# Why is uniqueness important for PDEs?

Every text on PDEs I come across will spend alot of time on showing the existence and uniqueness of solutions to a particular PDE. The importance of the existence of a solution to a PDE is obvious, but I can't see why so much time is spent on uniqueness. Why do we care whether a solution is unique or not as long as we know that there is a solution? Is uniqueness just shown for the sake of it, ie. the sake of completeness, or is there some deeper reason why it's considered important to show uniqueness?

• You would like to know all solutions. But if you want to use the equations to solve a physical problem, then you better be able to specify enough conditions so that there is only one solution--otherwise you wouldn't know which solution would would model what happens. Nothing says that our models are accurate. So, from a Mathematical perspective, you look for PDEs and conditions that lead to unique solutions (i.e., there exists a solution, and only one solution.) Jul 19 '16 at 12:57
• Jul 19 '16 at 16:43
• A great example of why we want to be sure that a differential equation has a unique solution is Norton's dome. Jul 20 '16 at 15:01
• @Rodrigo: When creating a new tag you need to weigh a few things. For example the scope of the tag and how likely it is to be misused. Since uniqueness is not a notion unique to to PDE, this means that the scope of the tag is going to be extremely ill-defined, and thus very open to misuse if you do think about it as a tag for ODE/PDE related questions. So either this is too broad of a tag, or it is too open for misuse. In any case, I'd be happier if you had laid off the new tags for a while and instead joined into the meta discussion about tagging to get the feel for when to add new tags. Jul 24 '16 at 15:19
• @Rodrigo: Yes, one can search for more than one tag. One can use more than one tag. Nevertheless, everything I wrote about the uniqueness tag is relevant. And if you were to join the discussion about the usage of tags on meta, you'll learn what are the directives that the people who have invested a lot of time and effort into the current tagging "philosophy" follow. Whether you agree or disagree with them, if you want to actively help the tagging effort, you should at least respect them and see what the fuss is all about on the meta site. Jul 24 '16 at 19:32

Why do you want to solve the PDE in the first place? Usually because it relates to the solution to a real world problem.

Say, for example, that you're studying the temperature in a heated metal. You have to understand if the temperature will stay below a certain value, otherwise the metal will collapse and your bridge with it. You find out that the temperature solves a PDE with certain boundary conditions, you solve it and find a value for the maximum temperature.

Now, if the solution to the PDE is unique, then you are done. But if the solution is not unique, then who's to say that the solution you found is what will actually happen in the real world?

In the real world there is only one solution! You then have to specify additional boundary conditions until you're left with a unique solution. To do all this you need some theory that tells you exactly when your boundary conditions are enough, what can go wrong, where you should look for exceptions, etc.

In general this is important not only for real world applications; uniqueness hugely simplifies problems. Say that you have to solve for your abstract math problem a complex PDE; if you can "guess" the solution, and it works, then it works! You're done, that is the only solution and you go forward. If there were multiple solutions you would probably need to check if the solution you found is "the right one" for the problem at hand. It is useful in all sorts of context. Just think how useful is the fact that we know that a polynomial of degree $n$ has exactly $n$ complex roots (counted with their multiplicity). It makes life much easier in a whole lot of ways

• On the flip side, it may be physically meaningful that a problem has multiple solutions, because the addition of additional information "selects" one of those solutions. For example, if you have an evolution PDE with unique solutions, it may have multiple equilibria. That's fine, but it raises the additional question of which initial conditions wind up at which equilibria.
– Ian
Jul 19 '16 at 20:15

Let's think practically.

PDE's is a field used primarily for physical modelling. Almost all introductory texts talk all about physics models using PDE's: Heat equation, transport equation, wave equation, etc.

In a real life scenario you see a behaviour and you want to model it by using math. For example, you heat up an insulated rod and you want to predict the temperature at any time using math. If you did not have uniqueness we would have two possible end results. For instance, at time $t=3$ seconds we could have that the temperature of the rod is $32^o C$ or $-76^o C$ based on the results of two solutions. This is of course physically ridiculous. This means that the model we have proposed does not replicate the reality of what is happening. So we must pose a new model with a unique solution as we only expect one result in testing.

• I do not agree with "This is of course physically ridiculous." Consider the PDE for location $x$ based on constant speed $v$. Not knowing the real starting location $x_0$, we can find different solutions $x'_0 + vt$ that predict different locations where we might be at time $t = 3$. But this is not "physically ridiculous" - this simply expresses that a lot is possible, and that we did not consider all the information necessary to determine the unique solution. In short, it is not physically ridiculous to say "if you move at speed $v$, in 3 seconds you might be here or there" - it depends!
– bers
Jul 19 '16 at 20:24
• @bers : Yes many problems can be under-determined. We may know a bit that needs to be true for a solution to make sense, but there may still be lots of freedom left to design solutions which satisfy those constraints. If we get "ridiculous" solutions it is because we do not have enough information available or thought about enough of which constraints are necessary. Jul 20 '16 at 7:36
• For physical systems with an unstable initial state, it definitely is possible to have two possible distinct end states. Consider a ball approximately at the top of a parabola, which can roll either left or right. QM tells us the position is fundamentally uncertain, so we can have two definitely distinct outcomes that are both physically allowable. Jul 20 '16 at 9:54
• You folks are missing the point of physical laws. The idea is that if the current state is completely known, the system should evolve according to the specific equations, and we should be able to know the state of the system at a later time. Since we know there is only one state at any given time in the universe, it is ridiculous for our physical universe to be in two states. And I'm not just talking classically. Yes, there are different states in the quantum mechanical system, but the state of the system as a whole, that is the spectrum and wavefunctions, are unique. Aug 11 '16 at 17:21

For P/LDEs, recurrences, etc. uniqueness theorems are powerful tools for proving equalities. Let's consider some very simple examples.

$e^{ix}$ and $\, \cos(x) + i\,{\rm sin}(x)\,$ are solutions of $\ y'\! = i y,\ y(0) = 1\,$ so they are equal by uniqueness.

$\sum_{k=1}^n k\cdot k! = (n\!+\!1)!-1\,$ since both satisfy $\,f(n\!+\!1)-f(n) = (n\!+\!1)(n\!+\!1)!,\ f(0) = 0.\,$ Here the uniqueness proof for the first-order difference equation (recurrence) has a trivial one-line inductive proof (sometime called the Fundmaental Theorem of Difference Calculus). See also here for a vivid $2$-dimensional version using differences of rectangles.

$\prod_{i=1}^{n-1}\left(1+\frac{1}{i} \right)^{i} = \frac{n^{n}}{n!}\$ since both satisfy $\,f(n\!+\!1) = (1+1/n)^n f(n),\,\ f(1) = 1.$

Binet's formula for fibonacci numbers $\, {f}_n = (\phi^n-\bar\phi^n)/(\phi-\bar\phi),\ \phi = (1+\sqrt 5)/2\,$ is very easily proved by showing the RHS satisfies $\,f_{n+2} = f_{n+1}+f_n,\,\ f_0 = 0,\ f_1 = 1.$

Combining such uniqueness theorems with the group-theoretic machine for special functions, some Lie theory (symmetry and separation of variables), and work on holonomic functions, one obtains very powerful algorithms for effectively computing with special functions. These lie at the core of many computer algebra systems such as Macsyma, Maple, Mathematica.

• Good answer! $\;\!$ Jul 20 '16 at 7:35

• You can use whatever method you want to solve the PDE (which in practice is almost always separation of variables), and you're guaranteed to get the right answer. You don't need to do a proof that the solution you found is the correct one. Point of comparison: Sometimes with non-linear ODE's, or even in solving non-linear algebraic equations, you have to take a square root, and then you need to think hard about which branch of the square root gives the "correct" solution. In PDE's, the headache of non-uniqueness is exponentially more troublesome.

• There is always a trade-off between complexity and functionality---the more complicated the behavior of a system, the more patents you can make out of it. To this end, sometimes non-uniqueness of PDE is really, really useful. Example: A piece of sheet metal with fixed endpoints might have more than one stable equilibrium, which is described by a PDE with two or more solutions (for given boundary conditions). This is how bimetalic circuit breakers in your house operate (modulo details).

• Because physicists and engineers tell them they should.
• It's an interesting question in its own right, which is why mathematicians are in this business in the first place. (Right?)

One other reason to add (which is of interest to physicists, engineers, AND mathematicians) is that sometimes non-uniqueness or non-existence tells you that your PDE is ignoring a physically important effect. For example, the inviscid Burger's equation is an example of a PDE which does not have solutions that exist for all time. There are ways of propagating the solution beyond the time where it usually fails to exist by adding new terms to the PDE that "smooth out" the solution (this process is called regularization), but there is not a unique way to do this (e.g. a second order regularization term leads to a different solution than a third order one). The non-uniqueness tells you something important, namely that the way the solution evolves at late times depends in an essential way on the microscopic details (i.e. the particular nature of that extra term you added) of the system being modeled by Burger's equation. So in order to predict how the system will evolve, you must know about its microscopic details, or conversely if you do an experiment and learn empirically how the system evolves you have learned something about the microscopic details. Either way, this is interesting and important to know.

As we can see most answers are legitimately citing the real world applications, as we wouldn't want our solution of the problem to be like "after 2 hours the velocity of this thing inside a nuclear reactor will be $$1.03*10^2$$... or wait a minute, I have another branch to my solution which says it will be $$3.001*10^{14}$$!!"

But, other than that,

uniqueness is closely related to approximation.

We would like for our PDE to produce for nearby initial data, or nearby parameters in the PDE (which come from experimental physical observation), to give nearby solutions. Otherwise, we will never be able to trust any computer calculations. But, closeness has uniqueness in it:

Assume $$S(u)$$ is the "map" that assigns a solution to given initial data $$u$$. In appropriate norms we like to have $$\|S(u_1)-S(u_2)\| \leq C\|u_1-u_2\|$$ Now, if we have nonuniqueness, this won't hold because very close $$u$$'s will exist while their corresponding solutions will be faaaar apart.

Suppose you show that $$y=e^x \rightarrow \left(\frac{dy}{dx} = y \wedge y(0) = 1\right).$$

Have you solved the IVP? In my opinion, no: its not solved until you've also proved that

$$\left(\frac{dy}{dx} = y \wedge y(0) = 1\right) \rightarrow y=e^x.$$

That's what uniqueness gives you; it lets you derive the second formula from the first, via this principle of logic that I still don't have a name for.

A simple example of non-uniqueness is the constant you're used to adding every time you solve for $f$ in the differential equation $f' = g$ given some specified function $g$.

This constant would have physical meaning if you're asking the question, "what does my hike look like, given that I know the steepness of the path at every point?". The constant reflects that the hike can start at any altitude and still be a solution to the equation. Knowing that the solution is unique up to a constant means you know that once you have the altitude of a single specified point on the hike, this boundary condition is enough to constrain the altitude of every point on the hike.

Of course pde's in multiple variables, when used for modelling, can model something way more complicated than that. Depending on the problem, uniqueness of the solution "up to some freedom" might tell you something interesting about the additional boundary conditions you need in order to fully describe the physical situation you're modelling. When modelling heat flow in some particular medium with no sources or sinks, we know what the solution looks like in the limit as time tends to infinity (constant temperature), but the initial distribution of heat is a boundary condition on which everything else depends. Proving that our solution is unique given an initial distribution just means our model makes a single prediction, which makes it more useful than a model that made (say) two predictions and refuses to tell us which will happen.

Lack of uniqueness might tells you that your physical model is not powerful enough to make a single prediction about the behaviour of the system you're modelling, and so you have more science to do, to rule out the solutions that don't happen in reality.

Or lack of uniqueness can tell you that your dynamic system has more than one different equilibrium, some or all of which can occur in reality. The "obvious" equations for a standing wave on some piece of string have infinitely many solutions with different frequencies, of which the first few might be readily achievable in the lab, before some deviation of the string from the ideal behaviour of the model makes it impossible. So you revise your model to describe the string more accurately until it has the right number of solutions.

Anyway, if we want to know what the model predicts can happen we need to find all the solutions (which means if there's only one we have to prove there is only one). And if we want to know what the model predicts definitely happens then there'd better be only one solution and we'd better prove it.

I think it's important because non-linear problems can have multiple, path-dependent solutions.

An easy example is shell buckling. Where you end up can be very sensitive to initial conditions and the path taken.

Perhaps it seems less important for linear problems that are well understood, but it's key for non-linear problems.