Is there solution for equation which is recursive?

Question

I have this following vector equation:$$\vec x = \vec x_0 + (1/2) * f(\vec x, t)t^2$$ where $\vec x$ has initial value when first fed into function $f$. All vectors are 3 dimensional and function $f$ returns an 3 dimensional vector. This function is intended to replace current recursive compution method, in order to get exact results instead as we cannot iterate over infinitely small numbers required for $t$.

Currently I start with $t = 0$ and I increment it by smallest possible value for each iteration. I first feed the $t = 0$ and $\vec x = \vec x_0$ ($\vec x_0$ being the starting value) into the function, and then I calculate the initial value for $f(\vec x, t)$ , and then I calculate the first output of above function. For next cycle I increment $t$ with very small constant value until it reaches the desired value in order to get $\vec x$ as function of $t$. This obviously leads into errors especially when used for bigger values of $t$.

Is there any way to reformulate this expression into solvable form, such that $\vec x$ depends on $t$, $\vec x_0$ and initial value of $\vec x$? I tried using differential equations to solve this, but I couldn't figure it out. I asked my maths teacher, but he too couldn't figure anything out. If this equation cannot be solved directly, then is there any other way to get exact result without using iteration? Any help is appreciated.

Answer 1

For $t=1$, the equation would be $$ x = x_0 + \frac{1}{2} f(x, 1) $$ This is a fixed point equation $$ x = \Phi(x) $$ which might be (depending on $f$) have zero, one or more solutions $x$.

For different $t$ values the function $\Phi$ changes, so we could write that as a familiy of equations:

$$ x = \Phi_t(x) $$

For general $f$, meaning you don't specify $f$ any further, nothing can be simplified, you will end up applying one of the methods to solve these kind of equations, like fixed-point iteration or some geometrical method.

Example

$$ f(x,t) = t \, x^2 $$ and $x_0 = 0$. So we have to solve the fixed-point equation $$ x = \frac{1}{2} x^2 $$ From the plot below

we take the solutions $x=0$ and $x = 2$ and because we know $f$ we are quite sure there are no more solutions around.

Example 2

Having higher dimensional vectors does not change the sitatuation dramatically. Let us go to 2D, because here we can still use a plot to spot solutions:

$$ f([x,y]^T,t) = t \, [x + y, x - y]^T $$

again for $t = 1$. We now have the vector equation $$ [x,y]^t = \frac{1}{2} [x+y, x-y]^T = \Phi([x,y]^T) $$

Answer 2

$\textbf{pseudo-code:}$

1. $t_n = t_{n-1} + dt$
2. find $x_n$ as the solution to $$F(x_n,t) = x_n - x_{0} - \frac{1}{2}f\left(x_n,t_n\right)t_n^2 = 0$$
3. repeat until you reach $t_N$

I apologize for the awful presentation.

To root solve you can use Newton method for multiple functions that equation to zero i.e. each component.