Which analogy between polynomials and differential equations did Rota have in mind in his TEN LESSONS?

Solution 1:

This might be a partial answer, but I will add later if need be.

There is quite a bit of theory underlying a systematic change of variables. The theory underlying it all is Noether's Theorem, which states for any symmetry there is a corresponding invariant. This invariant then is a useful substitution, as it will effectively reduce the order of a differential equation by 1, and if the order is already 1, the substitution makes the equation separable.

The standard example of this is with first order homogenous equations. In this case, we can see that the differential equation has symmetry under the transformation $x \rightarrow \lambda x$ and $y \rightarrow \lambda y$. An invariant corresponding to the symmetry is some quantity that does not change under the transformation; thus, $y/x$ is an invariant.

There are many other examples of such a change of variable. For example, take the differential equation $$x^{3/2}y''+\sqrt{x}y'+\frac{y^2}{x}=1$$ In this case, you can check that that the differential equation is symmetric with respect to the transformation $x \rightarrow \lambda^2 x$, $y \rightarrow \lambda y$. With this transformation, $y^2/x$ is an invariant of the transformation, and therefore the order of the equation can be reduced by 1 by substituting $u=y^2/x$. (specifically, it changes the equation into a differential equation that contains $u''$ and $u'$, so a substitution can be made to reduce the order).

There are however many other symmetries, and corresponding invariants, possible however, beyond these "scaling" symmetries. So the question then becomes how to find such substitutions. The most complete and systematic method is via Lie Groups to calculate the symmetries. Some references for doing so are here and here. The most comprehensive resource on this question is here. You will notice that these resources are generally graduate level however; I am not aware of systematic treatments of calculating these symmetries (beyond scaling symmetries) at a lower level.

Solution 2:

I'm adding my own answer because I figured out a small piece of the puzzle. Perhaps other answers can build on this and maybe this makes it clearer what kind of thing I'm looking for.

When we have a polynomial equation like $x^3 + bx^2 + cx + d = 0$ we can express symmetric polynomials of the roots as polynomials in the coefficients. For instance, if $(x - r_1)(x - r_2)(x - r_3) = x^3 + bx^2 + cx + d$ then $r_1 r_2 r_3 + r_1 + r_2 + r_3 = -(a+d)$. This is cool because the roots might be algebraic numbers, but any symmetric polynomial in them can be calculated explicitly, and will be a rational number if the coefficients of the original polynomial are rational.

Using the Laplace transform we can transform a differential equation like $f''' + bf'' + cf' + df = 0$ to the previous polynomial equation. This will give us three linearly independent $f_1 = \exp(r_1 t), f_2 = \exp(r_2 t), f_3 = \exp(r_3 t)$, provided the roots are distinct.

Suppose that we have a term $r_1^n r_2^m r_3^k$ in the symmetric polynomial. We then do $(D^n f_1)(D^m f_2)(D^k f_3) = r_1^n r_2^m r_3^k \exp((r_1 + r_2 + r_3) t)$. Hence if we have a symmetric "differential polynomial" in the $f_1,f_2,f_2$ then that will be some symmetric polynomial in the roots times $\exp((r_1+r_2+r_3)t)$. Since both the symmetric polynomial in the roots and the $r_1+r_2+r_3$ can be found as polynomials in the coefficients $b,c,d$, we can explicitly calculate any symmetric differential polynomial of the three solutions.

Now, there are many questions:

1. What if we picked another set of linearly independent solutions?
2. What about repeated roots?
3. What if the equation is not constant coefficients?

I think an anwser to these questions would explain what Rota meant by the first paragraph that I quoted.

Lie groups might indeed have been what he meant by the second paragraph, but I'd like to more clearly understand how they are analogous to classical invariant theory.

And then there's the third paragraph...

I don't know if I can award a bounty multiple times, but if I can, then I will award an additional 50 point bounty for an explanation for each of those three paragraphs.