# An issue with solving an IVP by Taylor Series

#### BobbyBear

Okay so suppose I have the Initial Value Problem:

$$\left. \begin{array}{l} \frac {dy} {dx} = f(x,y) \\ y( x_{0} ) = y_{0} \end{array} \right\} \mbox{IVP}$$

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

If $$f(x,y)$$ is analytic at x0 and y0 then that means that we can construct its Taylor Series centered around the point (x0,y0) and that the Taylor Series will have a positive radius of convergence, and also that within this radius of convergence the function $$f(x,y)$$ will equal its Taylor Series.

$$f(x,y)$$ being analytic at x0 and y0 also means that the IVP has a unique solution in a neighbourhood of the point x0, as follows from The Existence and Uniqueness Theorem (Picard–Lindelöf). Let y(x) be the function that satisfies the IVP in a neighbourhood of x0.

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

$$\frac {dy} {dx} = f(x,y) \; \; \rightarrow \; \; y'(x_0)=f(x_0,y(x_0))=f(x_0,y_0)$$

$$\frac {d^2y} {dx^2} = \frac {d} {dx} f(x,y)= \frac {\partial f} {\partial x} \frac {dx} {dx} \; + \; \frac {\partial f} {\partial y} \frac {dy} {dx} \\ \indent \rightarrow \; \; y''(x_0)= \frac {\partial f} {\partial x}|_{(x_0,y(x_0))} \; + \; \frac {\partial f} {\partial y}|_{(x_0,y(x_0))}\cdot f(x_0,y_0)$$

etc.

Obviously we can do this because if f is analytic, all the partial derivatives of f exist at (x0,y0), and by the relationship given by the ODE (which we know y(x) satisfies at least in a neighbourhood of x0), all the derivatives of y exist at x0, that is,
$$y(x) \in C^\infty _x (x_0)$$

Okay, so we can construct the Taylor Series of y(x) at x0:
$$\sum \frac{y^{(n)}(x_0)}{n!}x^n$$

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 at x0, but how do we know that this function is indeed the function y(x) which is what we called the solution to the IVP? I mean, maybe the solution to the IVP is not analytic at x0 (even though it is infinitely derivable at x0) and thus its Taylor series (the one we constructed) does not represent it in any neighbourhood of x0. After all, the existence and uniqueness theorem does not require or state that the solution be analytic at all.
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 any x near x0).

#### BobbyBear

Sometimes I feel I speak in Martian, especially when my posts are long ones:P Did anyone actually get was I was on about? Just curious, as sometimes I myself don't lol :P

#### BobbyBear

Yay I think I might be able to (partially) answer my own question xD

-gingerly tries to demostrate-

Let
$$T_{y,x_0} (x) = \sum_{n=0}^\infty \frac{y^{(n)}(x_0)}{n!}x^n$$

be the Taylor series of y(x) that I've constructed using the initial condition and the relationship
given by the o.d.e. If it has a positive radius of convergence R>0 (unfortunately I'll only be able to know
this if I can deduce an expression for the general term of the series), then within its radius of convergence,
it converges to an analytic (at x0) function, lets say u(x):

$$u(x) = T_{y,x_0} (x) = \sum_{n=0}^\infty \frac{y^{(n)}(x_0)}{n!}x^n = T_{u,x_0} (x) = \sum_ {n=0}^\infty \frac{u^{(n)}(x_0)}{n!}x^n \ \ \ \ \ \ \ \ \ \mbox{for} \ \ |x-x_0|<R$$

And!! And!! If f(x,y) is analytic at (x0,y0), then

$$u(x) = T_{y,x_0} (x) = \sum_{n=0}^\infty \frac{y^{(n)}(x_0)}{n!}x^n$$

is solution to the I.V.P. in a certain neighbourhood of x0 !! Because, within the Radius of Convergence, you can
differentiate u(x) by differentiating each term of the series and what not :P, so:

$$u'(x) = \frac {d} {dx} (T_{y,x_0} (x)) = \frac {dy} {dx}(x_0) + \frac {d^2y} {dx^2}(x_0) \cdot (x-x_0) + \frac {d^3y} {dx^3}(x_0) \cdot \frac {(x-x_0)^2}{2!} + ... \ \ \ \ \ \ \ \ \ \mbox{for} \ \ |x-x_0|<R$$

and, as f(x,y) is analytic at (x0,y0), then in some neighbourhood of the point (x0,y0), lets say |x-x0|,|y-y0|< R* :

$$f(x,u) = f(x_0,y_0) \ + \ \frac {\partial f} {\partial x} (x_0,y_0) \cdot (x-x_0) \ + \ \frac {\partial f} {\partial u} (x_0,y_0)\cdot (u-y_0) +$$
$$\ \ \ \frac {1}{2!} \left[ \frac {\partial ^2f} {\partial x^2}(x_0,y_0) \cdot (x-x_0)^2 + \ 2 \frac {\partial^2 f} {\partial x \partial u } (x_0,y_0) \cdot (x-x_0)(u-y_0) + \frac {\partial ^2f} {\partial u^2}(x_0,y_0) \cdot (u-y_0)^2 \right] + ... \ \ \ \ \mbox{for} \ \ |x-x_0|,|u-y_0| <R^*$$

$$= f(x_0,y_0) \ \ + \ \ \frac {\partial f} {\partial x} (x_0,y_0) \cdot (x-x_0) + \frac {\partial f} {\partial u} (x_0,y_0)\cdot \left( \frac {du} {dx}(x_0) \cdot (x-x_0) + \frac {1}{2!} \frac {d^2u} {dx^2}(x_0) \cdot (x-x_0)^2 + ...\right) +$$
$$\frac {1}{2!} \left[ \frac {\partial ^2f} {\partial x^2}(x_0,y_0) \cdot (x-x_0)^2 + 2 \frac {\partial^2 f} {\partial x \partial u } (x_0,y_0) \cdot (x-x_0)\left( \frac {du} {dx}(x_0) \cdot (x-x_0) + \frac {1}{2!} \frac {d^2u} {dx^2}(x_0) \cdot (x-x_0)^2 + ...\right)+$$
$$\left. \frac {\partial ^2f} {\partial u^2}(x_0,y_0) \cdot \left( \frac {du} {dx}(x_0) \cdot (x-x_0) + \frac {1}{2!} \frac {d^2u} {dx^2}(x_0) \cdot (x-x_0)^2 + ...\right)^2 \right] + ... \ \ \ \ \ \ \ \ \ \mbox{for} \ \ |x-x_0|< min(R, R^*)$$

$$= f(x_0,y_0) \ \ + \ \ \left[ \frac {\partial f} {\partial x} (x_0,y_0)+ \frac {\partial f} {\partial u} (x_0,y_0)\cdot\frac {du} {dx}(x_0)\right]\cdot (x-x_0) \ \ +$$
$$\frac {1}{2!} \left[ \frac {\partial f} {\partial u}(x_0,y_0) \cdot\frac {d^2u} {dx^2}(x_0) + \frac {\partial ^2f} {\partial x^2}(x_0,y_0) + 2 \frac {\partial^2 f} {\partial x \partial u } (x_0,y_0) \frac {du} {dx}(x_0) + \frac {\partial ^2f} {\partial u^2}(x_0,y_0)\left(\frac {du} {dx}(x_0)\right)^2 \right] \cdot(x-x_0)^2 + ... \ \ \ \ \ \ \ \ \ \mbox{for} \ \ |x-x_0|< min(R, R^*)$$

$$= f(x_0,y_0) \ \ + \ \ \frac {df} {dx}(x_0)\cdot(x-x_0) \ \ + \ \ \frac {d^2f} {dx^2}(x_0)\cdot\frac {(x-x_0)^2}{2!} \ \ + ... \ \ \ \ \ \ \ \ \ \mbox{for} \ \ |x-x_0|< min(R, R^*)$$

(which, by construction of y(n)(x0) )

$$= \frac {dy} {dx}(x_0) + \frac {d^2y} {dx^2}(x_0) \cdot (x-x_0) + \frac {d^3y} {dx^3}(x_0) \cdot \frac {(x-x_0)^2}{2!} + ... \ \ = \ \ \frac {d} {dx} (T_{y,x_0} (x)) \ \ = \ \ u'(x)$$

that is,

$$T_{y,x_0} (x) \ = \ u(x)$$

satisfies the o.d.e. for |x-x0|< min(R,R*), and it evidently satisfies the initial condition as well, so it is a solution to the I.V.P, and by the uniqueness of the solution, it is the solution to the I.V.P: u(x)=y(x) which also means that the solution to the I.V.P. is also analytic at x0.

So, in summary, if f is analytic at (x0,y0), then provided the Taylor Series of the solution to the I.V.P. y(x) has positive radius of convergence R, it represents the solution to the I.V.P within a neighbourhood of x0, as I have (I hope:P) shown.

I still would like to know whether there is any result that can affirm that if f is analytic at (x0,y0) (ie. there exists a neighbourhood of (x0,y0), R*, in which the Taylor Series of f represents f), then the solution is analytic too in a certain neighbourhood R of x0. Is there a relationship between R* and R?

The issue of the radius of convergence remains an important aspect, I think, because if there's no guarantee that the Taylor series of the solution that one constructs actually converges, then how can one use a finite No. of terms of the series as an approximation to the solution? In general one won't be able to deduce an expression for the general term of the series to be able to determine its radius of convergence:S

Any comments/refutations/insights or the like would be welcome xD
Thank you for your time :P

#### BobbyBear

Wikipedia says if the a function is analytic then it is locally given by a convergent power series.

http://en.wikipedia.org/wiki/Analytic_function
Fanku, John :)
I've always gone by that definition of analicity (for real functions anyhow, which are what concern me). That is why I equated f(x,u) to its power series (Taylor series) under the assumption that it was analytic xD.

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