Given the differential equation $t'=g(x,t)$
Use the improved Euler method by analysing the local error and its stability for the equation $t'=\delta t$, where $\delta$ is a complex
number.For $t_{n+1}=t_{n-1}+2kg(x_n,t_n)$, analyze its truncation error. Also, similar to part $1$, study its stability for $t'=\phi t$, where
this time $\phi$ is a real number.
How will I go about solving this question? Any help or hints will be greatly appreciated.
I know for a differential equation problem $y'(t)=f(t,y(t))$ I will use the approximation of the derivative $$y'(t)=\frac{y(t+h)-y(t)}{h}$$ for some small $h$. Then if I let $t_j=t_0+jh$ then $$\frac{y(t_{j+1})-y(t_j)}{h}=f(t_j,y(t_j))$$ leading to $y(t_{j+1})=y(t_j)+hf(t_j,t(t_j))$, which is Euler's method for an ordinary differential equation.
Best Answer
For the regular Euler method, as you say, you have $y(t_{j+1})=y(t_j)+hf(t_j,t(t_j))$ If you expand $y(t)$ in a Taylor series around $t_0$, you have $y(t)=y(t_0)+(t-t_0)y'(t_0)+\frac 12 y''(t-t_0)^2+\ldots$. Euler's method will give you $y(t_{j+1})=y(t_j)+hy'(t_j)$. For small steps the error is the lowest order term, $\frac {h^2}2y''(t_0)$
For stability for $y'=\delta t$, this will give $y(t_{j+1})=y(t_j)+(t_{j+1}-t_j)\delta y(t_j)=y(t_j)(1+\delta(t_{j+1}-t_j))$. If all the steps are the same size, $h$, this becomes $y(t_{j+1})=y(t_j)(1+\delta h)$ with solution $y(t_0+n\Delta t)=y(t_0)(1+\delta h)^n$ If $\delta$ is negative, the solution should decay to zero, but if $h \gt \frac 2{\delta}$ this blows up.
Can you follow this through for the improved Euler method?