# Can two solutions of a PDE with different initial conditions coincide after finite time?

Consider an ODE $$\dot x=f(x),\quad x(0)=x_0,$$ where $$f$$ is a smooth function. It is well-known that if $$y$$ is another solution to this ODE with different initial condition $$y_0\neq x_0$$, then the trajectories of $$x(t)$$ and $$y(t)$$ do not intersect - in particular, for any finite $$t$$ we have $$x(t)\neq y(t)$$.

I wonder if the same is true for PDEs. I'm interested in the following simple example: $$\partial_t u=\partial_{xx} u+f(t,(u(t,x)),$$ with periodic boundary conditions. Here $$x$$ lives on a circle $$[0,1]$$, $$t\ge0$$ and $$f$$ is a nice function. Is it true that if $$u,v$$ are two solutions of this PDE with $$u(0)\neq v(0)$$, then $$u(t)\neq v(t)$$ for any $$t$$? (equivalent statement: if $$u(1)=v(1)$$ is it true that $$u(0)=v(0)$$?) How can we prove this?

UPD: the case $$f\equiv0$$ can be found inn Evans' book and it involves taking second derivative of the energy. Is it possible to extend this technique to the case $$f\neq0$$?

• Have a look at this document which discusses techniques for proving uniqueness (energy, maximum principle, viscosity). However note that even the case where $f=0$ (the usual heat equation) does not guarantee uniqueness with all boundary conditions. The boundary conditions are critical, and you have not given any. Commented Apr 13, 2020 at 14:30
• thanks @NinadMunshi For simplicity, let's assume that we are on a torus, so no boundary conditions needed :)
– Oleg
Commented Apr 13, 2020 at 14:32
• I don't think you understand me, a boundary condition is always required. Just because it takes a weird form does not mean it doesn't exist. Without it there is no chance for uniqueness. Also realize that temporal boundaries are also boundaries. Anyway, you have the tools to evaluate when uniqueness might be possible. Commented Apr 13, 2020 at 14:38
• I also don't have here any issues with uniqueness - let's suppose that we know that this equation has a unique solution. How can we show that two different solutions can't collapse?
– Oleg
Commented Apr 13, 2020 at 14:38
• How can you not see that what you are asking for is uniqueness? You are asking whether two solutions with the same initial condition have to take the same path, which is just the contrapositive of what you have asked Commented Apr 13, 2020 at 14:41

Tychonoff gives an infinite family of solutions of the heat equation on $$(x,t) \in \Bbb{R} \times \Bbb{R}$$, all of which are $$0$$ everywhere at time $$t \leq 0$$ and different subsequently. These solutions are also discussed on MSE.

Such solutions are backwards from what you request -- they are nonunique at a prior time, and then subsequently become unique. However, by the choice $$f(t, u(t,x)) = 2\partial_t u(t,x)$$, time reversal is implemented. Tychonoff's solutions are then distinct on $$t <0$$ and identically zero on $$t \geq 0$$. It would be no great feat to adapt these to $$(x,t) \in S^1 \times \Bbb{R}$$.

• Thank you for your answer, but I think that for the Tychonoff solution to be non trivial it is crucial that the domain is unbounded. On the other hand, my domain is bounded (it's 1d torus as we discussed) and thus any solution being a continuous function is also bounded. Thus, my equation does have forward uniqueness and your idea won't work.
– Oleg
Commented Apr 13, 2020 at 17:34
• No, boundedness of the domain is essential - in a bounded domain you always have uniqueness - check Evans' PDE book, Section 2.3.4. Note that Tychonoff solution won't work on a circle: indeed $u(1)$ will coincide with $u(-1)$ (because the power series is even) but $du/dx (1)$ will be equal to $- du/dx (-1)$ (again because the power series is even).
– Oleg
Commented Apr 13, 2020 at 21:12
• Is your claim that you are wasting everyone's time with your question or is your claim that you have a proof that Tychonoff's counterexample cannot be adapted to a compact spatial domain? Commented Apr 13, 2020 at 21:15
• There is a result from Evans book (or any other lecture notes) saying that heat equation on a bounded domain has a unique solution. Therefore there are no nontrivial solutions. Check e.g. Theorem 1.1. from this MIT lecture notes math.mit.edu/~jspeck/18.152_Fall2011/Lecture%20notes/…
– Oleg
Commented Apr 13, 2020 at 21:21
• @Oleg : The proof you link requires $C^{1,2}$ regularity. Tychonoff's solution is not $C^{1,2}$. In fact, Tychonoff's solution is an excellent example of how uniqueness is hopeless if growth of solutions is not sufficiently regulated. Your link also shows how to adapt Tychonoff's solutions. Replace $x$ with $\cos x$ in Tychonoff's power series. Commented Apr 13, 2020 at 21:24