System of First-Order ODES

[evinda]

Hello! (Wave)

I am looking at the following exercise:

Consider the initial value problem

$\left\{\begin{matrix}
x''(t)=x(t)\\
x(0)=a\\
x'(0)=b
\end{matrix}\right.$

Write it as a system of First-Order ODES with suitable initial values and show that Euler method can get unstable for a great step $(h)$.

That is the solution that the assistant of the prof gave us:
$$y_1=x \Rightarrow y_1'=x'=y_2 | y_1(0)=x(0)=a \\ y_2=x' \Rightarrow y_2'=x''=y_1 | y_2(0)=x'(0)=b $$

$$\binom{y_1}{y_2}'=\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix} \binom{y_1}{y_2} \text{ with } \binom{y_1(0)}{y_2(0)}=\binom{a}{b}$$Euler method:

$$\binom{y_1^{n+1}}{y_2^{n+1}}=\binom{y_1^n}{y_2^n}+h \begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix} \binom{y_1^n}{y_2^n}=\binom{y_1^n+hy_2^n}{y_2^n+hy_1^n} \\ \binom{y_1^{n+1}}{y_2^{n+1}}= \begin{pmatrix}
1 & h\\
h & 1
\end{pmatrix} \binom{y_1^n}{y_2^n}$$$$b=-a$$

The exact solution of the initial value problem is ($t^n=nh$)

$\binom{y_1(t)}{y_2(t)}=e^t \binom{a}{-a} \Rightarrow \binom{y_1(t^n)}{y_2(t^n)}=e^{-t^n} \binom{a}{-a}=e^{-nh} \binom{a}{-a}$, $h$ constant, $n \to +\infty, y \to 0$.

$y=e^{-t}$Euler method

$\binom{y_1^1}{y_2^1}=\begin{pmatrix}
1 & h\\
h & 1
\end{pmatrix} \binom{y_1^0}{y_2^0}=\begin{pmatrix}
1 & h\\
h & 1
\end{pmatrix} \binom{a}{-a}=(1-h) \binom{a}{-a}$

$\binom{y_1^2}{y_2^2}=\begin{pmatrix}
1 & h\\
h & 1
\end{pmatrix} \binom{y_1^1}{y_2^1}=(1-h)^2 \binom{a}{-a}$

$\dots \dots \dots$

$\binom{y_1^n}{y_2^n}=(1-h)^n \binom{a}{-a}$

$|1-h|>1 \Rightarrow h>2$.

First of all, how do we find that $\binom{y_1^1}{y_2^1}=\begin{pmatrix}
1 & h\\
h & 1
\end{pmatrix} \binom{y_1^0}{y_2^0}=\begin{pmatrix}
1 & h\\
h & 1
\end{pmatrix} \binom{a}{-a}=(1-h) \binom{a}{-a},\binom{y_1^2}{y_2^2}=\begin{pmatrix}
1 & h\\
h & 1
\end{pmatrix} \binom{y_1^1}{y_2^1}=(1-h)^2 \binom{a}{-a}$, $\dots $,$\binom{y_1^n}{y_2^n}=(1-h)^n \binom{a}{-a}$?

I found the following:$\binom{y_1^1}{y_2^1}=\begin{pmatrix}
1 & h\\
h & 1
\end{pmatrix} \binom{a}{b}, \binom{y_1^2}{y_2^2}=\begin{pmatrix}
1+h^2 & 2h\\
2h & h^2+1
\end{pmatrix} \binom{a}{b}, \binom{y_1^3}{y_2^3}=\begin{pmatrix}
1+3h^2 & h(h^2+3)\\
h(h^2+3) & 1+3h^2
\end{pmatrix} \binom{a}{b}$

Am I wrong? (Worried) If not, how could we find the general formula? Also, don't we find the real solution of the system of the First-Order ODES as follows?$\begin{vmatrix}
-\lambda & 1\\
1& -\lambda
\end{vmatrix}=0 \Rightarrow \lambda^2=1 \Rightarrow \lambda=\pm 1$.

Now we are looking for the eigenvectors.

For $\lambda=1$:

$\begin{pmatrix}
-1 & 1 \\
1 & -1
\end{pmatrix} \binom{u}{w}=\binom{0}{0} \Rightarrow \left\{\begin{matrix}
-u+w=0\\
u-w=0
\end{matrix}\right. \Rightarrow \left\{\begin{matrix}
w=u\\
u=w
\end{matrix}\right. \overset{\text{ we set } u=1}{\Rightarrow } u=w=1$

For $\lambda=-1$:

$\begin{pmatrix}
1 & 1 \\
1 & 1
\end{pmatrix} \binom{u}{w}=\binom{0}{0} \Rightarrow u+w=0 \Rightarrow u=-w \overset{\text{ we set w=-1}}{\Rightarrow } u=1=-w$So the solution is of the form:

$\binom{y_1}{y_2}=c_1 \binom{1}{1} e^{t}+ c_2 \binom{-1}{1} e^{-t}$

Using the initial conditions, I got the following:

$\binom{y_1}{y_2}= \frac{a+b}{2} \binom{1}{1} e^t+ \frac{b-a}{2} \binom{-1}{1}e^{-t}$How could we show that for a great step $h$ the method can get unstable? (Thinking)

 

[jvicens]

Hello, thank you for bringing this exercise to our attention. I can provide some insights on the questions you have raised.

Firstly, to answer your question on how we find the general formula for $\binom{y_1^n}{y_2^n}$, we can use the power iteration method. This method involves repeatedly multiplying a vector by a matrix and taking the limit as the number of iterations approaches infinity. In this case, the matrix is $\begin{pmatrix}
1 & h\\
h & 1
\end{pmatrix}$ and the vector is $\binom{a}{b}$. This method allows us to find the general formula as $(1-h)^n\binom{a}{b}$.

Moving on to your question on finding the real solution of the system of First-Order ODES, the method you have used is correct. However, there is a slight error in your calculation for the eigenvector corresponding to $\lambda=-1$. It should be $\binom{1}{-1}$ instead of $\binom{-1}{1}$. This will result in the correct solution that you have mentioned.

To show that for a great step $h$ the Euler method can get unstable, we can use the concept of stability of numerical methods. A numerical method is considered stable if small changes in the initial conditions or parameters do not result in large changes in the solution. In this case, for a great step $h$, the value of $(1-h)^n$ will become very small and close to 0. This means that even a small error in the initial conditions or parameters will result in a large error in the solution. This shows that the Euler method can get unstable for a great step $h$.

I hope this helps answer your questions. Please let me know if you have any further doubts or concerns. I am always happy to help clarify any scientific concepts. Good luck with your exercise!