How find two differential operator $A$ and $B$ such $A\circ \dfrac{d}{dx}=B\circ x$

Question

Question:

show that there exists differential operators $A$ and $B$ where $$A=\sum_{k=0}^{n}a_{k}(x)\dfrac{d^k}{dx^k}\neq 0,B=\sum_{k=0}^{n}b_{k}(x)\dfrac{d^k}{dx^k}\neq 0$$ ($a_{k}(x),b_{k}(x)$ are infinitely differentiable functions), such that $$A\circ \dfrac{d}{dx}=B\circ x$$

We define: $F\circ G(f)=F(G(f))$

this problem is from a book,and the book only post answer :

for example $$A=x^2,B=x\dfrac{d}{dx}-1$$

But I don't know why this example is such two condition,can you explain detailed? Thank you

Answer 1

Restating the requirements, with $D = \dfrac{d}{dx}$ for convenience, we want $$ \sum_k a_k(x) D^k(D(f)) = \sum_k b_k(x) D^k(x f) \tag 1$$ for any function $f$.

Now by Leibniz's rule, $D(x f) = f + x D(f)$, and by induction $D^k(x f) = k D^{k-1}(f) + x D^k(f)$, so that (1) becomes

$$\eqalign{ \sum_k a_k(x) D^{k+1}(f) &= \sum_k b_k(x) (k D^{k-1}(f) + x D^k(f)\cr \sum_{k} a_{k-1}(x) D^k(f)&= \sum_k ((k+1) b_{k+1}(x) + x b_k(x)) D^k(f) \cr}$$ There is no $k=0$ term on the left, so we need $$0 = b_1(x) + x b_0(x) \tag 2$$ and for all $k \ge 1$, $a_{k-1}$ is defined by $a_{k-1}(x) = (k+1) b_{k+1}(x) + x b_k(x)$.

The example given has $b_0(x) = -1$, $b_1(x) = x$ (which does satisfy (2)) and all other $b_k(x) = 0$, and then $a_0(x) = 2 b_2(x) + x b_1(x) = x^2$, all other $a_k(x) = 0$. But there are plenty of other solutions. $B$ can be any differential operator of order $n$ (i.e. the highest $k$ such that $b_k \ne 0$ is $n$) that satisfies (2), and $a$ will be an operator of order $n-1$.

Answer 2

For existence, one can often find an example. The book has given an example. Here is how it works:

$A(\frac{d}{dx}(f))=x^2f'$

and

$B(x(f))=(x\frac{d}{dx}-1)(xf)=x(\frac{d}{dx}xf) - xf=x(f+xf')-xf=x^2f'+xf-xf =x^2f'$

And so both are equal.

If I have misunderstood you, please say.