Order of global truncation error

Question

I was reading something,here is something I stuck with the following integral is approximated using numerical methods \begin{align} \int_{0}^{s(t)} v\,dx\ = \int_{0}^{kh} v\,dx\ + \int_{kh}^{s(t)} v\,dx\ \end{align} where $$ k=\lfloor s(t)/h \rfloor,$$ this integral is approximated using composite trapezoidal and interpolation, the first integral on the R.H.S. is solved using composite trapezoidal rule and the second is solved using interpolation. Here is the approximation of second one \begin{align} \int_{kh}^{s(t)} v\,dx\ = \left (\delta-\frac{\delta^2 }{2h}\right)v_k ~~+ ~~ \left(\frac{\delta^2 }{2h}\right)v_{k+1} \end{align} $$ \delta = s(t)-kh $$ I have two doubts here, the first one is, I did not get this approximation using interpolation and second doubt is that this complete approximation is given with global truncation error of $O(h^2),~~$ this is also not clear to me. I know composite trapezoidal rule has truncation error of order $O(h^2)$ but for the interpolation part it is not clear to me. Please help me out with this and please correct me if I have said anything wrong. Thanks.

Answer 1

This looks to be a part of a procedure where the values $v_k$ are some fixed, pre-computed table of function values. Essentially you integrate the piecewise linear approximation of $v$. For that you split the input $s(t)=kh+δ$. On each interval you approximate $$ v(s)\approx v_j+\frac{s-jh}{h}(v_{j+1}-v_j)\text{ for } jh\le s<(j+1)h $$ For the full intervals the integal results in the trapezoidal formula $\frac h2(v_j+v_{j+1})$, for the last partial interval $$ \int_{kh}^{kh+δ}v_k+\frac{s-kh}{h}(v_{k+1}-v_k)ds=\Bigl[sv_k+\frac{(s-kh)^2}{2h}(v_{k+1}-v_k)\Bigr]_{kh}^{kh+δ}v_k=δv_k+\frac{δ^2}{2h}(v_{j+1}-v_j) $$ which is your last interpolation formula.

The distance of the piecewise linear interpolation to the original function is $O(h^2)$ if the function $v$ is smooth enough, thus in the integral the error is of the same size, more specifically $O(s(t)h^2)$.