Numerical differential equation handout

In common real-life situations, the differential equation that models the problem is toocomplicated to solve exactly, so the approach is taken to approximate the solution.. Once the appr

NUMERICAL SOLUTIONS FOR ORDINARY

DIFFERENTIAL EQUATIONS

E LECTRONIC VERSION OF LECTURE

Dr Lê Xuân Đại

HoChiMinh City University of Technology Faculty of Applied Science, Department of Applied Mathematics

Email: ytkadai@hcmut.edu.vn

HCMC — 2019.

1 I NITIAL -V ALUE P ROBLEMS

2 B OUNDARY -V ALUE P ROBLEMS

Differential equations are used to modelproblems in science and engineering thatinvolve the change of some variable withrespect to another.

Most of these problems require the solution

(

y0(x) = f (x,y(x)), a É x É b, y(a) = y0

(1)

y(x) at x = a.

In common real-life situations, the differential equation that models the problem is too

complicated to solve exactly, so the approach is taken to approximate the solution.

The approximation methods give more accurate results and realistic error information.

points, in the interval[a, b].

Once the approximate solution is obtained at the points, the approximate solution at other points

in the interval can be found by interpolation.

We divide [a, b] into n subinterval of equal

x0= a, x k = x0+ kh, k = 0, 1, 2, , n, x n = b. The

y(x k+1) =

y(x k ) + y0(x k )(x k+1 − x k ) + y00(ξ k)(x k+1 − x k)2

where ξ k ∈ (x k , x k+1). Since y = y(x) satisfies

GEOMETRIC MEANING OF EULER’S METHOD

tangent line to the curve, which intercepts the line

x = x k+1aty k+1 that is the approximate value of the

EXAMPLE 1.1

Using Euler’s method to approximate the

solution to the initial-value problem

With n = 10 we have

h = 2 − 0

10 = 0.2, x k = 0.2k, y0 = 0.5. Using Euler’smethod, we have

where the coefficientsA1, A2, , A n;α2 ,α3 , ,α n;β21 ,β31 , ,β n,n−1

are defined by the following method Let

Runge-Kutta methods have the high-orderlocal truncation error of the Taylor methodsbut eliminate the need to compute and

EXAMPLE 1.2

Use the Runge-Kutta method of order 4 with

h = 0.2,n = 10 to obtain approximations to the solution of the initial-value problem

Withn = 10we haveh = 2 − 0

10 = 0.2, x k = 0.2k, y0 = 0.5 Moreover,K1k = hf (x k , y k ) = 0.2(y k − 0.04k2+ 1),

In this lecture, we study how to

approximate the solution to

boundary-value problems, differentialequations with conditions imposed atdifferent points

For first-order differential equations,

only one condition is specified, so there

is no distinction between initial-valueand boundary-value problems We will

be considering second-order equationswith two boundary values

The two-point boundary-value problems inthis lecture involve a second-order

differential equation of the following formtogether with the boundary conditions

We divide the interval [a, b] into n

subinterval by the mesh points

x0= a, x k = x0+ kh, k = 1, 2, , n − 1, x n = b

n .

formula

y0(x k) ≈ y(x k+1 ) − y(x k−1)

2h

y00(x k) ≈ y(x k+1 ) − 2y(x k ) + y(x k−1)

= y k+1 − 2y k + y k−1

h2

The use of centripetal difference

formulas results in the equation

p k y k+1 − 2y k + y k−1

h2 + q k

y k+1 − y k−1 2h + r k y k = f k,

∀k = 1, 2, , n − 1 where

p k = p(x k ), q k = q(x k ), r k = r(x k) and f k = f (x k).

With the boundary conditionsy0= α, y n = β,we can the define the system of linear equations

Y = [y1, y2, , y n−1]Tand

Matrix Ais the tridiagonal matrix In order

to solve this system, we can use LU

The Doolittle method gives us

·

We can have the linear system

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