Past exam of the numerical analysis course of the University of Oxford / 2021 / 3 / b

For $f(x,y)=-x{y}^2$, the explicit Euler method evaluates the right-hand side at the known point $(x_n,y_n)$. ThereforeThis update requires no root finding.

The Backward Euler method uses the new point:Writing $a=h(x_n+h)>0$ gives the quadratic equation $a{y}_{n+1}^2+y_{n+1}-y_n=0$. Starting from $y_n>0$, its two polynomial roots have opposite signs. The positive root, which continues the solution as $h\to 0$, isThe other root is negative and does not approach $y_n$ as $h\to 0$. Rationalizing the numerator gives an equivalent form that avoids cancellation error when the step size is small:The defining quadratic equation gives $0<y_{n+1}<y_n$ for this root, so positivity is maintained inductively from $y_0=1$. The branch choice is part of specifying the Backward Euler method here: the algebraic update alone has two solutions.

With $f(x,y)=-x{y}^2-y^5$, write $t=x_{n+1}=x_n+h$ and $z=y_{n+1}$. The Backward Euler method requiresor the monic quintic polynomial equationConstruct its $5\times 5$ companion matrixIts characteristic polynomial is $\det (zI-C_n)=p(z)$. Thus the polynomial roots needed for this update are exactly the eigenvalues of $C_n$.

At each step, assemble $C_n$, compute its eigenvalues, for example with the QR algorithm, and choose its unique positive real eigenvalue as $y_{n+1}$. To justify this selection, letIf $y_n>0$, then $F(0)=-y_n<0$, $F(y_n)=h{y}_n^5+ht{y}_n^2>0$, andThe intermediate value theorem and strict monotonicity prove there is exactly one positive polynomial root, located in $(0,y_n)$. Starting at $y_0=1$ therefore makes every update well defined, positive and decreasing. In numerical arithmetic, select the nearly real eigenvalue in this interval and check the polynomial residual $|F(z)|$; a final Newton root-finding iteration can refine it. This reduces each update to the requested eigenvalue problem.