I have shown that for the given ODE system, that when we apply the forward Euler method to something like

$\begin{array}{rl}{\mathbf{y}}^{\prime}& =A\mathbf{y}\\ \mathbf{y}({t}_{0})& ={y}_{0}\\ t& \in ({t}_{0},T]\end{array}$

where $A$ is diagonalizable; $A=PD{P}^{-1}$, that the solution given by forward Euler is

$\begin{array}{r}{u}_{n}=P(I+hD{)}^{n}{P}^{-1}{u}_{0}\end{array}$

where we consider the Forward Euler method to be ${y}_{k+1}={y}_{k}+h{y}^{\prime}({x}_{k},{y}_{k}).$ N

Now, I want to show that if the real parts of ${\lambda}_{k}$ (the eigenvalues of $A$ and thus the entries of $D$) are negative, then $h<\frac{-2Re{\lambda}_{k}}{|{\lambda}_{k}{|}^{2}}$ for all $k$ implies that ${u}_{n}\to 0$.

Now, in order to show this, based on what we have obtained for the formula for ${u}_{n}$, we need only show that for each $|1+h{\lambda}_{k}|<1$. I am having a really tough time doing this, and have mostly just tried to play around with AM-GM stuff.

$\begin{array}{rl}{\mathbf{y}}^{\prime}& =A\mathbf{y}\\ \mathbf{y}({t}_{0})& ={y}_{0}\\ t& \in ({t}_{0},T]\end{array}$

where $A$ is diagonalizable; $A=PD{P}^{-1}$, that the solution given by forward Euler is

$\begin{array}{r}{u}_{n}=P(I+hD{)}^{n}{P}^{-1}{u}_{0}\end{array}$

where we consider the Forward Euler method to be ${y}_{k+1}={y}_{k}+h{y}^{\prime}({x}_{k},{y}_{k}).$ N

Now, I want to show that if the real parts of ${\lambda}_{k}$ (the eigenvalues of $A$ and thus the entries of $D$) are negative, then $h<\frac{-2Re{\lambda}_{k}}{|{\lambda}_{k}{|}^{2}}$ for all $k$ implies that ${u}_{n}\to 0$.

Now, in order to show this, based on what we have obtained for the formula for ${u}_{n}$, we need only show that for each $|1+h{\lambda}_{k}|<1$. I am having a really tough time doing this, and have mostly just tried to play around with AM-GM stuff.