Past exam of the numerical solution of differential equations course of the University of Oxford / 2010 / A / 1

Substitute a smooth exact solution into the linear multistep method and expand about $x_n$. Its residual isConsistency of a numerical method requires the coefficients of $y$ and $h{y}^{\prime }$ to vanish. Therefore $1+a+b=0$ and $3+a=1$, giving the unique choiceFor this choice the next coefficient is $9/2-1-2=3/2\ne 0$, so the residual is $\frac{3}{2}{h}^2{y}^{\prime \prime }+O(h^3)$. The normalized local truncation error, obtained by dividing the residual by $h$, is $O(h)$. Hence the method has order one.

A zero-stable has a uniformly bounded response to small perturbations of its starting values and accumulated residuals on a fixed time interval as $h\to 0$. Equivalently, its first characteristic polynomial satisfies the root condition for a multistep method: every root has modulus at most one, and each root on the unit circle is simple.

For the consistent parameters, the recurrence for $y^{\prime }=0$ is $y_{n+3}-2y_{n+1}+y_n=0$, with characteristic polynomialLet $r=(-1-\sqrt{5})/2$. Choose $y_1=r$ and $y_2=r^2$; together with $y_0=1$ these generate exactly $y_n=r^n$. Since $|r|=(1+\sqrt{5})/2>3/2$,This parasitic mode violates the root condition and grows without a bound independent of the number of steps. Scaling its starting amplitudes gives the same unbounded amplification for arbitrarily small initial perturbations. Thus the only consistent method in the family is not zero-stable.

A method has numerical convergence if it approximates the exact solution uniformly at mesh points on every fixed finite interval as $h\to 0$, whenever its starting approximations converge to the corresponding exact starting values, for sufficiently smooth solutions of a well-posed initial-value problem.

The Dahlquist equivalence theorem states that a fixed-coefficient linear multistep method is convergent if and only if it is consistent and zero-stable, under the usual Lipschitz condition on the differential equation and convergent starting data. In this family numerical consistency forces the parameters found in part (a), and part (b) proves that choice is not zero-stable. All other parameter choices are inconsistent. Therefore no member of the family is convergent.