A full SQP step is rejected by the L1 merit test and repaired by SOC

At $x=(c,s)$, $h(x)=0$, $Dh(x)=2(c,s)$, and $\mathrm{\nabla }f(x)=4(c,s)-(1,0)$. The unit tangent $t=(-s,c)$ satisfies $t^T\mathrm{\nabla }f(x)=s$. The QP constraint gives $p\in \mathrm{span}(t)$; minimizing $\mathrm{\nabla }f(x{)}^{T}p+\frac{1}{2}\mathrm{\parallel }p{\mathrm{\parallel }}_2^2$ on this line gives $p=-st=(s^2,-sc)$. Its QP multiplier is $-(4-c)\mathrm{/}2$, since $p+\mathrm{\nabla }f(x)=(4-c)(c,s)$.

Here $Dh(x)Dh(x{)}^T=4$, so [F3] gives $w=(-c{s}^2\mathrm{/}2,-s^3\mathrm{/}2)$ and $\mathrm{\parallel }w{\mathrm{\parallel }}_2=s^2\mathrm{/}2$. Direct substitution into the quadratic constraint yields $h(x+p+w)=\frac{s^4}{4}\mathrm{.}$ The corrected first coordinate is $c+s^2(1-c\mathrm{/}2)$, so the corrected objective change is $-s^2(1-c\mathrm{/}2)+s^4\mathrm{/}2$. Adding the penalty contribution gives ${\psi }_{\rho }(x+p+w)-{\psi }_{\rho }(x)=-s^2(1-c\mathrm{/}2)+s^4(1\mathrm{/}2+\rho \mathrm{/}4)\mathrm{.}$

Since $x^{T}p=0$ and $\mathrm{\parallel }p{\mathrm{\parallel }}_2^2=s^2$, $h(x+p)=s^2$. Also $\mathrm{\nabla }f(x{)}^{T}p=-s^2$, so [F2] gives $D_{\rho }=s^2$. The objective changes by $s^2$, while the equality violation changes from zero to $s^2$; hence the full-step merit change is $(1+\rho )s^2>0$, which fails the Armijo right side $-\sigma s^2<0$.

Divide the corrected merit change plus $\sigma D_{\rho }$ by $s^2$. With $\sigma =1\mathrm{/}4$, the result is $-1+c\mathrm{/}2+s^2(1\mathrm{/}2+\rho \mathrm{/}4)+1\mathrm{/}4$, which tends to $-1\mathrm{/}4$ as $\theta \to 0$. It is therefore negative for all sufficiently small nonzero $\theta$, proving the SOC Armijo test passes. The explicit $O(s^2)$ correction norm and $O(s^4)$ residual satisfy the quadratic-size and cubic-residual bounds of [F4].