# Unique solution of a 1-D ODE with a bounded positive right-hand-side

Consider the initial value problem $$\dot x(t) = F(t,x), \quad t \in (0,T)$$ with given initial datum $$x(0) = x_0 \in \mathbb R.$$ More precisely we consider the integral equation $$x(t)=x(0)+\int_0^t F(s,x(s))ds.$$

$$F$$ may be discontinuous, but let us assume that $$0 < m < F(t,x) < M.$$

The common counter-examples to uniqueness (or existence) of ODEs (or their associated integral equations) seem to rely on $$F$$ switching sign, or being close to $$0$$. This motivates the above assumption and the following questions.

Question 1: Is it true that there exist a solution under the assumptions above?

Question 2: Can we also prove uniqueness?

Question 3: Would it help to get a positive answer to the questions above if we also assumed $$F$$ to be autonomous (that is, no $$t$$ dependence)?

Non-autonomous case

Most of the time, either the few usual tricks work, or the conjecture is true. In the non-autonomous case, unfortunately, the answer is the former. This is the first counterexample I ever learnt to uniqueness (and I bet it's true for a lot of people):

$$x' = 2 \max(x,0)^{1/2}$$

That has two solutions, if $$x(0) = 0$$. One is $$x(t) = 0$$ and one $$x(t) = t^2\cdot 1_{t>0}$$. Because you want to assume boundedness (as you will see it's not really important, let's consider the bounded counterpart

$$x' = 2\cdot \begin{cases} 0 & \text{if }t\le 0 \\ \sqrt{x} & \text{if } t\in [0,1]\\ 1 & \text{if }t\ge 1 \\ \end{cases}$$

Let $$x(t)$$ be a solution to the problem above, and choose an $$v>0$$. What equation does $$z(t) = x(t)+vt$$ solve? A direct computation shows:

$$z' = 2\cdot \begin{cases} v & \text{if }z-vt\le 0 \\ v + \sqrt{z -vt} & \text{if } z-vt\in [0,1]\\ 1+v & \text{if }z-vt\ge 1 \\ \end{cases}$$

So you have a counterexample for nonautonomous solutions. From a philosophical perspective this boils down to the following:

In the non-autonomous case, you have time and space reparametrization symmetries. (At least you reparametrize by smooth changes of coordinates). If you want to show that a statement holds for a certain class of objects, you should look for a class that is invariant under reparametrizations.

Autonomous case

Now, in the autonomous case, the class you have given is invariant by the (reduced set of) symmetries of your problem, at least locally, and that gives hope. (I found an easier solution that shows uniqueness assuming existence). Assume that a solution $$x'(t)=F\circ x(t)$$ exists (in the mild/integral sense). Then $$t\mapsto x(t)$$ must be a bi-liptschitz map by the hypohtesis, and therefore must have a (bi-liptschitz inverse) $$x \mapsto t(x)$$. By the inverse function theorem, it must be that

$$t(x) = \int_0^x 1/F(x) dx$$

assuming wlog that $$x(0)=0$$. This tells you who $$t(x)$$ is, so uniqueness is set. You might, however, ask for well-posedeness of the solution. This will be a bit harder, but we'll also get an existence result from it (by density). Define three spaces:

$$\begin{split} \mathcal F &= (\{f\in L^1([-M,M]), f(x) \in [1/M,1]\}, \|\cdot\|_{L^1})\\ \mathcal T &= (\{f\in L^{1,1}([-M,M]), f(x) \in [-M^2,M^2], f(0) = 0, f'(x) \in [1/M,M]\}, \|\cdot\|_{W^{1,1}})\\ \mathcal X &= (\{f\in L^{1,1}([-1,1]), f(x) \in [-M,M], f(0) = 0, f'(x) \in [1,M]\}, \|\cdot\|_{L^{\infty}})\\ \end{split}$$ As the names already tell, the first is the space where $$F$$ lives (it will actually be where $$1/F$$ lives), the second where $$x\mapsto t(x)$$ lives, and the third one where $$x\mapsto x(t)$$ lives.

We will define two continuous maps, $$\int:\mathcal F \to \mathcal T$$ the indefinite integral. It is continuous by construction of the spaces. The magic is that there is a unique continuous map $$I:\mathcal T \to \mathcal X$$ defined implicitly by $$\tau(I(\tau)(t)) = t$$. In other words $$x = I(\tau)$$ is the left inverse of $$\tau$$, and therefore a solution to the equation by the inverse function theorem.

Let's now show that the map $$I$$ is continuous. Let $$\tau, \tau'\in \mathcal T$$. Then $$\chi = \tau^{-1}, \chi' = {\tau'}^{-1}$$ exist and are Lipschitz. Let $$(x,t)$$ be a point in $$(x,\tau(x))$$. By the mean value theorem for bi-Lipschitz functions applied to $$\tau'$$ at the points $$x,\chi'(x)$$, we see $$\frac{|t-\tau'(x)|}{|x-\chi'(t)|} \in [1/M,1]$$ (Here we use the hypothesis!). In particular $$\|\chi-\chi'\|_\infty < M\|\tau-\tau'\|_\infty.$$

We have then shown that the data-to-solution map is well defined and bi-Liptschtiz in the given metrics. However, we have "lost" a derivative in the process. We should be able to do better. (I think the map $$I$$ is not continuous if you endow $$\mathcal X$$ with the $$W^{1,1}$$ topology, so this proof won't work)

The "Riemann Sum" solution that I was proposing before is a discrete version of this proof, which gets messier because you're discretizing everything, and weaker.

General existence

You might be able to get existence under very mild assumptions using Schauder's fixed point, which doesn't ask for a contraction.

• Thanks for your answer. But in the counterexample $f$ is not bounded from below away from zero. About the autonomous case, could you add more details about the uniqueness of the approximate solutions and about the convergence?
– Riku
Sep 26 '20 at 11:44
• Sure I'll write more later! In the counterexample F(x,t)>v, and you can choose v to be as large as you want. Sep 26 '20 at 11:50
• I guess you are solving the ODE in a weak sense (a.e or in an integral sense). The Cauchy problem $x'=f(x), x(0)=0$ with $f(x)=2$ for $x \geq 0$ and $f(x)=1$ for $x < 0$ has no classical solution around $0$. Sep 26 '20 at 15:56
• Indeed! I think what I'm doing (the integral form) is usually called a mild solution, instead of solving in a weak sense (dualizing against test functions). Sep 26 '20 at 16:22