Linear Differential Operator order

tom_rylex
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[Solved] Linear Differential Operator order

Homework Statement


I'm misunderstanding something basic about how this works:

Give examples of linear differential operators L and M for which it is not true that L(M(u)) = M(L(u)) for all u.


Homework Equations


Since it's arbitrary, I made two first order differential functions of two variables:
\overline{x} = {x,y}
u=u(x,y)
L(u) = a(\overline{x})u + b_1(\overline{x})u_x + b_2(\overline{x})u_y
M(u) = c(\overline{x})u + d_1(\overline{x})u_x + d_2(\overline{x})u_y
Where a,b,c, and d are coefficients.

The Attempt at a Solution


When I expand L(M(u)), it seems to look like M(L(u)):

L(M(u)) = a(\overline{x})c(\overline{x})u+a(\overline{x})d_1(\overline{x})u_x +a(\overline{x})d_2(\overline{x})u_y + b_1(\overline{x})c(\overline{x})u_x+b_2(\overline{x})c(\overline{x})u_y

M(L(u)) = c(\overline{x})a(\overline{x})u+c(\overline{x})b_1(\overline{x})u_x +c(\overline{x})b_2(\overline{x})u_y + d_1(\overline{x})a(\overline{x})u_x+d_2(\overline{x})a(\overline{x})u_y

I'm stuck, since it seems that all roads lead back to a commutative relationship.
 
Last edited:
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Linear differential operators with constant coefficients are commutative.

Try using linear differential operators with variable coefficients. That is, something like (d/dx+ 3x)y or (xd/dx)u
 
[Solved] Linear Differential Operator order

Thanks. I just needed a nudge to get in the right direction. Some parts of DiffEq have been a while for me. I set up the linear differential operators

L = \frac {d}{dx}
M = \frac {d}{dx} + x

to show that L(M(u)) \neq M(L(u)) for all u. I think you helped me become 2 points smarter (don't ask me the scale...).
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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