1
$\begingroup$

What conditions are necessary and sufficient for a smooth vector field $v: \mathbb R^2 \to \mathbb R^2$ to induce an ODE $y' = f(x,y)$?

By "induce" I mean: If we start at any point on the plane, and follow the trajectory of the vector field, we trace out a curve $y=f(x)$ with $y'$ a function of the vector field at that point.

Motivation: In $\mathbb R^2$, slope fields look a lot like vector fields, and integral curves look a lot like trajectories, but there are differences between them (most notable, slope fields are $\mathbb R^2 \to \mathbb R$, whereas vector fields are $\mathbb R^2 \to \mathbb R^2$; and slope fields define a family of functions of $\mathbb R \to \mathbb R$, whereas vector field trajectories are not.) To better understand their relationship, I developed the question above and conjecture below.

My conjecture is below. Is it correct? Can we prove (or fix) it?


I conjecture that a single condition is both necessary and sufficient: $v(x,y)$ is never orthogonal to $\hat i$ (where $\hat i = \begin{bmatrix}1 \\ 0 \end{bmatrix}$).

If that condition is met, we can define $$f(x,y) = \frac {v(x,y) \cdot \hat j}{v(x,y) \cdot \hat i}.$$

Then $f$ is a smooth slope field, and by the Existence and Uniqueness Theorem, given any point $(x_0, y_0)$ there is exactly one function $F_0: \mathbb R^2 \to \mathbb R$ such that $F_0'(x) = f(x, F_0(x))$ and $F_0(x_0) = y_0$.

Conversely, if there exists $x, y$ such that $v(x,y) \perp \hat i$, then $v$ cannot induce an ODE, because $y'$ would be undefined at this point. (This part needs a better proof!)

A simple corollary to the above is that if a smooth vector field is never orthogonal to $\hat i$, then trajectories of the vector field can never "turn around": that is, a trajectory which crosses a vertical line can never cross it again.

Another corollary: Multiplying part of a vector field by a non-zero scalar $k$ does not change the resultant ODE or slope field.

Note that the above condition assumes the ODE is required to define $y$ as a function of $x$. If that is not required, we can relax the condition to simply be: There exists a constant vector $u$ such that $v(x,y)$ is never orthogonal to $u$.


Is the above correct? If not: What are the correct conditions? If yes: How can this be fully proven?

$\endgroup$
1
  • $\begingroup$ @Sal By "induce" I mean: If we start at any point on the plane, and follow the trajectory of the vector field, we trace out a curve $y = f(x)$ with $y'$ a function of the vector field at that point. $\endgroup$ Mar 27 at 3:39

1 Answer 1

3
$\begingroup$

Too long for a comment.

A vector field $v$ always induces an autonomous first order ODE in two dimensions $$\tag{1} \pmatrix{\dot x\\\dot y}=v(x,y)\,. $$ I am not sure why that seems not what you want but if you absolutely want a non autonomous ODE in one dimension $$\tag{2} y'=f(x,y) $$ you could first write this as a 2d system \begin{align}\tag{3} \cases{ x'=1\,,\\[2mm] y'=f(x,y)} \end{align} that gives you a hint what $v$ hast to be to be of that form. Perhaps one can also make a variable transformation to have a bit more flexibility.

Edit: ODEs of the form (2) are also called slope fields. It is obvious that every slope field gives rise to a vector field via (3).

What follows is in my own words what you seem to think already:

Given a vector field such that $$ t\mapsto x(t) $$ is invertible at least locally then we can locally parametrize the solution of (1) by $x$ instead of $t\,.$ To be on the safe side let's assume $\dot x=v_1(x,y)\not=0$ for $x,y$ in an open set. Then, from $$ y'=\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{\dot y}{\dot x}=\frac{v_2(x,y)}{v_1(x,y)} $$ we get locally a slope field.

I think the only vector fields for which this is not possible are those whose trajectories don't have a locally invertible $x\,,$ that is, those for which $x$ is constant. These are the vector fields with $v_1\equiv 0\,.$

$\endgroup$
2
  • $\begingroup$ Thanks. I'm trying to better understand the relationships between vector fields / trajectories and slope fields / integral curves, and realized the above. So my goal is not so much "I have a vector field. Is there an ODE describing it?" but rather "I notice that slope fields look a lot like vector fields, integral curves look a lot like trajectories, but there are still differences. How can I understand their relationship?". I see that should have been in the OP, so I'll add it in now. $\endgroup$ Mar 27 at 16:37
  • $\begingroup$ Given a slope field, the correct vector field and corresponding integral curves are given by the system in Kurt's answer. $\endgroup$
    – whpowell96
    Mar 27 at 16:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .