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$$\begin{matrix} x'=-y \\ y'=-x' \end{matrix} \begin{matrix} x(0)=1\\ y(0)=0 \end{matrix}$$ Calculate with the explicit Euler's method the expression $A=(x_n)^2+(y_n)^2$, where $x_n, y_n$ are the approximations at the step $n$.

The problem can be written as $$\binom{x}{y}'=\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\binom{x}{y} $$

Applying the explicit Euler's method we get $$\binom{x_{n+1}}{y_{n+1}}=\binom{x_n}{y_n}+h\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\binom{x_n}{y_n}=\begin{pmatrix} 1 & -h\\ -h & 1 \end{pmatrix}\binom{x_n}{y_n} $$

So we have $$ (x_{n+1})^2+(y_{n+1})^2=(x_n-hy_n)^2+(-hx_n+y_n)^2=(1+h^2)[(x_n)^2+(y_n)^2]-4hx_ny_n$$

How can we find a general formula for $(x_n)^2+(y_n)^2$ ?

For $n=0$ we have $(x_1)^2+(y_1)^2=1+h^2$.

For $n=1$ we have $(x_2)^2+(y_2)^2=(1+h^2)^2-4xhx_1y_1$.

evinda
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1 Answers1

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Still using matrix: $$\begin{pmatrix}x_{n+1}\\y_{n+1}\end{pmatrix}=\begin{pmatrix} 1 & -h\\ -h & 1 \end{pmatrix}^{n+1}\binom{x_0}{y_0}$$

You can find the eigenvalues of this matrix. There are two real ones. Then find its diagonalization to find the power. That would give you an explicit expression.

KittyL
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  • I found that $\binom{x^n}{y^n}= \binom{\frac{1}{2} (1+h)^n+ \frac{1}{2} (1-h)^n}{-\frac{1}{2} (1+h)^n+ \frac{1}{2}(1-h)^n}$. $$$$ So I found $\binom{x^2}{y^2}=\binom{5}{-4}$. Am I right? – evinda Jun 19 '15 at 10:17
  • @evinda: My second component is $\frac{1}{2} (1+h)^n- \frac{1}{2}(1-h)^n$. – KittyL Jun 19 '15 at 10:23
  • @evinda: Also, did you mean $x_n,y_n$ and $x_2,y_2$? – KittyL Jun 19 '15 at 10:28
  • Isn't it the same, with the only difference that we took other coefficients when finding the igenvectors? Yes, that's what I meant... $$$$

    I will take a look again at my calculations.

    $$$$ Could you maybe also take a look at my other question?

    http://math.stackexchange.com/questions/1331318/smallest-possible-number-of-steps

     – evinda Jun 19 '15 at 10:30
  • @evinda: I don't think they are the same, because they give different answers. And from my calculation, $x_2=1+h^2, y_2=2h$. How did you get $5, -4$?? – KittyL Jun 19 '15 at 10:43
  • Could you maybe take a look at my question: http://math.stackexchange.com/questions/1505467/show-upper-bound ? – evinda Oct 31 '15 at 18:03