Finding the absolute stability

Question

I am trying to find whether the following is stable absolutely using the improved Euler and the Adams-Bashforth 2 scheme, $u'=\begin{bmatrix} -20&0&0\\ 20&-1&0\\0&1&0\end{bmatrix}u=Au$, where the timestep is $\frac{1}{2}$.

From my notes we know that for $u'=\lambda u$ is absolutely stable iff $|1+h\lambda|<1$. But I am confused as to what the `$\lambda$' is. Is it simply the eigenvalue(s) of the problem, which then confuses me if we are using other methods, or is it just a scalar? If it is the eigenvalues, then how to do you derive the absolute stability for the system using the Adams-Bashforth 2 scheme?

So far I found, for Euler, that $\lambda=-20,-1,-0.1$ which is not stable for our timestep. However for deriving the stability for the Adams-Bashforth I got as far as,

$$u^{n+1}=u^nA(1-\frac{A}{2}+\frac{A^2}{4})$$ but struggle as to how to proceed. Any advice would be appreciated.

Answer 1

In the improved Euler method, your ODE gives the step \begin{align} k_1&=Au_n\\ k_2&=A(u_n+hk_1)=(A+hA^2)u_n\\ u_{n+1}&=u_n+\frac h2 (k_1+k_2)=(I+hA+\frac{h^2}2A^2)u_n \end{align} What you want for stability is that the matrix factor is contracting if all eigenvalues of $A$ have negative part. That means you need to find $h$ so that $|1+(1+hλ)^2|<2$ for all relevant eigenvalues.

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In the most simple Adams-Bashford method $u_{n+1}=u_n+\frac h2(3Au_n-Au_{n-1})$ you actually need to solve the linear recursion problem. Considering the eigenspaces this again reduces to an equation for the eigenvalues, the characteristic equation of the recursion for an eigenvalue $λ$ is $$ q^2=q+\frac h2(3λq-λ) $$ which has roots around $q=1+hλ$ and $q=\dfrac{hλ}{2+3hλ}$, compute the exact values with the solution formula, which both need to have absolute value smaller $1$.

Answer 2

With the improved Euler method we have

$$ u_{k+1}=\left(I_3+hA + \frac h2 A^2\right)u_k = M u_k $$

this schema is contractive if $\max|\text{eigenvalues}(M)| < 1$ but calculating we get

$$ M = \left( \begin{array}{ccc} 200 h^2-20 h+1 & 0 & 0 \\ 20 h-210 h^2 & \frac{h^2}{2}-h+1 & 0 \\ 10 h^2 & h-\frac{h^2}{2} & 1 \\ \end{array} \right) $$

and $\text{eigenvalues}(M) = \left\{1,\frac{1}{2} \left(h^2-2 h+2\right),200 h^2-20 h+1\right\}$ so no stability chance. This is due mainly to the null last column in matrix $A$.