Question

Local versus Global Error. In deriving formula (4) for Euler’s method, a rectangle was used to approximate the area under a curve (see Figure 3.14). With

\({\bf{g(t) = f(t,f(t))}}\), this approximation can be written as \(\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt}} \approx {\bf{hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \)where \({\bf{h = }}{{\bf{x}}_{{\bf{n + 1}}}}{\bf{ - }}{{\bf{x}}_{\bf{n}}}\) .

1. Show that if g has a continuous derivative that is bounded in absolute value by B, then the rectangle approximation has error\(\left( {\bf{O}} \right){{\bf{h}}^{\bf{2}}}\); that is, for some constant M, \(\left| {\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt - hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} } \right| \le {\bf{M}}{{\bf{h}}^{\bf{2}}}\).This is called the local truncation error of the scheme. [Hint: Write \(\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt - hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} {\bf{ = }}\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {\left[ {{\bf{g(t)dt - g(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \right]{\bf{dt}}} \). Next, using the mean value theorem, show that\(\left| {{\bf{g(t)dt - g(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \right| \le {\bf{B}}\left| {{\bf{t - }}{{\bf{x}}_{\bf{n}}}} \right|\) . Then integrate to obtain the error bound\(\left( {\frac{{\bf{B}}}{{\bf{2}}}} \right){{\bf{h}}^{\bf{2}}}\).]
2. In applying Euler’s method, local truncation errors occur in each step of the process and are propagated throughout the further computations. Show that the sum of the local truncation errors in part (a) that arise after n steps is (O)h. This is the global error, which is the same as the[ss1] [m2] convergence rate of Euler’s method.

Short answer

1. The required result gets by the mean value theorem i.e., \(\left| {\int\limits_{{x_n}}^{{x_{n + 1}}} {g\,\left( t \right)\,dt - hg\,\left( {{x_n}} \right)} } \right| \le M\,{h^2}\).
2. The sum of the local truncation error arises after n steps i.e., \(O\left( h \right)\).

Step-by-step solution

Important hint.

For the solution apply mean value theorem.

Find the integral value.

Here \(\int\limits_{{x_n}}^{{x_{n + 1}}} {g\left( t \right)dt \approx hg\left( {{x_n}} \right)} \) where \(h = {x_{n + 1}} - {x_n}\)

Assume that g has a continuous derivative that is bounded by B.

\(\left| {g'\left( y \right)} \right| \le B\).

According to the mean value theorem

\(\begin{array}{c}\frac{{g\left( t \right) - g\left( {{x_n}} \right)}}{{t - {x_n}}} = g'\left( y \right)\\g\left( t \right) - g\left( {{x_n}} \right) = g'\left( y \right).\left( {t - {x_n}} \right)\\\left| {g\left( t \right) - g\left( {{x_n}} \right)} \right| \le B\left| {t - {x_n}} \right|\end{array}\)

Apply mode on an integral part.

Now,

\(\begin{array}{c}\left| {\int\limits_{{x_n}}^{{x_{n + 1}}} {g\left( t \right)dt - hg\left( {{x_n}} \right)} } \right| = \left| {\int\limits_{{x_n}}^{{x_{n + 1}}} {\left[ {g\left( t \right)dt - g\left( {{x_n}} \right)} \right]dt} } \right|\\ \le \int\limits_{{x_n}}^{{x_{n + 1}}} {\left| {g\left( t \right)dt - g\left( {{x_n}} \right)} \right|dt} \\ \le \int\limits_{{x_n}}^{{x_{n + 1}}} {B\left| {t - {x_n}} \right|dt} \\ = \left[ {\frac{{B\left( {t - {x_n}} \right)\left| {t - {x_n}} \right|}}{2}} \right]_{{x_n}}^{{x_{n + 1}}}\end{array}\)

\(\begin{array}{c} = - \left[ {\frac{{B\left( {{x_n} - {x_{n + 1}}} \right)\left| {{x_n} - {x_{n + 1}}} \right|}}{2}} \right]\\ = \frac{{B{h^2}}}{2}\end{array}\)

Let \(M = \frac{B}{2}\) then

\(\left| {\int\limits_{{x_n}}^{{x_{n + 1}}} {g\left( t \right)dt - hg\left( {{x_n}} \right)} } \right| \le M{h^2}\)

use the concept of part (a).

Let\(a = {x_n},b = {x_{n + 1}}\)

From part (a) the local truncation error of the scheme is \(\frac{{\left( B \right){h^2}}}{2}\) and \(h = \frac{{b - a}}{n}\).

Finding the result apply the result on n steps.

After n steps the result is

\(\begin{array}{c} = h\frac{{B\left( {b - a} \right)}}{2}\\ = O\left( h \right)\end{array}\)

Hence, the sum of the local truncation error arises after n steps i.e., \(O\left( h \right)\).