When do differential operators commute?

Question

Given that the equation of motion of a particle placed on the apex of Norton's Dome is $$\frac{d^2 r}{dt^2}=r^{1/2}\qquad\longleftarrow\text{as proved in this previous question}\tag{1}$$ Solve this non-linear equation by multiplying both sides by $$2\frac{dr}{dt}$$ and integrating to find that $$r(t)=\frac{\left(t-T\right)^4}{144}$$ for any $T\ge 0$.

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Pre-multiplying both sides of $(1)$ by $$2\frac{dr}{dt}$$ gives $$2\color{red}{\frac{dr}{dt}}\color{blue}{\left(\frac{d^2r}{dt^2}\right)}=2\frac{dr}{dt}r^{1/2}\tag{2}$$

Rewriting $(2)$ $\bbox[yellow]{\text{with the order of the red and blue factors switched}}$ gives:

$$2\color{blue}{\frac{d^2r}{dt^2}}\color{red}{\left(\frac{dr}{dt}\right)}=2\frac{dr}{dt}r^{1/2}\tag{3}$$ The reason for the switch is because the blue factor in $(3)$ is the derivative of the red factor (I had to write it this way in order to integrate by inspection). Integrating both sides gives $$\left(\frac{dr}{dt}\right)^2=\frac{4}{3}r^{3/2}+C\tag{4}$$

In obtaining $(4)$ from $(3)$ I used the chain rule on the RHS, namely $$\frac{d}{dt}=\frac{d}{dr}\cdot\frac{dr}{dt}$$

Finally, my question is all about the validity of the sentence highlighted in yellow. I am aware that differential operators do not commute in general. So is it plausible for me to simply switch the order of the red and blue factors as I did in getting from $(2)$ to $(3)$?

Many thanks.

Answer 1

About Switching Operators

I hope that I am not missing something! :)

I think what you are asking is the same as that the multiplication of two real numbers is commutative or not!

The answer is YES, as you know that one can write

$$a\cdot b=b\cdot a$$

where $a$ and $b$ are two real numbers.

In fact, in your problem no operator switching is involved

$$\color{red}{\left(\frac{dr}{dt}\right)} \color{blue}{\left(\frac{d^2r}{dt^2}\right)}=ab=ba=\color{blue}{\left(\frac{d^2r}{dt^2}\right)}\color{red}{\left(\frac{dr}{dt}\right)}$$

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Another Way to Solve

There is another way for solving $(1)$ that you do not need to multiply by $2\frac{dr}{dt}$!

I will give you a hint. Consider the following

$$\ddot r = {{{d^2}r } \over {d{t^2}}} = {d \over {dt}}\left( {{{dr } \over {dt}}} \right) = {d \over {dr }}\left( {{{dr } \over {dt}}} \right){{dr } \over {dt}} = \dot r {d \over {dr }}\left( {\dot r } \right) = {d \over {dr }}\left( {{1 \over 2}{{\dot r }^2}} \right)$$

and hence your ode in $(1)$ becomes

$${d \over {dr }}\left( {{1 \over 2}{{\dot r }^2}} \right)=r^{1/2}$$

Answer 2

Actually, the question of commutation of two ordinary differential operators (ODO) is not a trivial one. If two ODOs commute, then they necessarily satisfy a polynomial relationship (by work of Burchnall and Chaundy). In particular, their eigenvalues lie on an affine algebraic curve (its projective completion is called spectral curve). The mutual eigenfunctions (so-called Baker-Akhiezer functions) of commuting ODOs form a vector bundle over the spectral curve.

When BA bundle's rank $r=1$, Krichever has given an explicit expression for the coefficients of operators in terms of the $\theta$-function of the Jacobian of the spectral curve.

In the case $r>1$, the problem gets more difficult and, to the best of my knowledge, is far from being completely solved. Krichever and Novikov found an explicit expression for operators with elliptic spectral curve ($g=1$) with $r=2$. Mokhov has treated the case $g=1, \: r=3$. Mironov has given examples of commuting differential operators of $r=2$ with hyperelliptic spectral curve of arbitrary genus. Zuo's given an example of operators with $g=2, \: r=3$.