Solving the Initial Value Problem of Ordinary
Differential Equations by Polynomials
Maitree Podisuk and Chinda Chaichuay
Department of Mathematics and Computer Science
King Mongkut’s Institute of Technology Chaokhuntaharn Ladkrabang
Ladkrabang Bangkok 10520
THAILAND
Abstract: In this paper, we will use the polynomials of degree 10 to approximate the solution of the initial value problem of the ordinary differential equations. We will use them to find the numerical solutions of some examples and compare these results with the known 2 points, 3points and 4 points Runge-Kutta method.
Key-Words: Polynomial Runge-Kutta Iintial-Value-Problem Ordinary Differential-Equation
1 Introduction
The initial value problem of the ordinary differential equation may be written in the form
(1)
with the initial condition
(2) .
If we divide the close interval into n subintervals at the points , where for then some known numerical formulas for finding the numerical solution of the above equations at these points are;
(3)
where
which is 2 points Runge-Kutta formula which we shall denote by RK2,
(4)
where
which is 3 points Runge-Kutta formula which we shall denote by RK3,
(5)
where
which is 4 points Runge-Kutta formula which we shall denote by RK4.
2Problem Formulation
At the present, the computer is very powerful. Thus we should use this advantage to find the numerical solution of the initial value problem of the ordinary differential equations. We shall call the method of our new idea as “Polynomial Approximation Method” which we shall denote by “PA”. This idea is that for each interval , we will construct a polynomial of degree k,
(6)
to be an approximated polynomial for the unknown function of (1)-(2). We have
(7) .
Now substitute and in (1) we obtain
(8)
Put the value in (6) we get
(9) .
Divide the interval in to k subintervals at the points with equal distance where . Put the values in (8) we get k equations. Thus we have the total of k+1 equations and k+1 unknowns. We use the Gauss Elimination Method to solve for the value of , then put these values in (6) and use it to find the approximated value of . We continue this process to approximate all values of the function at the points .
3Examples
There will be 3 examples in this section. We will use the above polynomial approximation method with and the above three formulas of the Runge-Kutta method to find the numerical solutions of these 3 examples.
3.1Example 1
Find the numerical solution of the equation
(10)
with the initial condition
(11)
The analytical solution of the above equation is
.
The 10 exact values at the points 2.1, 2.2,…, 3.0 are in table 1.
The numerical results of RK2, RK3, RK4 and the polynomial approximation method are in the following table 2, 3, 4, and 5 for , and table 6, 7, 8 and 9 for
x / y(x)2.1 / 0.92714269117
2.2 / 0.86205666957
2.3 / 0.80342706706
2.4 / 0.75021647205
2.5 / 0.70159948464
2.6 / 0.65691448474
2.7 / 0.61562761392
2.8 / 0.57730554900
2.9 / 0.54159668830
3.0 / 0.50820507334
Table 1
RK2x / calculated y / error
3.0 / /
Table 2
RK3x / calculated y / error
3.0 / /
Table 3
RK4x / calculated y / error
3.0 / /
Table 4
PAX / calculated y / error
3.0 / /
Table 5
RK2X / Calculated y / error
/ 0.5084380476 /
Table 6
RK3X / Calculated y / error
/ 0.5082006498 /
Table 7
RK4x / calculated y / error
/ 0.5082051700 /
Table 8
PAx / calculated y / error
/ 0.5082050731 /
Table 9
3.2Example 2
Find the numerical solution of the equation
(12)
with the initial condition
(13)
The analytical solution of the above equation is
.
The 10 exact values at the points 1.1, 1.2,…, 2.0 are in table 10.
The numerical results of RK2, RK3, RK4 and the polynomial approximation method are in the following table 11, 12, 13, and 14 for , and table 15, 16, 17 and 18 for
x / y(x)1.1 / 0.1048419779
1.2 / 0.2187858682
1.3 / 0.3410735438
1.4 / 0.4710611313
1.5 / 0.6081976622
1.6 / 0.7520058068
1.7 / 0.9020680268
1.8 / 1.0580159968
1.9 / 1.2195223837
2.0 / 1.3862943611
Table 10
RK2X / calculated y / error
2.0 / /
Table 11
RK3X / calculated y / error
2.0 / /
Table 12
RK4X / calculated y / error
2.0 / /
Table 13
PAx / calculated y / error
2.0 / /
Table 14
RK2x / calculated y / error
Table 15
RK3x / calculated y / error
Table 16
RK4X / calculated y / error
Table 17
PAX / calculated y / error
Table 18
3.3Example 3
Find the numerical solution of the equation
(13)
with the initial condition
(14)
The analytical solution of the above equation is
.
The 10 exact values at the points 0.2, 0.4,…, 2.0 are in table 19.
The numerical results of RK2, RK3, RK4 and the polynomial approximation method are in the following table 20, 21, 22, and 23 for , and table 24, 25, 26 and 27 for .
x / y(x)0.2 / 0.0455992472
0.4 / 0.2068823619
0.6 / 0.5249861257
0.8 / 1.0459560438
1.0 / 1.8186613473
1.2 / 2.8914086293
1.4 / 4.3064449601
1.6 / 6.0955466472
1.8 / 8.2659274270
2.0 / 10.793782018
Table 19
RK2x / calculated y / error
0.4 / /
0.8 / /
1.2 / /
1.6 / /
2.0 / /
Table 20
RK3x / calculated y / error
0.4 / /
0.8 / /
1.2 / /
1.6 / /
2.0 / /
Table 21
RK4x / calculated y / error
0.4 / /
0.8 / /
1.2 / /
1.6 / /
2.0 / /
Table 22
PAX / calculated y / error
0.4 / /
0.8 / /
1.2 / /
1.6 / /
2.0 / /
Table 23
RK2X / calculated y / error
0.2 / /
0.4 / /
0.6 / /
0.8 / /
1.0 / /
1.2 / /
1.4 / /
1.6 / /
1.8 / /
/ / 1.4040772088
Table 24
RK3X / calculated y / error
0.2 / /
0.4 / /
0.6 / /
0.8 / /
1.0 / /
1.2 / /
1.4 / /
1.6 / /
1.8 / /
2.0 / /
Table 25
RK4x / calculated y / error
Table 26
PAx / calculated y / error
Table 27
4Conclusion
The polynomial approximation method works quite well when we compare to the classical known Runge-Kutta method. However in this paper we use the polynomial of degree 10 for each step. If we do not need very close approximation then we may reduce the degree of polynomial to be less than 10. In the other hand if we need more accuracy then we must use the polynomial of degree greater than 10 which will have no problem with most computer.
Reference:
[1] S.D. Conte and C. de Boor, Elementary Numerical Analysis second edition, McGraw-Hill Book Company, 1972.