Differential Operator squared

Question

If I have the differential operator $$L = \dfrac{d}{dx} + c v(x)$$ what will be equal to $L^2$?

To $L^2 v(x)?$

What is the meaning of $L^2$? What are my issues to compute it?

In my task I have the formula $$L^N v(x) + \sum\limits_{j=1}^N a_j(x)(L^{j - 1} v(x)) + a_0(x) = 0,$$ where $N$ is the order of equation. So I need to find $L^N v(x)?$ somehow, but I don't know if I correctly work with the differential operator.

What have I tried?

I tried to get the result by substituting $N=2$: $$L^2 v(x) + \sum\limits_{j=1}^2 a_j(x)(L^{j - 1} v(x)) + a_0(x) = 0,$$ $$(\dfrac{d}{dx} + c v(x))^2 v(x) + a_1(x) v(x) + a_2(x)((\dfrac{d}{dx} + c v(x)) v(x)) + a_0(x) = 0,$$ $$\dfrac{d^2v}{dx^2} + c^2 v^3(x) + 2c v(x) \dfrac{dv}{dx} + a_1(x) v(x) + a_2(x)\dfrac{d}{dx} + a_2(x)c v^2(x) + a_0(x) = 0,$$ $$\dfrac{d^2v}{dx^2} + (a_2(x) + 2c v(x)) \dfrac{dv}{dx} + c^2 v^3(x) + a_1(x) v(x) + a_2(x)c v^2(x) + a_0(x) = 0,$$

But the actual result is: $$\dfrac{d^2v}{dx^2} + (a_2(x) + \color{red}{3}c v(x)) \dfrac{dv}{dx} + c^2 v^3(x) + a_1(x) v(x) + a_2(x)c v^2(x) + a_0(x) = 0,$$ where somehow I get $\color{red}{3}c v(x)$. So I guess I work incorctly with differential operator $L$.

Welcome to MSE. Nice question! What is the meaning of $L^2$? What are your issues to compute it? What have you tried? Please update your question with those elements. — mathcounterexamples.net, Apr 25 '23 at 10:32
See: https://math.stackexchange.com/questions/732473/squaring-an-operator for a similar question. — Viktor VN, Apr 25 '23 at 10:53
In the math.SE community, we enforce quality standard. One of the most crucial rules is to provide context when presenting a problem, including any work you have done so far, and other relevant details. Therefore, a question that lacks proper context is not in line with our standards. — Sangchul Lee, Apr 25 '23 at 10:58
Thank you very much for your comments. I've updated the description of task accordingly to yor advices. @SangchulLee — Andrew, Apr 25 '23 at 11:14
@mathcounterexamples.net please, look at the quastion again. Is it now fully correct? — Andrew, Apr 25 '23 at 11:23

score 1 · Answer 1 · answered Apr 25 '23 at 11:05

In general the operator $L^2$ is $$L^2=\left(\dfrac{\mathrm{d}}{\mathrm{d}x}+cv(x)\right)^2=\dfrac{\mathrm{d}^2}{\mathrm{d}x^2}+cv(x)\dfrac{\mathrm{d}}{\mathrm{d}x}+c\dfrac{\mathrm{d}}{\mathrm{d}x}v(x)+c^2v(x)^2$$ So when you apply it to $v(x)$ you have $$\left(\dfrac{\mathrm{d}^2}{\mathrm{d}x^2}+cv(x)\dfrac{\mathrm{d}}{\mathrm{d}x}+c\dfrac{\mathrm{d}}{\mathrm{d}x}v(x)+c^2v(x)^2\right)v(x)$$ $$v''(x)+2cv(x)v'(x)+c^2v(x)^3$$

score -1 · Answer 2 · edited Apr 26 '23 at 05:49

And I have figured it out by mythelf. This should be $$L^2 v(x) = L(L v(x)) = L\left(\dfrac{dv}{dx} + cv(x)^2\right) = \left(\dfrac{d}{dx} + c v(x)\right)\left(\dfrac{dv}{dx} + cv(x)^2\right) = \dfrac{d^2v}{dx^2} + 2c v(x) \dfrac{dv}{dx} + c v(x) \dfrac{dv}{dx} + c^2 v(x)^3$$

Thanks a lot to https://math.stackexchange.com/users/628839/viktor-vn who adviced me to check Squaring an operator. The simular question.

Differential Operator squared

2 Answers2