I asked this question before but did not have any luck with an answer. It might be a student level question but I need to understand that with possibly some help. I am considering the hyperbolic equation of the form $$u_t+\frac{x}{T-t}u_x=0$$ and some initial data $u(x,0)=u_0(x)$. I would like to claim that this is a well posed problem in $\mathbb{L}^2$ on $[0,T]$. For that existence, uniqueness and stability needs to be established. So I have the following steps to achieve that.

Step 1: there is a change of variables I can employ as $z=x*(T-t)$. Then for $v(z(t,x),t)=u(x,t)$ I can find $$u_t=v_t+v_z(-x),\; u_x=v_z(T-t).$$Plug them into original equation I obtain an equivalent formulation to the original problem(up to a change of variables): $$v_t=0,\; v(z,0)=u_0(z/T,0).$$ The alternative formulation is clearly trivially has a unique solution for all $t\in [0,T]$, thus I claim the original problem does as well. Moreover, $u(x,T)=v(0,T)=u_0(0,0)$, so even though it initially looks like a not defined p.d.e. at that time it does have a solution.

Step 2: show stability via energy methods. For an arbitrary hyperbolic equation $w_t-a(y)w_y=0, \; y \in [0,1]$: \begin{align*} \frac{d||w||^2}{dt}&=(w_t,w)+(w,w_t)=(a(y)w_y,w)+(w,a(y)w_y)\\ &=(a(y)w_y,w)-(w_y,a(y)w)-(w,a_y(y)w)+a(y)ww|^1_0\\ &\leq max_{y\in [0,1]}|a_y(y)|||w||^2+a(y)ww|^1_0 \end{align*} Thus, the energy estimate boils down to the bound of $a_y$ provided boundary conditions are bounded. I have to estimate the stability of the original equation for $u(x,t)$ and it is not equivalent to the stability of $v(z,t)$. But from the energy estimate above $a_x=\frac{1}{T-t}$ which is not bounded as $t \to T$. Does that imply that is not well posed or energy methods did not work? What are other approaches to prove stability here as I can explicitly find the solution, so it perfectly stable on $[0,T]$? In this case everything is solvable, but I am considering the general case where the trajectories do have this kind of asymptotic behavior.


1 Answer 1


It is certainly not true that just because one way to prove a result doesn't work, the result is actually false. In other words, just because this way doesn't work doesn't mean that the equation isn't stable.

In your case, you can express the solution $u(x,T)$ using the method of characteristics as $u(x,T)=u_0(y)$ with some $y=y(x)$ (you find $y$ by backtracing the characteristic starting at $x,t=T$). Using this, you will immediately get the estimate $$\|u(\cdot,T)\|_{L^p(0,1)} \le \|u_0(\cdot)\|_{L^\infty(0,1)}.$$

If one follows this a bit, one notices that one can rewrite the equation as $$ \left( 1 \atop \frac{x}{T-t} \right) \cdot \left( \partial_t \atop \partial_x \right) u(x,t) = 0. $$ In other words, the characteristics turn horizontal in the $x-t$ diagram as $t\rightarrow T$; following back all characteristics we find that every characteristic ending in $x,T$ in fact started at $0,0$. In other words, we do indeed find that $u(x,T)=u(0,0)$ as you have already noticed. This yields a trivial stability bound.

What you see is that there are often multiple ways of proving a particular statement. That there are additional ways that do not work does not invalidate the result.

  • $\begingroup$ thanks for the answer. In fact, I'd like to have $$\|u(\cdot,T)\|_{L^p(0,1)} \le \|u_0(\cdot)\|_{L^p(0,1)}$$ for $p=2$, for example. If $p=\infty$ the inequality is trivial, but if I try to write two integrals on both sides in ${L}^2(0,1)$ and do change of variables in one of the integrals, than a Jacobian again creates this singularity and I get stuck with a proof. What are the tools to show the relation between two norms in ${L}^2(0,1)$? $\endgroup$
    – Kamil
    Commented Jan 1, 2013 at 0:37
  • $\begingroup$ Well, you can't find a way to prove the result because it's not true :-) You have already proved (and the method of characteristics confirmed) that $u(x,T)=u_0(0)$. So think of a sequence of functions $u_0^{(n)}(x)=(1-x)^n$. For all of these, $\|u^{(n)}(\cdot,T)\|_{L^p}=1$ but $\|u_0^{(n)}(\cdot)\|_{L^p}\rightarrow 0$ as $n\rightarrow \infty$. In other words, there isn't even a $C>0$ so that $\|u^{(n)}(\cdot,T)\|_{L^p} \le C \|u_0^{(n)}(\cdot)\|_{L^p}$ for $p<\infty$. Note that this is true here even for continuous initial conditions; it's only going to be worse for discontinuous ones. $\endgroup$ Commented Jan 2, 2013 at 0:01
  • $\begingroup$ I see your point. However, can I claim that the problem is stable in $L^2(0,1)$ for $t \in [0,T)$ and stable in $L^{\infty}(0,1)$ for $t\in [0,T]$? And second question, when I solve it using methods of characteristics with a quadratic spline interpolation I observe the $O(h^2)$ convergence rate at $t=T$ when I measure discrete $L^2$ norm. But even though the problem is not $L^2$ well posed there this is just a luck I see that norm to be $O(h^2)$? Am I supposed to measure discrete $L^{\infty}$ then? $\endgroup$
    – Kamil
    Commented Jan 2, 2013 at 2:04
  • $\begingroup$ I imagine you can prove $L^p$ estimates with a constant that depends on both $p$ and $(t-T)$. As for the convergence rate -- I don't know. $\endgroup$ Commented Jan 2, 2013 at 22:42
  • $\begingroup$ As a postscript: what do you mean when you say that you solve the equation using the method of characteristics with a quadratic spline interpolation? Convergence rates for hyperbolic equations are notoriously difficult to prove in the general case, and one would need to start with defining what exactly your proposed method does :-) $\endgroup$ Commented Jan 2, 2013 at 23:18

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