Suppose I have two differential operators $S, T$ with the property that $\ker S \subseteq \ker T$. Is it possible to exploit this property to write $T$ in terms of $S$ in some compact way?

I would be happy with an answer just for this specific example: Let $S$ be the Schwartzian derivative, $$S[f] := \frac{f'''}{f'} - \frac{3}{2}\left(\frac{f''}{f'}\right)^2,$$ and $T$ the operator $$T[f] := \frac{f''''}{f''} - \frac{4}{3} \left(\frac{f'''}{f''}\right)^2 .$$ The kernel of $T$ consists of the rational functions of the form $$x \mapsto \frac{a_2 x^2 + a_1 x + a_0}{b_1 x + b_0}$$ and the kernel of $S$ famously consists of the fractional linear transformations, i.e., the rational functions of the above form with $a_2 = 0$.

How can one write $T[f]$ in terms of $S[f]$ in a way that exploits the containment $\ker S \subset \ker T$?


I am not sure if this qualifies as a counter-example, but let me offer it anyway. I would say that the answer is no, there is no convenient way to do that.

Consider these differential operators on smooth functions $\mathbb R\to\mathbb R$: $$ Af=f''\\ Bf=(1+(f')^2)f''\\ Cf%=\frac{d}{dx}\left[(1+(f')^2)f''\right] =(1+(f')^2)f'''+2f'(f'')^2. $$ Now $\ker A=\ker B=\{x\mapsto a+bx;a,b\in\mathbb R\}$, but you cannot express $Af(x)$ in terms of $Bf(x)$ or vice versa with any local formula. They differ by something depending on $f'$, but there seems to be no way to write the first derivative in terms of $Af$ or $Bf$.

The operator $Cf$ can be written in terms of $Bf$ since $Cf=(Bf)'$. But it seems that $Cf$ cannot be written in terms of $Af$, since you would also need the first derivative. Note that $\ker B=\ker A\subsetneq\ker C$.

If you meant writing $Tf$ in terms of $Sf$ (and its derivatives) and derivatives of $f$ of order less than the degree of $Tf$, that might be possible — I assumed you meant using only $Sf$ and its derivatives. It certainly works for my example operators and also yours: It seems to be possible to write $Tf$ using $(Sf)'$, $f'$, and $f''$.

  • $\begingroup$ Thanks for the comment (and +1). I don't have a precise definition in mind for "writing in terms of", but it would be ideal if the role of the containment of kernels was apparent in such an procedure. Continuing your examples (and proceeding in definitely an ad hoc way), we have $Bf = (1 + (f')^2) Af$, and this makes it clear that $\ker A \subseteq \ker B$. $\endgroup$ – Travis Willse Apr 7 '17 at 0:01
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    $\begingroup$ @Travis So you mean something like "writing $T$ in terms of $S$ so that kernel inclusion is obvious"? I guess a function $F$ satisfying $Tf=F((Sf)',Sf,f''',f'',f',f)$ and $F(0,0,\cdot,\cdot,\cdot,\cdot)\equiv0$ would be a reasonable formulation. I'll get back to you if I have more ideas. My only offer now for your concrete example is $Tf=(Sf)'f'/f''+4Sf+3(f''/f')^2-\frac43(f'''/f'')^2$, but this does make the kernel inclusion visible at all. $\endgroup$ – Joonas Ilmavirta Apr 7 '17 at 0:19

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