limits_{s\rightarrow-\infty}v(s)=u_l,\lim\limits_{s\rightarrow \infty}v(s)=u_r $$ for $u_l>u_r$, if and only if $$ \sigma=\frac{F(u_l)-F(u_r)}{u_l-u_r}. $$

(c). Let $u^\varepsilon$ denote the above traveling wave solution of (1) for $a=\varepsilon$, with $u^\varepsilon(0,0)=\frac{u_l+u_r}{2}$. Compute $\lim\limits_{\varepsilon\rightarrow 0}u^\varepsilon$ and explain your answer.

My try:

(a). It is easy to check $u$ solve the PDE.

(b). By definition of $v$, we have $$ F(v(s))-\sigma v(s)+b=av'(s). $$ If we have $v'(\pm\infty)=0$, then $$ F(u_l)-\sigma u_l+b=0=F(u_r)-\sigma u_r+b $$ which implies $\sigma=\frac{F(u_l)-F(u_r)}{u_l-u_r}$. But I don't know how to show that $v'(\pm\infty)=0$.

(c). I have no much idea about it.

EditPiAf
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Q-Y
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  • I would have thought that, if a function converges to a a finite value in the limit to infinity i.e $u_{l}, u_{r}$, then the gradient of that function in the limit is necessarily zero because the function converging to a finite value would make that value an asymptote. I could be wrong though. – Matthew Cassell Jan 28 '20 at 09:32
  • @mattos The derivative must have finite value at infinity too in order to say so. As a counter-example, consider for instance $x \mapsto \sin(x^3)/x$. – EditPiAf Jan 28 '20 at 10:21
  • @Harry49 Thanks, I never was good at analysis. – Matthew Cassell Jan 28 '20 at 10:38

1 Answers1

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This post is related.

(b) Let us inject the traveling wave Ansatz $u(x,t) = v(s)$ with $s = x-\sigma t$ in the PDE. Then, we integrate the second-order ODE so-obtained once, so that $$ F(v) - \sigma v + b = a v' , $$ where $b$ is an integration constant*. One notes that if $v$ has a finite limit at $\pm \infty$, then $v'$ too. By virtue of the mean value theorem, $v'$ vanishes at $\pm \infty$. Thus, the limits of $v$ must satisfy $$ \sigma u_l - F(u_l) = b = \sigma u_r - F(u_r) \, , $$ i.e. $\sigma\, (u_l - u_r) = F(u_l) - F(u_r)$. Injecting the expression of $\sigma$ so-obtained in the differential equation, we have $$ a v' = F(v) - F(u_r) - \sigma (v - u_r) \leq 0 , \qquad \forall\; v \in [u_r, u_l] $$ due to the convexity of $F$ (the graph of a convex function lies below its chords). It is therefore possible to connect $u_l$ and $u_r$ with a smooth decreasing function $v$ if $u_l > u_r$, and if $\sigma$ represents the rate of increase of $F$ over $[u_r, u_l]$.

(c) Now, we are left with $u^\varepsilon(x,t) = v(s)$ and $s=x - \sigma t$, where $$ s = \int_{c}^{v(s)} \frac{\varepsilon}{F(z)-\sigma z + b} dz \, , \qquad \sigma = \frac{F(u_l) - F(u_r)}{u_l - u_r} . $$ The constant $b$ equals simultaneously $\sigma u_l - F(u_l)$ and $\sigma u_r - F(u_r)$. Evaluating the implicit integral equation at $s=0$, we may choose $c = \frac12(u_l + u_r)$.

On the one hand, the numerator of the integrand vanishes as $\varepsilon \to 0$. On the other hand, we know from (b) that the denominator of the integrand has a constant (negative) sign for $z \in ]u_l, u_r[$, and that it vanishes as $z\to u_l$ or $z\to u_r$. Therefore, to keep the integral equal to a given $s\neq 0$ as $\varepsilon \to 0$, we must either increase the value of $v(s)\to u_l$ if $s<0$, or decrease the value of $v(s)\to u_r$ if $s>0$. Finally, we obtain a piecewise constant solution, with a single discontinuity located at $x=\sigma t$. It is a shock wave that satisfies the Rankine-Hugoniot condition. This solution corresponds to the vanishing viscosity solution by definition.

*This differential equation is separable as $$ \text{d}s = \frac{a\, \text{d}v}{F(v)-\sigma v + b} $$ which provides the implicit formula in OP.

EditPiAf
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  • Would you mind explain a bit more on why $v’$ vanish at infinity? Since you also gave a counter example above that a function goes to 0 with derivative does not converge. – Q-Y Jan 28 '20 at 13:18
  • @whereamI If $v$ has a finite limit at $\pm \infty$, then the differential equation on $v$ tells us that $v'$ has a finite limit too (say, $v' \to \ell$ as $|x|\to\infty$). According to the mean value theorem, we have $(v(x+\delta) - v(x))/\delta \to \ell$ for fixed $\delta > 0$ as $|x|\to\infty$. Hence, since $v$ has a finite value at $\pm\infty$, we must have $\ell = 0$. – EditPiAf Jan 28 '20 at 14:08
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    So we can claim that if a function v and its derivative converges as $x\rightarrow\infty$, then the derivative must be $0$, right? – Q-Y Jan 28 '20 at 22:15
  • And would you mind explain a bit more on "Therefore, to keep the integral finite, the denominator must vanish too"? I can't see why we can obtain a piecewise constant solution. – Q-Y Jan 28 '20 at 22:45
  • Thank you very much! – Q-Y Jan 29 '20 at 05:36