# Solutions of $(D_1\circ D_2)f(x)=0$ for commuting $D_1, D_2$.

What am I doing in the following? Is there a name for it? Please be polite.

Say we know, for linear differential operators $$D_1,D_2$$ that the following two equations are true

$$\,\,(D_1\circ D_2)f(x)=0 \tag{1}$$ $$\quad\quad\quad\quad \quad\quad(D_1\circ D_2)f(x)=(D_2\circ D_1)f(x) \tag{2}$$ So that $$D_1$$ and $$D_2$$ commute.

We then say that $$f(x)$$ can be written as $$f(x)=f_1(x)+f_2(x) \tag{3}$$ where $$D_1f_1(x)=0 \tag{4}$$ $$D_2f_2(x)=0 \tag{5}$$

This appears true because substituting (3) into the left-hand side of (1), we actually satisfy (1), we have \begin{align} (D_1\circ D_2) (f_1(x)+f_2(x) )&=(D_1\circ D_2) f_1(x)+(D_1\circ D_2) f_2(x) \\ &=(D_2\circ D_1) f_1(x)+(D_1\circ D_2) f_2(x) \\ &=0+0 \end{align} Where I have used first, linearity, then commutativity, then (4) with (5) in the last step. Is there a name for what's going on here?

Can the solutions of (1) always be split up as in (3)?

Other Information.

The material of this question is relevant to fluid mechanics, to the theory of the instability of jets.

• This is not true for all pairs $D_1$ and $D_2$. However, if $D_1=D_2=\frac{d}{dx}$, then solutions to $D_1D_2f=0$ are linear polynomials $a+bx$, which cannot always be written as $f_1+f_2$ with $D_1f_1=0$ and $D_2f_2=0$. You can say that $f=f_1+f_2$ is a solution to $D_1D_2f=0$, but you cannot say that any solution $f$ is a sum $f_1+f_2$. – user593746 Nov 4 '18 at 14:35
• On the other hand, if there exists a polynomial $p(u,v)$ whose coefficients are some nice functions in $x$ such that $p(0,0)=0$ and $p(D_1,D_2)$ is the identity operator, then you can say that every solution $f$ to $D_1D_2f$ satisfies $f=f_1+f_2$. – user593746 Nov 4 '18 at 14:42