# Justifying the "Physicist's method" for ODEs using differential forms

I need some help in untangling and solving the following exercise:

Let the curve $$c:[a,b] \to \mathbb{R}^2, t \mapsto (t, y(t))$$ be a solution for the ODE $$y'(x) = f(x, y(x)).$$ Justify the "Physicist's method" (no offense intended) of rearranging the equation $$\frac{dy}{dx} = f(x,y)$$ through formal multiplication of $$dx$$ to $$dy = f(x,y)dx,$$ by showing that both differential forms agree in every point $$c(t)$$ on the tangent space.

As far as I understand the situation, we are considering the two differential forms $$$$dy: \mathbb{R}^2 \to {\bigwedge}^1(T_p\mathbb{R}^2)\\ f(x,y)dx: \mathbb{R}^2 \to {\bigwedge}^1(T_p\mathbb{R}^2)\\$$$$ Here $$dy$$ is a constant differential form, in the sense that $$dy(p)(x,y) = y$$ independent of $$p=(v,w) \in \mathbb{R}^2$$. However, in general $$f(x,y)dx$$ is not a constant differential form. Now, if we choose a point on the curve $$c$$, say $$p = c(t_0) \in c([a,b])$$ then we have $$f(p)dx = f(c(t_0))dx = y'(t_0)dx,$$ since $$c$$ is a solution to the ODE given above. I don't know how to proceed from this point. How can we argue that this equals $$dy$$?

The equality in $$dy = f\, dx$$ is very misleading, because strictly speaking it's not true. To see why, note that $$dy$$ is a differential $$1$$-form defined on $$\Bbb{R}^2$$, which means for each $$p \in \Bbb{R}^2$$, $$dy_p : T_p \Bbb{R}^2 \to \Bbb{R}$$ is linear. Similarly, $$dx$$ is also a differential $$1$$-form on $$\Bbb{R}^2$$. Let's for the sake of concreteness say $$f: \Bbb{R}^2 \to \Bbb{R}$$ is defined on all of $$\Bbb{R}^2$$, so that $$f \, dx$$ is still a $$1$$-form on $$\Bbb{R}^2$$.

So, if we just write $$dy = f \, dx$$, this means that the $$1$$-form on the LHS must equal the $$1$$-form on the RHS. But this is just not the case, because it amounts to saying that $$dy$$ and $$dx$$ are linearly-dependent over the module $$C^{\infty}(\Bbb{R}^2)$$. Just to really drive this point home, let's fix a point $$p \in \Bbb{R}^2$$, then, if that equality were true, it would mean $$dy_p = f(p)\, dx_p$$, where the equality is as elements in $$T_p^*(\Bbb{R}^2)$$ (the dual of the tangent space; i.e the cotangent space). But this is of course absurd, because if you evaluate both sides on the tangent vector $$\dfrac{\partial}{\partial y}\bigg|_{p} \in T_p\Bbb{R}^2$$, you'll get the absurd equality $$1 = 0$$. Yet again, the statement $$dy = f \, dx$$ is kind of like saying the row vector $$(0 , 1)$$ equals $$\lambda \cdot (1,0)$$ for some $$\lambda \in \Bbb{R}$$... which is plain wrong.

Now that I've hopefully convinced you that the equation taken literally is false, how do we interpret it? Well, the last sentence of your question gives a clue it says

"... by showing that both differential forms agree in every point c(t) on the tangent space."

But the tangent space of what? $$\Bbb{R}^2$$? Clearly not, as I've just shown above. What is actually meant is that these two differential forms agree at every point $$c(t) \in \Bbb{R}^2$$, when restricted to the (one-dimensional) subspace $$T_{c(t)}\left(\text{image}(c) \right) \subset T_{c(t)} \Bbb{R}^2$$. But what is the tangent space to the image of $$c$$? It shouldn't be too hard to convince yourself that if you write $$c(t) = (t, c_2(t))$$ then the tangent space to the image equals the linear span of the (non-zero) vector \begin{align} \xi_{c(t)} :=\dfrac{\partial}{\partial x}\bigg|_{c(t)} + c_2'(t) \dfrac{\partial}{\partial y}\bigg|_{c(t)} \in T_{c(t)} \Bbb{R}^2 \end{align} (i.e $$c(t) = (t, c_2(t))$$ implies $$c'(t) = (1, c_2'(t))$$ so the tangent space is just the span of this vector).

So, we have to show that for all $$t \in [a,b]$$ and for all $$\zeta_{c(t)} \in T_{c(t)} \left( \text{image}(c)\right)$$, \begin{align} dy_{c(t)}(\zeta_{c(t)}) &= f(c(t)) \cdot dx_{c(t)}(\zeta_{c(t)}) \end{align} But notice that since the tangent space to the image is one-dimensional, it suffices to verify equality when evaluated on the basis vector $$\xi_{c(t)}$$ defined above; i.e it's enough to prove \begin{align} dy_{c(t)}(\xi_{c(t)}) &= f(c(t)) \cdot dx_{c(t)}(\xi_{c(t)}). \end{align} This is straight forward: \begin{align} dy_{c(t)}(\xi_{c(t)}) &= dy_{c(t)}\left( \dfrac{\partial}{\partial x}\bigg|_{c(t)} + c_2'(t) \dfrac{\partial}{\partial y}\bigg|_{c(t)}\right) \\ &= c_2'(t) \\ &= f(c(t)) \tag{c solves the ODE} \\ &= f(c(t)) \cdot 1 \\ &= f(c(t))\cdot dx_{c(t)}(\xi_{c(t)}). \end{align}

So, this completes the proof.

Note that another way of stating the equality is that $$c^*(dy) = c^*(f \, dx)$$; i.e when you pull-back the two $$1$$-forms on $$\Bbb{R}^2$$ via the curve $$c$$, you get two $$1$$-forms , but now defined on $$[a,b]$$; and it is these forms which are equal.

• Very illuminating, thank you! There's one thing I don't quite understand. Why is $dy$ a $2$-Form? Up until now I thought a $2$-Form over $\mathbb{R}^2$ has to be of the form $\alpha(x,y) dx \wedge dy$, but in $dy$ there's no exterior product to be seen? Mar 28 '20 at 13:35
• @jazzinsilhouette you're right, it's a 1-form on $\Bbb{R}^2$... lol I got confused because there were two of them, $dy$ and $dx$ (and they were both defined on $\Bbb{R}^2$ so my brain registered $2$ everywhere ;) ). I'll edit now Mar 28 '20 at 13:36

I think I may be the 'physicist' in question, but I'll give it a go.

$$dy$$ is a one-form on $$\mathbb{R}^2$$, so: $$dy_p: T_p \mathbb{R}^2\to \mathbb{R}$$, i.e. it takes 1-simplexes defined in space isomorphic to $$T_p \mathbb{R}^2$$ and gives real values (via integration). In what is to follow $$p\in\mathcal{D}$$ is a fixed point and $$\mathcal{D}\subseteq\mathbb{R}^2$$ is the subspace where the solution of $$y'=f$$ is defined. Same defnitions apply to $$\left(fdx\right)_p$$

Next, since linear functionals can be linearly combined, we can define $$dq_p=\left(dy-fdx\right)_p$$. We want to prove that this new functional is zero.

Now, consider integral along the sufficiently short line-segment, 1-simplex, $$\sigma_p=(p_0,p_1)$$, that contains $$p$$. Let $$\phi_p:[0,1]\to\mathcal{D}$$ be the push-forward such that (with slight abuse of notation) $$\phi_p\left([0,1]\right)=\sigma_p$$. Then:

$$\int_{\sigma_p} dq_p=\int_{\phi_p[0,1]} dq_p=\int_0^1\, \phi_p^*dq_p=\int_0^1 \left(\frac{d\bar{y}}{ds}-f\left(\phi\left(s\right)\right)\frac{d\bar{x}}{ds}\right)ds$$

where $$\bar{y}\left(s\right)=y\left(\phi\left(s\right)\right)$$ and the same for $$\bar{x}$$. The good thing is that now we are no longer dealing with forms, so the simple chain rule applies:

$$f\left(\phi\left(s\right)\right)\frac{d\bar{x}}{ds}=\frac{dy}{dx}\bigg|_{\phi\left(s\right)}\frac{d}{ds}x\left(\phi\left(s\right)\right)=\frac{d}{ds}\bar{y}$$

so:

$$\int_{\sigma_p} dq_p=0$$

Thus we have a functional that gives zero for any 'vector' (simplex) we apply it to (via integration). It must be that $$dq_p=0$$ (how else would you define a zero-functional), which proves what you wanted.