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Consider the family of linear one-step methods defined by $$y_n = y_{n-1} + h(\theta f_n + (1 - \theta)f_{n-1})$$ where $0\leq \theta \leq 1$.

Relevant definition- A difference method is $A$-stable if its region of absolute stability contains the entire left half-plane of $z = h\lambda$.

Show that the methods associated with $0\leq \theta < 1/2$ cannot be $A$-stable.

Using the advice from user LutzL we have:

Let $$f_k = \lambda y_k$$ then $$y_n(1 - \theta\lambda h) = y_{n-1}(1 + (1 - \theta)\lambda h)$$ sorting the inequality $$2Re(z) < (2\theta - 1)|z|^2$$ which requires $(2\theta - 1)\geq 0$ to get the condition for $A$-stability, hence the methods associated with $0\leq \theta < 1/2$ cannot be $A$-stable.

Nick
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Wolfy
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1 Answers1

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Just insert $$ f_k=λy_k $$ to get $$ y_n·(1-θλh)=y_{n−1}·(1+(1−θ)λh) $$ and thus the stability condition $$ \left|\frac{1+(1−θ)z}{1-θz}\right|<1. $$

Lutz Lehmann
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  • So this shows that the methods cannot be $A$-stable? I can't understand why I do not see it – Wolfy Nov 29 '15 at 16:42
  • You'd have to solve it to get a condition on $z$, first is using the squared equation $1+2(1−θ)·Re(z)+(1−θ)^2·|z|^2<1-2θ·Re(z)+θ^2·|z|^2$. – Lutz Lehmann Nov 29 '15 at 16:46
  • I assume this solution is also ugly? – Wolfy Nov 29 '15 at 16:54
  • So once we insert $f_k = \lambda y_k$ to get etc... then we use the squared equation? I am not sure if I completely understanding this still. – Wolfy Nov 29 '15 at 17:02
  • No, this is a simple straightforward calculation. The next and final step is to sort the inequality out to get $2Re(z)<(2θ-1)|z|^2$ which requires $2θ-1\ge0$ to get the condition for A-stability. – Lutz Lehmann Nov 29 '15 at 17:18
  • Ok, so starting from the stability condition you present, we sort the inequality out to get $2Re(z) < (2\theta - 1)|z|^2$? – Wolfy Nov 30 '15 at 09:25
  • I edited the answer using what you did, let me know if it is correct or if I need more information to explain the steps. Thanks! – Wolfy Nov 30 '15 at 10:15
  • For here you do not need more steps, in you exam that might be different. Note that for $θ\in[0,\frac12)$ the stability region is a circle in the negative halp-plane tangent to the imaginary axis. (And for the other values of $θ$ the complement is a circle in the positive half-plane.) – Lutz Lehmann Nov 30 '15 at 10:36