Solutions for diff equation in form of series

[Jhenrique]

Given the following diff equation: $(1 - D)y(x) = f(x)$, being D = d/dx, the "implicit" solution is: $y(x) = \frac{1}{1-D}f(x)$, so, for "explicit" the solution is necessary to expand the fraction 1/(1-D) by identity: $\frac{1}{1-x}=\sum_{0}^{\infty}x^n \Delta n$, but this infinity series is true only for |x|<1, for |x|>1 is necessary utilize the following series: $\frac{1}{1-x} = -\sum_{-\infty}^{-1} x^n \Delta n$. Happens that D isn't a number for I say that |D| is less or greater than 1. So, how can I interprate this form of solution correctly?

PS, this ideia from Operational Calculus (http://www.latp.univ-mrs.fr/~chaabi/ARTICLES%20IMPORTANTS/Autres%20articles%20interessant/H.-J.%20Glaeske,%20A.%20P.%20Prudnikov,%20K.A.%20Skornik%20-%20Operational%20Calculus%20and%20Related%20Topics%20%28Analytical%20Methods%20and%20Special%20Functions%29%20%28Chapman%20,2006%29.pdf )

 

[Mark44]

Given the following diff equation: $(1 - D)y(x) = f(x)$, being D = d/dx, the "implicit" solution is: $y(x) = \frac{1}{1-D}f(x)$,

No, this makes no sense. You are treating 1 - D as if it were multiplying y(x) - it isn't, any more than (d/dx)y means d/dx times y. Here 1 - D is an operator that operates on y(x). It does NOT multiply y(x).

so, for "explicit" the solution is necessary to expand the fraction 1/(1-D) by identity: $\frac{1}{1-x}=\sum_{0}^{\infty}x^n \Delta n$, but this infinity series is true only for |x|<1, for |x|>1 is necessary utilize the following series: $\frac{1}{1-x} = -\sum_{-\infty}^{-1} x^n \Delta n$. Happens that D isn't a number for I say that |D| is less or greater than 1. So, how can I interprate this form of solution correctly?

PS, this ideia from Operational Calculus (http://www.latp.univ-mrs.fr/~chaabi/ARTICLES%20IMPORTANTS/Autres%20articles%20interessant/H.-J.%20Glaeske,%20A.%20P.%20Prudnikov,%20K.A.%20Skornik%20-%20Operational%20Calculus%20and%20Related%20Topics%20%28Analytical%20Methods%20and%20Special%20Functions%29%20%28Chapman%20,2006%29.pdf )

 

[Jhenrique]

No, this makes no sense. You are treating 1 - D as if it were multiplying y(x) - it isn't, any more than (d/dx)y means d/dx times y. Here 1 - D is an operator that operates on y(x). It does NOT multiply y(x).

You are being much formal, look:
https://es.wikipedia.org/wiki/Transformada_de_Laplace#Perspectiva_hist.C3.B3rica

And ditto mark for my 1st question more one times...

 

[Xiuh]

Just a recommendation for you based on reading your other posts.. It seems that you have a lot of misconceptions. Maybe it wouldn't hurt if you mastered basic math before trying to do more advanced stuff :)

There is indeed something similar to what you're thinking. Suppose you have a bounded operator \textit{T} on a normed space \textit{X}. If the series \sum\limits_{k=0}^\infty T^k converges, then (1 - T) is invertible and then
\left(1 - T\right)^{-1} = \sum\limits_{k=0}^\infty T^k
If \textit{X} is Banach and \| T \| &lt; 1 (in operator norm) the convergence is guaranteed. The thing is, you have to define the function space you are working with. If you are working with polynomials then convergence is trivial (it is only a finite sum). On the other hand, if you are considering \frac{d}{dx}: C^1([0,1]) \rightarrow C([0,1]) the operator is not bounded.

 

[lurflurf]

There are ways to know in advance if the method will work, but it is often easier to go ahead with it and check the answer for correctness later. That is use the method to produce a possible solution to be verified afterward.

$$(1-\mathrm{D})\mathrm{y}(x)=\mathrm{f}(x) \\
\mathrm{y}(x)=(1-\mathrm{D})^{-1}\mathrm{f}(t) \\
\mathrm{y}(x)=-\mathrm{D}^{-1}(1-\mathrm{D}^{-1})^{-1}\mathrm{f}(t) \\
\mathrm{y}(x)=-\sum_{k=1}^\infty\mathrm{D}^{-k}\mathrm{f}(t) \\
\mathrm{y}(x)=-\mathrm{D}^{-1}\sum_{k=1}^\infty\frac{(x-t)^{k-1}}{(k-1)!}\mathrm{f}(t) \\
\mathrm{y}(x)=-\mathrm{D}^{-1}e^{x-t}\mathrm{f}(t) \\
\mathrm{y}(x)=-e^{x}\mathrm{D}^{-1}e^{-t}\mathrm{f}(t)
$$
where
$$\mathrm{D}^{-1}\{\cdot\}=\int_a^x \! \{\cdot\} \, \mathrm{d}t$$
Also observe with some intuition we could skip to the end
$$(1-\mathrm{D})\mathrm{y}(x)=\mathrm{f}(x) \\
(1-\mathrm{D})e^x e^{-x}\mathrm{y}(x)=\mathrm{f}(x) \\
-e^x\mathrm{D}e^{-x}\mathrm{y}(x)=\mathrm{f}(x) \\
\mathrm{y}(x)=-e^x\mathrm{D}^{-1}e^{-t}\mathrm{f}(t)$$

 

[Mark44]

You are being much formal, look:
https://es.wikipedia.org/wiki/Transformada_de_Laplace#Perspectiva_hist.C3.B3rica

The English version of that page doesn't have any of that factoring business in it that is in the Spanish page.

And ditto mark for my 1st question more one times...

 

[Mark44]

There are ways to know in advance if the method will work, but it is often easier to go ahead with it and check the answer for correctness later. That is use the method to produce a possible solution to be verified afterward.

$$(1-\mathrm{D})\mathrm{y}(x)=\mathrm{f}(x) \\
\mathrm{y}(x)=(1-\mathrm{D})^{-1}\mathrm{f}(t) \\
\mathrm{y}(x)=-\mathrm{D}^{-1}(1-\mathrm{D}^{-1})^{-1}\mathrm{f}(t) \\
\mathrm{y}(x)=-\sum_{k=1}^\infty\mathrm{D}^{-k}\mathrm{f}(t) \\
\mathrm{y}(x)=-\mathrm{D}^{-1}\sum_{k=1}^\infty\frac{(x-t)^{k-1}}{(k-1)!}\mathrm{f}(t) \\
\mathrm{y}(x)=-\mathrm{D}^{-1}e^{x-t}\mathrm{f}(t) \\
\mathrm{y}(x)=-e^{x}\mathrm{D}^{-1}e^{-t}\mathrm{f}(t)
$$

How are you going from the 2nd line above to the third?

where
$$\mathrm{D}^{-1}\{\cdot\}=\int_a^x \! \{\cdot\} \, \mathrm{d}t$$
Also observe with some intuition we could skip to the end
$$(1-\mathrm{D})\mathrm{y}(x)=\mathrm{f}(x) \\
(1-\mathrm{D})e^x e^{-x}\mathrm{y}(x)=\mathrm{f}(x) \\
-e^x\mathrm{D}e^{-x}\mathrm{y}(x)=\mathrm{f}(x) \\
\mathrm{y}(x)=-e^x\mathrm{D}^{-1}e^{-t}\mathrm{f}(t)$$

 

[Mark44]

PS, this ideia from Operational Calculus (http://www.latp.univ-mrs.fr/~chaabi/ARTICLES%20IMPORTANTS/Autres%20articles%20interessant/H.-J.%20Glaeske,%20A.%20P.%20Prudnikov,%20K.A.%20Skornik%20-%20Operational%20Calculus%20and%20Related%20Topics%20%28Analytical%20Methods%20and%20Special%20Functions%29%20%28Chapman%20,2006%29.pdf )

One other thing: if you cite a paper that is 400+ pages long, at least say what page or section you're referring to.

 

[lurflurf]

^distributive property

$$(1-\mathrm{D})^{-1}=[(-\mathrm{D})(-\mathrm{D}^{-1})+(-\mathrm{D})(1)]^{-1} \\
=[-\mathrm{D}(-\mathrm{D}^{-1}+1)]^{-1}=(-\mathrm{D})^{-1}(1-\mathrm{D}^{-1})^{-1}=-\mathrm{D}^{-1}(1-\mathrm{D}^{-1})^{-1}$$
In operators 1 should be considered the identity of the operator algebra and not a number.
Maybe it would be better to use I
For example D(1-D)=D-D^2 not -D^2

 

[Jhenrique]

One other thing: if you cite a paper that is 400+ pages long, at least say what page or section you're referring to.

Sorry, kkkkkkkkkk, the theory about operational calculus is in the first pages...

I know that the following expression works, $\frac{1}{1-D}=\sum_{0}^{\infty}D^n \Delta n$, and that the following too, $\frac{1}{1-D} = -\sum_{-\infty}^{-1} D^n \Delta n$. So, my question is: given a diff equation like: $y(x) = \frac{1}{1-D}f(x)$, which of the two I must to use, and why?

 

[Mark44]

Sorry, kkkkkkkkkk, the theory about operational calculus is in the first pages...

I know that the following expression works, $\frac{1}{1-D}=\sum_{0}^{\infty}D^n \Delta n$, and that the following too, $\frac{1}{1-D} = -\sum_{-\infty}^{-1} D^n \Delta n$. So, my question is: given a diff equation like: $y(x) = \frac{1}{1-D}f(x)$, which of the two I must to use, and why?

IMO it's misleading to write $\frac 1 {1 - D}$ instead of (1 - D)^-1; i.e., as a fraction instead of an inverse.

 

[Jhenrique]

IMO it's misleading to write $\frac 1 {1 - D}$ instead of (1 - D)^-1; i.e., as a fraction instead of an inverse.

But that difference this make?

PS: this link has more example of "operational calculus": http://rip94550.wordpress.com/2012/08/27/heavisides-operational-calculus/

So, again I reiterate my question:

I know that the following expression works, $\frac{1}{1-D}=\sum_{0}^{\infty}D^n \Delta n$, and that the following too, $\frac{1}{1-D} = -\sum_{-\infty}^{-1} D^n \Delta n$. So, my question is: given a diff equation like: $y(x) = \frac{1}{1-D}f(x)$, which of the two I must to use, and why?

 

[Jhenrique]

There is indeed something similar to what you're thinking. Suppose you have a bounded operator \textit{T} on a normed space \textit{X}. If the series \sum\limits_{k=0}^\infty T^k converges, then (1 - T) is invertible and then
\left(1 - T\right)^{-1} = \sum\limits_{k=0}^\infty T^k
If \textit{X} is Banach and \| T \| &lt; 1 (in operator norm) the convergence is guaranteed. The thing is, you have to define the function space you are working with. If you are working with polynomials then convergence is trivial (it is only a finite sum). On the other hand, if you are considering \frac{d}{dx}: C^1([0,1]) \rightarrow C([0,1]) the operator is not bounded.

I didn't undertand...