Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 341 / 1

Insert the exact solution and expand about $t=t_{n+2}$. Since $f(y)=y^{\prime }$ and $f^{\prime }(y)f(y)=y^{\prime \prime }$, the left side isThe coefficient of $h^j{y}^{(j)}$ on the right side isfor $j\ge 1$, with the last term absent for $j=0$. These coefficients agree through $j=3$; at $j=4$ the left side minus the right side is $1/10$. Thus the local truncation error is $h^4{y}^{(4)}/10+O(h^5)$ and the method has order three.

At $h=0$ the first characteristic polynomial isIts roots satisfy the root condition for a multistep method: the only unit-modulus root is the simple root $1$, and the other root has modulus $1/5$. The method is consistent because part a gives positive order. The Dahlquist equivalence theorem, in the form allowed by the question for this derivative-augmented method, therefore proves convergence.

For the test equation $y^{\prime }=\lambda y$, put $z=h\lambda$. The amplification roots satisfyThe quadratic Schur stability criterion shows that both roots lie in the closed unit disk whenTo check these inequalities on the closed left half-plane, write $z=-u+iv$ with $u\ge 0$. The right side of the second inequality isAfter squaring, the difference between its square and $16|5\overline{D}+1+z{|}^2$ is a polynomial in $u$ and $v^2$ with nonnegative coefficients:The first inequality follows from the same displayed positive expression. The unit-circle cases are semisimple, so the entire closed left half-plane is in the linear stability domain. Hence the method is A-stable.