I am tasked with the following exercise:
Suppose we have 2 cars, and that car 2 has experienced engine failure and is stationary at position $x_2(t) = \bar{x}$. Let $x_1(0) = > \bar{x} - d$. Introduce the variable $\widetilde{x}(t) = \bar{x} - > x_1(t)$ and derive the differential equation for $\widetilde{x}(t)$ under the assumption that $f(x) = \frac{v_{\text{max}}}{d} x$. What is the stability condition for the time step $h$ in discretization with forward Euler? Compare with the answer in a), which condition is more stringent?
$$x_1^{n+1} = x_1^n + hf \\ x_2^{n+1} = x_2^n + hg$$
We are interested in the ODE
$$\widetilde{x}^{n+1} = \widetilde{x}^n + hf \\$$ where $$f = \frac{v}{d} \cdot x_1 \quad \text{and} \quad x_1 = \bar{x} - \widetilde{x}$$ and v (=speed), d (=distance to car 2), $\bar{x}$ are constants.
My attempt so far:
$$\widetilde{x}^{n+1} = \widetilde{x}^n + h\left( -\frac{v}{d} \cdot \left( \bar{x} - \widetilde{x}^n \right) \right)$$
and then $$ -\frac{v}{d} \cdot \left( \bar{x} - \widetilde{x}^n \right) = -\frac{v}{d} \cdot \left( \frac{\bar{x}}{\widetilde{x}^n} - 1 \right) \cdot \widetilde{x}^n$$ so that we can write
$$ \widetilde{x}^{n+1} = \widetilde{x}^n \left( 1 + h\left(-\frac{v}{d} \left(\frac{\bar{x}}{\widetilde{x}^n} - 1\right)\right) \right) $$
I am trying to find an expression for which I can easily use the rule of thumb $\frac{2}{|\lambda|}$ but I fail to see how to express $\lambda$ in a simpler manner.
I can only get rid of ${\widetilde{x}^n}$ in the denominator by invoking initial conditions such that ${\widetilde{x(0)} = d}$, but even then I awkward answers, and in the attempts ive made at best I have gotten answers equal to the one they refer to in part a) (which was $h < 6$), and not a different answer which they seem to be eluding one ought to get.
Can anyone tell if I am on the right track or not? Thanks.
