The Cauchy Problem: Example-based (Arnold’s ODE)

This post is based on my reading of the section on the Cauchy problem in Arnold’s ODE.

Definition: Given a vector field $\alpha$ , the Cauchy problem for $L_{\alpha} u = 0$ is the problem of determining a function $u$ from its values on a given hyper-surface, $\gamma$ .

The first question we can ask ourselves is: when does the Cauchy problem not have a solution?

Recall that the first integral $u$ is constant on any characteristic. Thus, if $\gamma$ intersects a characteristic at 2 distinct points and the initial values assigned to these 2 points are different, then no such first integral solution exists.

Problem 1: Solve the Cauchy problem $u|_{x=0} = sin(y)$ for the equation $\frac{\partial u}{\partial x} - y \frac{\partial u}{\partial y} = 0$ .

The corresponding vector field is:

$\dot{x} = 1$

$\dot{y} = -y$

and the corresponding characteristics are $y = Ce^{-x}$ .

The first integrals are functions that are constant on $y = Ce^{-x}$ . Notice that one such function would be $u_0(x, y) = ye^x$ . As mentioned in the previous post, all first integrals are functions of $ye^x$ .

To satisfy the initial conditions, we also need $u(0, y) = sin(y)$ .

The function that we are looking for is:

$u(x, y) = sin(ye^x)$ .

Problem 2: Which points of the line $x = 1$ are non-characteristic for the equation $y\frac{\partial u}{\partial y} - x \frac{\partial u}{\partial y} = 0$ ?

The corresponding vector field is:

$\dot{x} = y$

$\dot{y} = -x$

and the characteristics are the family of circles centred at the origin.

Notice: the characteristic $x^2 + y^2 = 1$ is tangent to our initial surface $x = 0$ at the point $(1, 0)$ . All circles with radius greater than 1 intersect $x = 0$ at two points.

Functions of $u_0(x, y) = x^2 + y^2$ are first integrals (can we say the converse holds as well?).

We also need $u(x, y) = y^2$ .

Since $u(1, 0) = 0$ , this tells us that $u$ is equal to 0 on the unit circle.

One possible function is $u(x, y) = x^2 + y^2 - 1$ .

Are there any other first integrals that might work?

Yes, let’s consider things in terms of the polar coordinates. Then, notice that we only require:

$u(r, \theta) = r^2 - 1$ for all $r \geq 1$ .

We can smoothly extend $u(r, \theta)$ to $latexx 0 < r < 1$ in a way that $u(r, \theta) \neq r^2 - 1$ in this interval. So more that one possible answer exists.

Why does not uniqueness not hold in this case? $x = 1$ is NOT a Cauchy surface. The set of characteristics with $r < 1$ does not intersect with $x = 1$ . Hence, the initial conditions imposed does not determine the values on them.

Cauchy Problem for Inhomogeneous Linear Equations

Instead of studying equations of the form $L_{\alpha} u = 0$ , we now study equations where the LHS is a function of $x_1, ..., x_n$ , i.e. $L_{\alpha} u = b(x_1, ..., x_n)$ .

Theorem: In some sufficiently small neighbourhood of any non-characteristic point of the initial surface, the solution exists and is unique.

Key observation: Suppose $u(x_1, ..., x_n)$ to be a solution and $\phi(t)$ to be a phase flow. Then, $\frac{d}{dt} u \circ \phi = b$ . Why?

$\frac{d}{dt} u \circ \phi = \begin{bmatrix} \frac{\partial u}{\partial x_1} \\ . \\ . \\ . \\ \frac{\partial u}{\partial x_n} \end{bmatrix} \dot{\phi}$ (*)

By definition, $\dot{\phi}|_{p} = (a_1(p), ..., a_n(p)) =: \alpha(p)$ is the vector belonging to the vector field $\alpha$ at point p.

Hence,

(*) $= a_1 \frac{\partial u}{\partial x_1} + ... + a_n \frac{\partial u}{\partial x_n}$

$= b(x_1, ..., x_n)$ (by definition).

What does $\frac{d}{dt} u \circ \phi = b$ tell us?

Physically, this implies that the increment of $u$ along a segment of $\phi$ is equal to the integral of $b$ with respect to the time of motion along this segment, i.e.

$u \circ \phi(t) - p = \int_0^t \frac{d}{dt} u \circ \phi dt = \int_0^t b(\phi(t))dt$ .

We can simplify the RHS by supposing that within the neighbourhood of $p$ , $a_1 \neq 0$ (otherwise, we can just relabel. We know that at least one of the $a_i$ ‘s is not zero since $p$ is not a characteristic point.).