# When does a smooth vector field induce an ODE?

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?

• @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. Mar 27 at 3:39

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 $$v_1\equiv 0\,.$$

• 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. Mar 27 at 16:37
• Given a slope field, the correct vector field and corresponding integral curves are given by the system in Kurt's answer. Mar 27 at 16:47