The following methods were described in this semester:
- Rectangular method
- Trapezoidal rule
- Simpsons rule
- Gauss-Legendre method
- Gauss-Chebyshev method
2.2.1 Rectangular method
The most straightforward way to approximate the area under a curve is to divide up the interval along the x-axis between and into a number of smaller intervals, each of the same length. For example, if we divide the interval into subintervals, then the width of each one will be given by:
The approximate area under the curve is then simply the sum of the areas of all the rectangles formed by our subintervals:
The summary approximation error for intervals with width is less than or equal to
Thus it is impossible to calculate maximum of the derivative function, we can estimate integration error like value:
2.2.2 Trapezoidal rule
Trapezoidal rule is a way to approximately calculate the definite integral. The trapezium rule works by approximating the region under the graph of the function by a trapezium and calculating its area. It follows that
To calculate this integral more accurately, one first splits the interval of integration into n smaller subintervals, and then applies the trapezium rule on each of them. One obtains the composite trapezium rule:
The summary approximation error for intervals with width is less than or equal to:
2.2.3 Simpsons rule
Simpsons rule is a method for numerical integration, the numerical approximation of definite integrals. Specifically, it is the following approximation:
If the interval of integration is in some sense "small", then Simpsons rule will provide an adequate approximation to the exact integral. By small, what we really mean is that the function being integrated is relatively smooth over the interval . For such a function, a smooth quadratic interpolant like the one used in Simpsons rule will give good results.
However, it is often the case that the function we are trying to integrate is not smooth over the interval. Typically, this means that either the function is highly oscillatory, or it lacks derivatives at certain points. In these cases, Simpsons rule may give very poor results. One common way of handling this problem is by breaking up the interval into a number of small subintervals. Simpsons rule is then applied to each subinterval, with the results being summed to produce an approximation for the integral over the entire interval. This sort of approach is termed the composite Simpsons rule.
Suppose that the interval is split up in subintervals, with n an even number. Then, the composite Simpsons rule is given by
The error committed by the composite Simpsons rule is bounded (in absolute value) by
2.2.4 Gauss-Legendre method and Gauss-Chebyshev method
Since function values are given in fixed points then just two points Gauss-Legendre method can be applied. If is continuous on , then
The Gauss-Legendre rule G2( f ) has degree of precision . If , then
It should be noted that even in case of two points method we have to calculate values of the function in related to , , this values could be evaluated by linear interpolation (because it is necessary to avoid oscillations), so estimation of integration error become very complicated process, but this error will be less or equal to trapezoidal rule.
Mechanism of Gauss-Chebyshev method is almost the same like described above, and integration error will be almost the same, so there is no reason to use such methods for the current problem.
3.1 Problem definition
It is well known that the third order Runge-Kutta method is of the following form
Suppose that you are asked to derived a new third order Runge-Kutta method in the following from
Find determine the unknowns , , and so that your scheme is a third order Runge-Kutta method.
3.2 Problem solution
Generally Runge-Kutta method looks like:
where coefficients could be calculated by the scheme:
The error function:
Coefficients , , must be found to satisfy conditions , consequently we can suppose that for each order of Runge-Kutta scheme those coefficients are determined uniquely, it means that there are no two different third order schemes with different coefficients. Now it is necessary to prove statement.
For , :
Thus we have system of equations:
Some of coefficients are already predefined:
; ; ; ; ; ; ;
Obtained results show that Runge-Kutta scheme for every order is unique.
4.1 Problem definition
Discuss the stability problem of solving the ordinary equation , via the Euler explicit scheme , say . If , how to apply your stability restriction?
4.2 Problem solution
The Euler method is 1st order accurate. Given scheme could be rewritten in form of:
If has a magnitude greater than one then will tend to grow with increasing and may eventually dominate over the required solution. Hence the Euler method is stable only if or:
For the case mentioned above inequality looks like:
Last result shows that integration step mast be less or equal to .
For the case , for the iteration method coefficient looks like
As step is positive value of the function must be less then . There are two ways to define the best value of step , the firs one is to define maximum value of function on the integration area, another way is to use different for each value , where . So integration step is strongly depends on value of .
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- Ernst Hairer, Syvert Paul Nrsett, and Gerhard Wanner. Solving ordinary differential equations I: Nonstiff problems, second edition. Berlin: Springer Verlag, 1993. ISBN 3-540-56670-8.
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