Is (1-D)g(x) the Correct Solution to the Differential Equation?

[eljose]

let,s suppose we have the differential equation:

g(x)=y+Dy+D^{2}y+D^{3}y+.....

D=d/dx, then this is equal to:

g(x)= \frac{1}{1-D}y or inverting y=(1-D)g(x)

although this should be the solution it seems too easy to be true..but is that correct?..thanks.

 

[HallsofIvy]

Well, let me see. If you start with a differential equation that is meaningless (or which, at least, you didn't bother to define) then, I guess, any thing you want to say is just as meaningless as the equation!

 

[AKG]

He's defining a function g by

g(x) = y(x) + y'(x) + \sum_{k=2}y^{(k)}(x)

Assuming that series converges for each x, g is well-defined. Then, is it true that y(x) = g(x) - g'(x) (assuming g, as defined, is differentiable). This essentially asks if it is true that g'(x) = y'(x) + \sum_{k=2}y^{(k)}(x), i.e. is the rule that 'the derivative of a sum the sum of the derivatives' extendible to the case where the sum is one of infinitely many functions?

 

[AKG]

g'(x) = \lim _{h\to 0}\frac{\sum_{k=0}y^{(k)}(x+h) - \sum_{k=0}y^{(k)}(x)}{h}

= \lim _{h\to 0}\frac{\lim _{n\to\infty}\sum_{k=0}^ny^{(k)}(x+h) - \lim _{n\to\infty}\sum_{k=0}^ny^{(k)}(x)}{h}

= \lim _{h\to 0}\frac{\lim _{n\to\infty}\sum_{k=0}^n\left (y^{(k)}(x+h) - y^{(k)}(x)\right )}{h}

= \lim _{h\to 0}\lim _{n\to\infty}\sum_{k=0}^n\frac{y^{(k)}(x+h) - y^{(k)}(x)}{h}

= \lim _{n\to\infty}\lim _{h\to 0}\sum_{k=0}^n\frac{y^{(k)}(x+h) - y^{(k)}(x)}{h}

= \lim _{n\to\infty}\sum_{k=0}^n\lim _{h\to 0}\left (\frac{y^{(k)}(x+h) - y^{(k)}(x)}{h}\right )

= \lim _{n\to\infty}\sum_{k=0}^n\lim _{h\to 0}y^{(k+1)}(x)

= \sum _{k=1}y^{(k)}(x)

Are any of these steps unjustified? The only lines I'm not sure about are line 3 (does it need absolute convergence) and line 5.

 

[eljose]

Well, let me see. If you start with a differential equation that is meaningless (or which, at least, you didn't bother to define) then, I guess, any thing you want to say is just as meaningless as the equation!

Why should it be meaningless?..is a differential equation of infinite order..Euler Himself worked with this type of equations..for example to solve f(x+1)-f(x)=1 he make a Taylor expansion getting:

1= \sum_{n=0}^{\infty} \frac{y^{n}}{n!} (1)

for which he gets the solution (in a form of infinite series)

y= \sum_{n=0}^{\infty}a_{n} e^{i2\pi i x}

where he concludes that the a(n) must be chosen so the equation (1) is satisfied

 

[matt grime]

Euler was notoriously unbothered by convergence issues. You should not be. In what space are you asking for the differential series you give to be convergent? You need at least to specify that the sequence is absolutely convergent for all x, and then the result is easy. So the answer is, I suppose: yes when that is true it is trivially true and when it is false it is false. I.e. when you're allowed to rearrange the summation of g(x) it works when you're not it doesn't.

Of course, if the differential operator had norm less than 1 then it would be fine in the sense of Banach spaces.