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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 $f\neq0$?

Oleg
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  • 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. – Ninad Munshi Apr 13 '20 at 14:30
  • thanks @NinadMunshi For simplicity, let's assume that we are on a torus, so no boundary conditions needed :) – Oleg Apr 13 '20 at 14:32
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    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. – Ninad Munshi Apr 13 '20 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 Apr 13 '20 at 14:38
  • @NinadMunshi thanks! I've edited my question accordingly. But it's not a question of uniqueness, it's a completely different question, so I don't see how those techniques can be relevant here. – Oleg Apr 13 '20 at 14:41
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    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 – Ninad Munshi Apr 13 '20 at 14:41
  • They have different initial conditions: $u(0)\neq v(0)$! I am asking, is it possible that they will collapse and that u(1)=v(1)? – Oleg Apr 13 '20 at 14:42
  • Contrapositive! I cannot keep repeating this – Ninad Munshi Apr 13 '20 at 14:42
  • To show uniqueness is to show that if $u(0)=v(0)$ than $u(1)=v(1)$. Here the problem is different: I know that $u(1)=v(1)$ and I want to prove that $u(0)=v(0)$. This is not contrapositive at all, it's a very different statement. In PDE you cannot revert time, so I don't see how uniqueness would imply my statement. – Oleg Apr 13 '20 at 14:46
  • As you say it is on a torus, so there is a degeneracy with the fact we are dealing with periodic functions. Let's ignore that for now and suppose we are talking about two physically distinct points on the surface. It is certainly true that if we have uniqueness, forward time translating the equation would still have uniqueness. If you take $t=1$ to be the new $t=0$ then we are back to square one. – Ninad Munshi Apr 13 '20 at 14:50
  • No, we are not. At time 1 we have $u(1)=v(1)$ (and of course thus $u(2)=v(2)$). But my question is whether $u(0)=v(0)$. It's a different question. – Oleg Apr 13 '20 at 14:52
  • A contrapositive statement to uniqueness would be the following one: if $u(0)=v(0)$ is it possible that $u(1)\neq v(1)$. But I'm asking a different question, please understand it, or show how uniqueness is helpful here. – Oleg Apr 13 '20 at 14:57
  • What is your objection to the statement: Let $w = u(t+1)$ and $y=v(t+1)$, with $u,v$ being unique solutions.Then $w,y$ satisfy the PDE and also have uniqueness. – Ninad Munshi Apr 13 '20 at 15:01
  • I don't see what is the connection between this statement and my question. We have $u(1)=v(1)$. Thus, $w(0)=u(1)=v(1)=y(0)$. Why does this imply that $w(-1)=v(-1)$? – Oleg Apr 13 '20 at 15:06
  • @EricTowers thanks for your comment. $x$ is on torus and $t\ge0$. – Oleg Apr 13 '20 at 15:07
  • @EricTowers $x$ lives on 1D torus $[0,1]$. I've edited the question accordingly. – Oleg Apr 13 '20 at 15:09
  • @EricTowers I mean you can take $$\mathbb T^1= (\mathbb S^1)^1$$ if you want :) – Maximilian Janisch Apr 13 '20 at 15:11
  • @EricTowers There is an $n$-dimensional torus, but no $1$-dimensional torus? This is strange... https://en.wikipedia.org/wiki/Torus#n-dimensional_torus – Oleg Apr 13 '20 at 15:12
  • @EricTowers thanks for your remarks, I have edited the question accordingly. However this is not important at all: if you know how to solve this problem with Dirichlet or Neumann or any other boundary conditions, I'm happy to learn about it. – Oleg Apr 13 '20 at 15:18
  • @Oleg The point of my statement is that $u,v$ and $w,y$ are the same function. Nothing meaningful has happened at the temporal boundary of $t=0$ Uniqueness would have also applied at $t=1$ etc. since there is nothing special about $0$. Specifying any point in time would specify the trajectory, but only back up until $t=0$. – Ninad Munshi Apr 13 '20 at 15:32
  • @NinadMunshi Right, it's a simple time change. To prove my statement you need to deduce from the fact that $w(0)=y(0)$ the statement $w(-1)=y(-1)$. However any uniqueness statement gives you only "forward" uniqueness, but never backward uniqueness. So it is not very useful here. – Oleg Apr 13 '20 at 15:36
  • Because $w(0)$ does not occur at the "real" zero, it's just a trick. Another way to say this is that the equation coupled with $u(1) = $ something also uniquely defines a trajectory on $t\in[0,\infty)$ – Ninad Munshi Apr 13 '20 at 15:37
  • @NinadMunshi I'm confused: do you still claim that usual uniqueness is enough here? It's not true: uniqueness tells you that i $w(0)=y(0)$ then $w(t)=y(t)$ for any $t>0$ - check your document. It does NOT guarantee that $w(-1)=y(-1)$. For this, one needs uniqueness for a backward heat equation which is a different and a non-standard equation. I'm not aware of any uniqueness results related to backward heat equation. – Oleg Apr 13 '20 at 15:40
  • Right because the document assumes $u(0)$ is the furthest boundary value, that is not the case with $u(1)$. In such a situation negative time values are undefined. But $w(-1)$ is still in the domain because that is really $u(0)$. $w$ is defined on the interval $[-1,\infty)$. – Ninad Munshi Apr 13 '20 at 15:42
  • @NinadMunshi even if we start everything from $-1$, but we know that two solutions coincide only from time $0$, it does NOT imply that they coincided from $-1$ to $0$. All the statements in that document are like this: suppose that two solutions are the same now. Then they are the same in future. To go the past you need to study backward heat equation, which is something else. – Oleg Apr 13 '20 at 15:45
  • @NinadMunshi check for example this question which is the same as mine but for ODEs. https://math.stackexchange.com/questions/3591157/can-smooth-ode-converge-to-its-equilibrium-in-finite-time To show that x(T)=y(T) implies x(0)=y(0) people had to reverse time. For ODE it's fine, but here time reversal leads to a different PDE. – Oleg Apr 13 '20 at 15:49
  • @NinadMunshi Also I've just found a proof of what I want in a much simpler setup - https://users.math.msu.edu/users/yanb/847ch3.pdf Theorem 4.10 - and look how backward uniqueness proof (that's what i'm looking for) is different from forward uniqueness proof. I hope that I've convinced you by now that these are very different things. – Oleg Apr 13 '20 at 15:53
  • I'm convinced that I completely wasted my time in trying to help you. Of course, I think the medium matters, drawing pictures would have really helped, and that's something not easily done in a comment section. – Ninad Munshi Apr 13 '20 at 15:53
  • @NinadMunshi if you open Evans PDE book (you are from Berkeley, so you must know this guy), then at pages 63-64 he is discussing backward uniqueness for a simple heat equation - exactly what I'm asking for a more complicated PDE. You will see that the standard energy techniques which are used for uniqueness has to be significantly modified to show backward uniqueness. I can't believe that you are still insisting that forward and backward uniqueness are the same! – Oleg Apr 13 '20 at 21:04

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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 $(x,t) \in S^1 \times \Bbb{R}$.

Eric Towers
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  • 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 Apr 13 '20 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 Apr 13 '20 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? – Eric Towers Apr 13 '20 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 http://math.mit.edu/~jspeck/18.152_Fall2011/Lecture%20notes/18152%20lecture%20notes%20-%203.pdf – Oleg Apr 13 '20 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. – Eric Towers Apr 13 '20 at 21:24
  • Perfect, but here if I want to construct everything backwardly (so that two solutions coincide say at $t=1$), then I need C^{1,2} regularity around time point 1. Here people are discussing exactly my question but for a simpler heat equation https://math.stackexchange.com/questions/1284252/backwards-heat-equation-u-t-lambda2-u-xx I want exactly the same result but for a more complicated equation. Also the function $f$ for me depends only on solution but not in its derivative - though I don't think it is important here. – Oleg Apr 13 '20 at 21:28
  • Please check Theorem 5.23 here - it's lecture notes from UChicago http://math.uchicago.edu/~may/REU2014/REUPapers/Ji.pdf They show uniqueness for solutions to heat equation on bounded domain assuming only continuity on $\bar U$ (and of course C^{1,2} within U). You say that "In fact, Tychonoff's solution is an excellent example of how uniqueness is hopeless if growth of solutions is not sufficiently regulated" - this is true. But if the domain is bounded, then the solution is also bounded! It cannot grow to infinity as Tychonoff's solution. – Oleg Apr 13 '20 at 21:49
  • @Oleg : You keep quoting theorems that require fairly strict regularilty and then asserting this rejects a solution that does not satisfy that regularity condition. I've pointed out that you are making a clear logical error, but you don't choose to see it and keep citing inapplicable theorems. Tychonoff's solutions are not as smooth as required in any open neighborhood of $x = 0$, so the relevant derivatives are not bounded by "continuous function on compact domain" arguments. – Eric Towers Apr 13 '20 at 21:54
  • ok, let us say that I'm interested only in C^{1,2} solutions. How can I prove backward uniqueness? – Oleg Apr 13 '20 at 22:01
  • And by the way - if you replace $x$ by $cos x$ in Tychonoff solution, as you suggested, it won't be a solution any more - its second space derivative will be quite different from its first time derivative - check it! – Oleg Apr 13 '20 at 22:37