Question 1.9.1: Use the Euler formula to compute the numerical solution of x......

Solution First, the exact solution can be obtained by direct integration or by assuming a solution of the form $x(t)=A e^{\lambda t}$. Substitution of this assumed form into the equation $\dot{x}=-3 x$ yields $A \lambda e^{\lambda t}=-3 A e^{\lambda t }$, or $\lambda=-3$, so that the solution is of the form $x(t)=A e^{-3 t}$. Applying the initial conditions $x(0)=1$ yields $A=1$. Hence the analytical solution is simply $x(t)=e^{-3 t}$.

Next, consider a numerical solution using the Euler method suggested by equation (1.91). In this case the constant $a=-3$, so that $x_{i+1}=x_i+\Delta t\left(-3 x_i\right)$. Suppose that a very crude time step is taken (i.e., $\Delta t=0.5$ ) and the solution is formed over the interval from $t=0$ to $t=4$. Then Table 1.4 illustrates the values obtained from equation (1.91):

$x_{i+1}=x_i+\Delta t\left(a x_i\right)$     (1.91)

$\begin{aligned} & x_0=1 \\ & x_1=x_0+(0.5)(-3)\left(x_0\right)=-0.5 \\ & x_2=-0.5-(1.5)(-0.5)=0.25\\&\qquad\vdots \end{aligned}$

forms the column marked “Euler.” The column marked “Exact” is the value of $e^{-3 t}$ at the indicated elapsed time for a given index. Note that while the Euler approximation gets close to the correct final value, this value oscillates around zero while the exact value does not. This points out a possible source of error in a numerical solution. On the other hand, if $\Delta t$ is taken to be very small, the difference between the solution obtained by the Euler equation and the exact solution becomes hard to see, as Figure 1.41 illustrates. Figure 1.41 is a plot of $x(t)$ obtained via the Euler formula for $\Delta t=0.1$. Note that it looks very much like the exact solution $x(t)=e^{-3 t}$.