Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability (resit) / 4 / c

For $y=1+x^2$, the derivatives are $y^{\prime }=2x$ and $y^{\prime \prime }=2$. Its endpoint values are $y(0)=1$ and $y(2)=5$, so it satisfies both the ordinary differential equation and the boundary conditions. Moreover any solution of $y^{\prime \prime }=2$ has the form $x^2+Ax+B$; the boundary conditions force $B=1$ and $A=0$. Thus the unique solution of the boundary value problem is

The endpoint boundary conditions areThe finite difference equation can be written $y_{n+1}-2y_n+y_{n-1}=2h^2$. Set $d_n=y_n-y_{n-1}$. Then $d_{n+1}-d_n=2h^2$, so $d_n=d_1+2h^2(n-1)$. Summing from $1$ to $n$ givesUsing $y_N=5$ and $h=2/N$ yields $N{d}_1=4-h^{2}N(N-1)=4/N$, hence $d_1=4/N^2=h^2$. ThereforeThis also covers $N=1$, where there are no interior equations. Uniqueness follows because the difference of two solutions has zero second finite difference, hence is linear in $n$, and vanishes at both endpoints. This is an exact finite difference solution for constant second derivative: the method is exact at the nodes here because the centered second finite difference differentiates this quadratic polynomial exactly.