 # The linear system y′=Ay,y(0)=y_0, where A is a symmetric matrix, is solved by the Euler method. Let e_n=y_n−y(nh) Where y_n denotes the Euler approximation and y(nh) the exact solution (h is Euler step size). Prove that ||e_n||_2=||y_0||_2max|(1+h lambda)^n−e^(nh lambda)| Where lambda in sigma(A) where sigma(A) is the set of eigenvalues of A. asigurato7 2022-07-22 Answered
The linear system ${y}^{\prime }=Ay,y\left(0\right)={y}_{0}$, where $A$ is a symmetric matrix, is solved by the Euler method.
Let ${e}_{n}={y}_{n}-y\left(nh\right)$
Where ${y}_{n}$ denotes the Euler approximation and $y\left(nh\right)$ the exact solution ($h$ is Euler step size).
Prove that $||{e}_{n}|{|}_{2}=||{y}_{0}|{|}_{2}max|\left(1+h\lambda {\right)}^{n}-{e}^{nh\lambda }|$
Where $\lambda \in \sigma \left(A\right)$ where $\sigma \left(A\right)$ is the set of eigenvalues of $A$.
I have tried various approaches such as writing ${e}_{n}$ as the error bound of the Euler method and taking the norm, but I can't seem to get $||{y}_{0}|{|}_{2}$ in my answers.
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Given that A is symmetric it can be diagonalised. Therefore we can write down the solution of the system of ODEs directly, and simply subtract the discreteized solution front the exact.