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I'm asked to decide if I should solve the system $$\dot{y}=\begin{pmatrix} -600 & 400 \\ 400 & -600 \end{pmatrix}y, \quad t\in[t_0,t_e], \quad y(t_0) = y_0$$ with either the explicit Euler method or the implicit Euler method.

Using the explicit Euler method I would get the updating scheme $$y_{n+1}=\begin{pmatrix} 1-600h & 400h \\ 400h & 1-600h \end{pmatrix}y_n$$ where the eigenvalues of the driving matrix is $$\lambda_1 = 401-600h,$$ $$\lambda_2 = -399 - 600h.$$ For the solution to be stable these need to be less than one which gives the conditions $$h\geq\frac{4}{6}=\frac{2}{3},$$ $$h\geq -\frac{4}{6}=-\frac{2}{3}.$$ The last condition doesn't say anything but the first condition seems restrictive since I can't choose $h$ as small as possible.

If I instead were to use the implicit Euler method I would get the updating scheme $$y_{n+1} = \begin{pmatrix} 1-600h & 400h \\ 400h & 1-600h \end{pmatrix}^{-1}y_n.$$ Now I can't solve for the eigenvalues of this system but I've heard the implicit Euler is unconditionally stable so it shouldn't matter.

So is the answer that I should choose implicit Euler because it is unconditionally stable or am I missing something? The order of consistency of both is $1$ so that should not matter.

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Your computation of eigenvalues is wrong. Of the system matrix in $y'=Ay$ there is one eigenvalue $-200$ with eigenvector $\pmatrix{1\\1}$ and one eigenvalue $-1000$ with eigenvector $\pmatrix{1\\-1}$. These translate into eigenvalues $1-200h$ and $1-1000h$ of the "driving matrix" for the Euler method, requiring $h<0.002$ for stability.

For the implicit method there are no step size restrictions to get stability in the method, to get into the range where the error behaves like order 1 one still will need $500h\ll 1$.

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  • $\begingroup$ Ah shit used $400$ instead of $400h$ in the eigenvalue calculation. So the conclusion is that even though the order of consistency of both methods is 1, the explicit euler will require significantly more computations, in comparison to implicit euler, to achieve that? $\endgroup$ Commented Aug 18, 2019 at 18:31
  • $\begingroup$ Yes and no. If you measure accuracy, then both methods require about the same step size, within a factor of perhaps 2, to get similar errors. If you only care about the quantitative behavior, that the solution rapidly converges to zero, then the A-stability of implicit Euler allows much larger step sizes. $\endgroup$ Commented Aug 18, 2019 at 18:49

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