Existence of solution to IVP and Lipschitz condition using continuous derivatives.

Question

Today my professor showed that the ODE: $\ddot{y}(t)=\frac{y^{3/2}}{t^{1/2}}, y(0)=y_0 \neq 0, \dot{y}(0)=z_0$ has a local solution by restating the ODE as:

$$ \frac{d}{ds}\begin{pmatrix} u(s) & w(s) \end{pmatrix}^T = \begin{pmatrix} 4w^{3/2}(s) & su(s) \end{pmatrix} ^T, \begin{pmatrix} u(0) & w(0) \end{pmatrix} ^T = \begin{pmatrix} 2z_0 & y_0 \end{pmatrix} ^T $$

This has a solution since the rhs is locally Lipschitz-continuous, since its derivatives are continuous (and therefore bounded) on the closed set $\overline{(0,s) \times (y_0, u(s)) \times (2z_0,w(s))}$. (I believe it then follows by the mean value theorem)

I dont fully understand this because isn’t he assuming that $u(s), w(s)$ exist when defining the closed set? And if we assume they do, how do we know, that $u(s), w(s)$ don’t diverge on $\overline{(0,s)}$?

Answer 1

I'm unable to make out how the two versions of your ODE system are equivalent. However, this point is moot because your concern is more general than this particular system of ODEs. The root of your concern can be expressed like this:

Let $U$ be a convex, bounded open subset of $\mathbb R^n$, and let $f: U \to \mathbb R^n$ be a continuously differentiable function. We want to prove that $f$ is locally Lipschitz on $U$.

One line of thought - which is erroneous - goes as follows. Suppose that $\bar{f}$ is a continuously differentiable extension of $f$ to an open set $V \subset \mathbb R^n$ containing the closure $\bar{U}$. Since $\bar{U}$ is compact, the partial derivatives of $\bar{f}$ are bounded on $\bar{U}$. Therefore, using the mean value theorem, one can show that $\bar{f}$ is Lipschitz on $\bar{U}$.

This approach is problematic because it may not be possible to construct a continuously differentiable extension of $f$ on an open set containing the closure $\bar{U}$. For example, $f$ might tend to infinity towards the boundary of $\bar{U}$, making it impossible to construct such an extension. To make matters more concrete, consider the case where $U = (0, 1)$ and $f(x) = 1 / x$, where such an approach would clearly fail due to $f$ diverging at $x = 0$.

So what is the remedy?

Let $\mathbf x_0$ be an arbitrary point in $U$, and let $r > 0$ be a number such that the open ball $B(\mathbf x_0, 2r)$ lies within $U$. We will aim to prove that $f$ is Lipschitz on the smaller open ball $B(\mathbf x_0, r)$. This will show that $f$ is Lipschitz on a local neighbourhood of $\mathbf x_0$; since $\mathbf x_0$ is arbitrary, this will show that $f$ is locally Lipschitz on $U$.

So observe that $\overline{B(\mathbf x_0, r)}$ lies inside $B(\mathbf x_0, 2r)$ , which lies inside $U$. Since $\overline{B(\mathbf x_0, r)}$ is compact, the partial derivatives of $f$ are bounded on $\overline{B(\mathbf x_0, r)}$. By a mean value theorem argument, it follows that $f$ is Lipschitz on $B(\mathbf x_0, r)$.

Let's make things more concrete by exploring how the argument plays out in our example where $U = (0, 1)$ and $f = 1/x$.

• Take $x_0 = \tfrac 1 2 $ and $r = \tfrac 1 4$. The ball $B(x_0, 2r)$ lies inside $U$, but $f$ is not Lipschitz on $B(x_0, 2r)$, since $f$ tends to infinity at the left-hand edge of this ball. However, $f$ is Lipschitz on the smaller ball $B(x_0, r)$, with a Lipschitz constant of $1/(\tfrac 1 4)^2$.
• What if $x_0 = \tfrac 1 {100}$? If we take $r = \tfrac 1 {200}$, we see that $f$ is Lipschitz on $B(x_0, r)$, with a Lipschitz constant of $1/(\tfrac 1 {200})^2$. This Lipschitz constant is different from the Lipschitz constant we found for $x_0 = \tfrac 1 2, r = \tfrac 1 4$. But that's fine. We are only trying to show that $f$ is locally Lipschitz on $U$, so it's okay to use different Lipshitz constants on local balls around different points in $U$.

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Edit: Addressing the question in the comment below.

I'll run through the individual steps in the logic, making it clear that there is no circular logic.

Step 1: Defining the differential equation.

In this step, we specify an open set $U \subset \mathbb R^n$, a continuously differentiable function $\mathbf v: U \to \mathbb R^n$, and a point $\mathbf x_0 \in U$.

The $U$, the $\mathbf v$ and the $\mathbf x_0$ define the differential equation that we're going to study: $$ \frac{d\mathbf x}{dt}(t) = \mathbf v(\mathbf x(t)), \ \ \ \ \ \ \ \ \mathbf x(0) = \mathbf x_0.$$

To make this more concrete, let's imagine that this differential equation governs the motion of a duck in a pond. Then:

• $U$ is the pond.
• $\mathbf v(\mathbf x)$ is the velocity of the water at a position $\mathbf x$ in the pond. (From now on, I will call the function $\mathbf v$ the "water velocity field".)
• $\mathbf x(t)$ is the position of the duck in the pond at time $t$.
• $\mathbf x_0$ is the initial position of the duck.

I'll make this even more concrete for you. Let's say that the pond is square shaped. Let's say the water flows from left to right, and let's say that the speed of the water is twice as fast at the top of the pond than at the bottom of the pond. Let's say that the duck starts from the centre of the pond. Then

• $U = \{ (x, y) \in \mathbb R^n : 0 < x < 1, 0 < y < 1 \}$
• $\mathbf v(x, y) = (1 + y, 0)$
• $\mathbf x_0 = (\tfrac 1 2, \tfrac 1 2)$.

Let me emphasise that $U$, $\mathbf v$ and $\mathbf x_0$ are determined by the setup of the pond and the duck. Therefore, it doesn't make sense to ask questions like "How do we know that $U$ exists"?

Related to this, I should also emphasise that you don't need to have a duck trajectory function $\mathbf x(t)$ in order to construct the pond $U$ (which seems to be what is suggested in your question). The pond comes first!!!

Step 2: Verifying that the velocity field is locally Lipschitz continuous

In our example, the velocity field is $\mathbf v(x, y) = (1 + y, 0)$. The function $(x, y) \mapsto (1 + y, 0)$ is a differentiable function of $(x, y)$. It is therefore locally Lipschitz continuous.

Let me make something very clear. The claim I'm making is that the function $\mathbf v(\mathbf x)$ is differentiable. $\mathbf v(\mathbf x)$ is a function of $\mathbf x$: it is the function that tells us the velocity of the water at a given position $\mathbf x$ in the pond.

I'm not saying that $\mathbf v(\mathbf x(t))$ is differentiable. $\mathbf v(\mathbf x(t))$ is a completely different function! In fact, $\mathbf v(\mathbf x(t))$ is a function of $t$: it is the function that tells us the velocity of the duck at time $t$.

As you rightly pointed out, we can't make claims about $\mathbf v(\mathbf x(t))$ being differentiable, because at this point in the argument, we haven't yet constructed $\mathbf x(t)$! That would indeed be circular logic.

Step 3: Invoke the existence/uniqueness theorem for ODEs.

The theorem says something like this:

Suppose we're given a pond $U \subset \mathbb R^n$, a water velocity field $\mathbf v: U \to \mathbb R^n$, and an initial position $\mathbf x_0 \in U$ for our duck. The motion of the duck is governed by the differential equation $$\frac{d\mathbf x}{dt}(t) = \mathbf v(\mathbf x(t)), \ \ \ \ \ \ \ \ \mathbf x(0) = \mathbf x_0$$ As long as we can verify that the water velocity field $\mathbf v (\mathbf x)$ is locally Lipschitz, then we can be mathematically certain that there exists a trajectory $\mathbf x(t)$ for our duck that is compatible with our differential equation, for $t $ in some time interval $(-\epsilon_1, \epsilon_2)$. Furthermore, this trajectory is unique.

I hope that gives you a sense of how the logic is organised in the context of a simple example. Going back to your second order ODE, those algebraic manipulations are a red herring - they don't fundamentally affect how the overall logic is structured.