Mixed approximation vs. full approximation for a power series expansion

[FranzS]

TL;DR

Mixed approximation vs. full approximation for a power series expansion: which works better?

Hi PF,

I'm trying to find an approximate solution of a differential equation that can't be solved in exact form.
The differential equation is of the form:
$$
f'(t)=g(f(t),t)
$$
I want to find the approximate solution in terms of a power series:
$$
f(t) \approx f(0) + f'(0) \cdot t + \frac{f''(0)}{2} \cdot t^2
$$
$f(0)$ is known, as is everything else in order to find $f'(0)$ directly from the differential equation.
The problem arises when calculating $f''(0)$, because $g(f(t),t)$ includes a square root term and its derivative will diverge for $t=0$.
Now, I found a good approximation for the square root term that has no divergence problems when differentiated.
My question is: for the power series expansion of $f(t)$, should I use the term $f'(0)$ that was calculated with the exact form of $g(f(t),t)$ even though the term $f''(0)$ comes from the approximation of $g(f(t),t)$ or should I use the latter to find a new $f'(0)$ as well? How would those two options differ numerically and conceptually?

Thanks.

 

[anuttarasammyak]

What actually is $g(f(t),t)$ in your case ?

 

[anuttarasammyak]

$$ f''=\frac{\partial g}{\partial f} f'+ \frac{\partial g}{\partial t} =\frac{\partial g}{\partial f} g+ \frac{\partial g}{\partial t} $$
Thus
$$ f''(0)=\frac{\partial g}{\partial f} |_{t=0}\ g(f(0),0)+ \frac{\partial g}{\partial t}|_{t=0} $$

If some term here diverges at t=0, why don't you try expansion around some $t \neq 0$ where no terms diverge ?

 

[FranzS]

What actually is $g(f(t),t)$ in your case ?

Oh well, it's pretty intricated. I'll write it by steps:
$$
\begin{align}
& f'(t)= - f(t) \cdot \
\frac{V_1'(t) + c_0 F(t)}{V_1(t)}
\\[0.5em]
& F(t) = 2 \sqrt{x(t) \big( 1-x(t) \big)}
\\[0.5em]
& x(t) = 1 - \frac{h(t)}{f(t)}
\\[0.5em]
& h(t) = \frac{c_1 c_2 - f(t) V_1(t)}{V_2(t)}
\\[0.5em]
& V_1(t) = c_3 - c_4 t
\\[0.5em]
& V_2(t) = c_5-c_6 t
\end{align}
$$

 

[anuttarasammyak]

I observe no divergences at t=0 in post #3. Denominators $V_1=0, V_2=0$ would matter.

Say $f(0)$ and function form $g(f(t),t)$ are given, we may think of t evolution by continuing iteration,i.e.,
$$f(\triangle t) \approx f(0)+g(f(0),0) \triangle t
$$
$$f(2\triangle t) \approx f(\triangle t)+g(f(\triangle t),\triangle t) \triangle t$$
$$f(3\triangle t) \approx f(2\triangle t) + g(f(2\triangle t),2\triangle t) \triangle t$$
...
$$f(N\triangle t) \approx f((N-1)\triangle t) + g(f((N-1)\triangle t),(N-1)\triangle t) \triangle t$$

A numerical solution chart by ChatGPT for $c_0=c_1=c_2=c_4=1,\ c_3=5,\ c_5=4,\ c_6=0.5,\ f(0)=1$.

 

[renormalize]

Oh well, it's pretty intricated. I'll write it by steps:
$$
\begin{align}
& f'(t)= - f(t) \cdot \
\frac{V_1'(t) + c_0 F(t)}{V_1(t)}
\\[0.5em]
& F(t) = 2 \sqrt{x(t) \big( 1-x(t) \big)}
\\[0.5em]
& x(t) = 1 - \frac{h(t)}{f(t)}
\\[0.5em]
& h(t) = \frac{c_1 c_2 - f(t) V_1(t)}{V_2(t)}
\\[0.5em]
& V_1(t) = c_3 - c_4 t
\\[0.5em]
& V_2(t) = c_5-c_6 t
\end{align}
$$

I don't see a function "$g$" defined anywhere in those lines.