Okay so suppose I have the Initial Value Problem:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

\left. \begin{array}{l}

\frac {dy} {dx} = f(x,y) \\

y( x_{0} ) = y_{0}

\end{array} \right\} \mbox{IVP}

[/tex]

NB. I am considering only real functions of real variables.

If [tex]f(x,y) [/tex] is analytic atxand_{0}ythen that means that we can construct its Taylor Series centered around the point_{0}(xand that the Taylor Series will have a positive radius of convergence, and also that within this radius of convergence the function [tex]f(x,y) [/tex] will equal its Taylor Series._{0},y_{0})

[tex]f(x,y) [/tex] being analytic atxand_{0}yalso means that the IVP has a unique solution in a neighbourhood of the point_{0}x, as follows from The Existence and Uniqueness Theorem (Picardâ€“LindelĂ¶f). Let_{0}y(x)be the function that satisfies the IVP in a neighbourhood ofx._{0}

The idea behind the Taylor Series method is to use the differential equation and initial condition to find the Taylor coefficients ofy(x):

[tex]

\frac {dy} {dx} = f(x,y) \; \; \rightarrow \; \;

y'(x_0)=f(x_0,y(x_0))=f(x_0,y_0)

[/tex]

[tex]

\frac {d^2y} {dx^2} = \frac {d} {dx} f(x,y)= \frac {\partial f} {\partial x} \frac {dx} {dx} \; + \; \frac {\partial f} {\partial y} \frac {dy} {dx} \\

\indent \rightarrow \; \; y''(x_0)= \frac {\partial f} {\partial x}|_{(x_0,y(x_0))} \; + \; \frac {\partial f} {\partial y}|_{(x_0,y(x_0))}\cdot f(x_0,y_0)

[/tex]

etc.

Obviously we can do this because iffis analytic, all the partial derivatives offexist at(x, and by the relationship given by the ODE (which we know_{0},y_{0})y(x)satisfies at least in a neighbourhood ofx), all the derivatives of_{0}yexist atx, that is,_{0}

[tex]

y(x) \in C^\infty _x (x_0)

[/tex]

Okay, so we can construct the Taylor Series ofy(x)atx:_{0}

[tex]

\sum \frac{y^{(n)}(x_0)}{n!}x^n

[/tex]

But!

1)How do we know that this series has a non-zero radius of convergence?

2)And secondly, if it does have a positive radius of convergence, how do we know it is equal to the solution of the IVP within its radius of convergence? That is, obviously within its radius of convergence, the series represents a certain function which is analytic atx, but how do we know that this function is indeed the function_{0}y(x)which is what we called the solution to the IVP? I mean, maybe the solution to the IVP is not analytic atx(even though it is infinitely derivable at_{0}x) and thus its Taylor series (the one we constructed) does not represent it in any neighbourhood of_{0}x. After all, the existence and uniqueness theorem does not require or state that the solution be analytic at all._{0}

Um, if you can affirm that the Taylor Series we have constructed is indeed a solution to the IVP within its radius of convergence, then by the uniqueness of the solution, the function it represents must be the solution of the IVP,y(x). But can this be affirmed?

If anyone has any ideas please help, I'm just not happy with this method coz yes, I can calculate a number of terms of the Taylor Series to approximate the solution of the IVP, but I really have no assurance that the series I am constructing is indeed solution of the IVP (not to mention whether it even does converge at all for anyxnearx)._{0}

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# An issue with solving an IVP by Taylor Series

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