Numerical Analysis/ODE in vector form Exercises

< Numerical Analysis

All of the standard methods for solving ordinary differential equations are intended for first order equations. When you need to solve a higher order differential equation, you first convert it to a system of first order of equations. Then you rewrite as a vector form and solve this ODE using a standard method. On this page we demonstrate how to convert to a system of equations and then apply standard methods in vector form.

Reduction to a first order system

(Based on Reduction of Order and Converting a general higher order equation.)

I want to show how to convert higher order differential equation to a system of the first order differential equation. Any differential equation of order n of the form

f\left(t, u, u', u'',\ \cdots,\ u^{(n-1)}\right) = u^{(n)}

can be written as a system of n first-order differential equations by defining a new family of unknown functions

y_i = u^{(i-1)}\quad\text{for}\quad i = 1, 2,... n\,.

The n-dimensional system of first-order coupled differential equations is then

\begin{array}{rclcl} y_1&=&u\\ y_2&=&u'\\ y_3&=&u''\\ &\vdots&\\ y_n&=&u^{(n-1)}.\\ \end{array}

Differentiating both sides yields

\begin{array}{rclclcl} y_1'&=&u' &=&y_2\\ y_2'&=&u''&=&y_3\\ y_3'&=&u'''&=&y_4\\ &\vdots&\\ y_{n}'&=&u^{(n)}&=&f(t,y_1,\cdots,y_n).\\ \end{array}

We can express this more compactly in vector form

\mathbf{y}'=\mathbf{f}(t,\mathbf{y})

where $\ y_{i+1} = f_i\left(t, \mathbf{y} \right)$ for $i < n$ and $\ f_n\left(t, \mathbf{y} \right)$ = $\ f\left(t, y_1, y_2, \cdots,y_n \right)\,.$

Exercise

Consider the second order differential equation $\ u''+u =0$ with initial conditions $\ u{(0)}=1$ and $\ u'{(0)}=0$ . We will use two steps with step size $\ h = \frac{\pi}{8}$ and approximate the values of $\ u{(\frac{\pi}{4})}$ and $\ u'{(\frac{\pi}{4})}.$

Since the exact solution is $u(t)=\cos(t)$ we have $\ u{(\frac{\pi}{4})}=0.707106781$ and $u'{(\frac{\pi}{4})}=-0.707106781$ .

Exercise 1: Convert this second order differential equation to a system of first order equations.

Solution:

We have second order differential equation $\ u''{(t)}= -u{(t)}$ with initial conditions $\ u{(0)}=1$ and $\ u'{(0)}=0$ .

Let $y_1 = u$ and $y_2 = u'$ . Differentiating $y_1$ and $y_2$ gives

\ y_1' = u' = y_2

\ y_2' = u'' = -u =-y_1.

Thus we have a system of first order equation in vector form

\left[\begin{array}{c} y_1' \\ y_2' \end{array}\right] = \left[\begin{array}{c} y_2 \\ -y_1 \end{array} \right] = f\left(t, \left[\begin{array}{c} y_1 \\ y_2 \end{array} \right] \right)

Exercise 2: Apply the Euler method twice.

Solution:

By the Euler's method, $y_{n+1} = y_n + h f\left(t_n , y_n \right)$ .

I. First apply to get $y_1$

y_1 = y_0 + h f\left(t_0 , y_0 \right)

Now we convert $\ y_1$ as a vector form, where $\ t_0 = 0 ,h=\frac{\pi}{8}$ and $\left[\begin{array}{c} y_{0,1} \\ y_{o,2} \end{array}\right] = \left[\begin{array}{c} 1 \\ 0 \end{array} \right]$ , then we have

\begin{align} \left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array}\right] & = \left[\begin{array}{c} y_{0,1} \\ y_{0,2} \end{array} \right] + h f\left(t_0, \left[\begin{array}{c} y_{0,1} \\ y_{0,2} \end{array}\right]\right) \\ & = \left[\begin{array}{c} y_{0,1} \\ y_{0,2} \end{array} \right] + h\left[\begin{array}{c} y_{0,2} \\ -y_{0,1} \end{array}\right] \\ & = \left[\begin{array}{c} 1 \\ 0 \end{array} \right] + h \left[\begin{array}{c} 0 \\ -1 \end{array}\right] \\ & = \left[\begin{array}{c} {1+0} \\{ 0-h} \end{array} \right] \\ & = \left[\begin{array}{c} 1 \\ {-h} \end{array} \right] \\ & = \left[\begin{array}{c} 1 \\ {- \frac{\pi}{8}} \end{array} \right] \\ & = \left[\begin{array}{c} 1 \\ {-0.392699082} \end{array} \right]\,. \end{align}

2. Second apply to get $\ y_2$

y_2 = y_1 + h f\left(t_1 , y_1 \right)

Similarly, we convert $y_2$ as a vector form, where $t_1 = t_0 + h =0+h =h$ , where $h=\frac{\pi}{8}$ and $\left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array}\right] = \left[\begin{array}{c} 1 \\ -h \end{array} \right]$ , then we have

\begin{align} \left[\begin{array}{c} y_{2,1} \\ y_{2,2} \end{array}\right] & = \left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array} \right] + h f\left(t_1, \left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array}\right]\right) \\ & = \left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array} \right] + h\left[\begin{array}{c} y_{2,1} \\ -y_{1,1} \end{array}\right] \\ & = \left[\begin{array}{c} 1 \\ {-h} \end{array} \right] + h \left[\begin{array}{c} { -h} \\ { -1} \end{array}\right] \\ & = \left[\begin{array}{c} {1-h^2} \\{ -h-h} \end{array} \right] \\ & = \left[\begin{array}{c} {1-h^2} \\ -2h \end{array} \right] \\ & = \left[\begin{array}{c} {1 - \frac{\pi^2}{8^2}}\\ {- 2 \frac{\pi}{8}} \end{array} \right] \\ & = \left[\begin{array}{c} 0.8457874321 \\ {-0.785398163} \end{array} \right]\,. \end{align}

Exercise 3: Apply the Backward Euler method twice.

Solution:

By the Backward Euler's method, $\ y_{n+1} = y_n + h f\left(t_{n+1} , y_{n+1} \right).$

I. First apply to get $\ y_1$

\ y_1 = y_0 + h f\left(t_1 , y_1 \right)

Now we convert $y_1$ as a vector form, where $t_1=t_0+h=0+h=h$ , where $h=\frac{\pi}{8}$ and $\left[\begin{array}{c} y_{0,1} \\ y_{o,2} \end{array}\right] = \left[\begin{array}{c} 1 \\ 0 \end{array} \right]$ , then we have

\begin{align} \left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array}\right] & = \left[\begin{array}{c} y_{0,1} \\ y_{0,2} \end{array} \right] + h f\left(t_1, \left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array}\right]\right) \\ & = \left[\begin{array}{c} y_{0,1} \\ y_{0,2} \end{array} \right] + h\left[\begin{array}{c} y_{1,2} \\ -y_{1,1} \end{array}\right] \\ & = \left[\begin{array}{c} 1 \\ 0 \end{array} \right] + \left[\begin{array}{c} {hy_{1,2}} \\ { -hy_{1,1}} \end{array}\right] \\ & = \left[\begin{array}{c} {1+ hy_{1,2} } \\ { -hy_{1,1} } \end{array} \right]\,. \end{align}

Now we have to solve a system of linear equations

\ \left\{ \begin{array}{l l} y_{1,1}=1+ hy_{1,2}& \quad \\ y_{1,2}= -hy_{1,1} \quad \\ \end{array} \right.

That is,

\ \left\{ \begin{array}{l l} y_{1,1} -hy_{1,2}=1& \quad \\ hy_{1,1}+ y_{1,2}= 0 \quad \\ \end{array} \right.

Set up augmented matrix to solve this system,

\left( \begin{array}{cc|c} 1 & -h & 1 \\ h & 1 & 0 \\ \end{array} \right) \Rightarrow \left( \begin{array}{cc|c} 1 & -h & 1 \\ 0 & 1+h^2 & -h \\ \end{array} \right)

Thus, the solution is $\ y_{1,1}=\frac{1}{1+h^2}, y_{1,2}=-\frac{h}{1+h^2}.$

Plugging in $\ h=\frac{\pi}{8}$ , then we have

\left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array}\right] = \left[\begin{array}{c} 0.8457874321 \\ {-0.785398163} \end{array} \right]\,.

2. Second apply to get $\ y_2$

\ y_2 = y_1 + h f\left(t_2 , y_2 \right)

Similarly, we convert $\ y_2$ as a vector form, with $\ t_2 = t_1 + h =h+h =2h$ , where $\ h=\frac{\pi}{8}$ and $\left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array}\right] = \left[\begin{array}{c} {\frac{1}{1+h^2}} \\ {-\frac{h}{1+h^2} } \end{array} \right]$ , then we have

\begin{align} \left[\begin{array}{c} y_{2,1} \\ y_{2,2} \end{array}\right] & = \left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array} \right] + h f\left(t_2, \left[\begin{array}{c} y_{2,1} \\ y_{2,2} \end{array}\right]\right) \\ & = \left[\begin{array}{c} {\frac{1}{1+h^2}} \\ {-\frac{h}{1+h^2}} \end{array} \right] + h f \left(2h, \left[\begin{array}{c} y_{2,1} \\ y_{2,2} \end{array}\right]\right) \\ & = \left[\begin{array}{c} {\frac{1}{1+h^2}} \\ {-\frac{h}{1+h^2}} \end{array} \right] + h \left[\begin{array}{c} y_{2,2} \\ {- y_{2,1} } \end{array}\right] \\ & = \left[\begin{array}{c} {\frac{1}{1+h^2} + hy_{2,2} } \\ {-\frac{h}{1+h^2} - hy_{2,1} } \end{array} \right]\,. \end{align}

Now we have to solve a system of linear equations

\ \left\{ \begin{array}{l l} y_{2,1}= \frac{1}{1+h^2} + hy_{2,2}& \quad \\ y_{2,2}= -\frac{h}{1+h^2} - hy_{2,1} \quad \\ \end{array} \right.

That is,

\ \left\{ \begin{array}{l l} y_{2,1} -hy_{2,2}=\frac{1}{1+h^2}& \quad \\ hy_{2,1}+ y_{2,2}= -\frac{h}{1+h^2} \quad \\ \end{array} \right.

Set up augmented matrix to solve this system,

\left( \begin{array}{cc|c} 1 & -h & \frac{1 }{1+h^2}\\ h & 1 & -\frac{h}{1+h^2} \\ \end{array} \right) \Rightarrow \left( \begin{array}{cc|c} 1 & -h & \frac{1}{1+h^2} \\ 0 & 1+h^2 & -\frac{2h}{1+h^2} \\ \end{array} \right)

Thus, the solution is $\ y_{2,1}=\frac{1-h^2}{(1+h^2)^2}, y_{2,2}=-\frac{2h}{(1+h^2)^2}.$

Plugging in $\ h=\frac{\pi}{8}$ , then we have

\left[\begin{array}{c} y_{2,1} \\ y_{2,2} \end{array}\right] = \left[\begin{array}{c} 0.63487705 \\ {-0.589546795} \end{array} \right]\,.

Exercise 4: Apply the Midpoint method twice.

Solution:

By the Midpoint method, $y_{n+1} = y_n + h f\left(t_n +\frac{1}{2} h , y_n + \frac{1}{2} h f\left(t_n, y_n\right)\right).$

I. First apply to get $y_1$

y_1 = y_0 + h f\left(t_0 +\frac{1}{2} h , y_0 + \frac{1}{2} h f\left(t_0, y_0\right)\right).

Now we convert $\ y_1$ as a vector form, where $\ t_0 = 0 ,h=\frac{\pi}{8}$ and $\left[\begin{array}{c} y_{0,1} \\ y_{0,2} \end{array}\right] = \left[\begin{array}{c} 1 \\ 0 \end{array} \right]$ , then we have

\begin{align} \left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array}\right] &= \left[\begin{array}{c} y_{0,1} \\ y_{0,2} \end{array} \right] + h f\left(t_0 +\frac{1}{2} h , \left[\begin{array}{c} y_{0,1} \\ y_{0,2} \end{array} \right] + \frac{1}{2} h f\left(t_0, \left[\begin{array}{c} y_{0,1} \\ y_{0,2} \end{array} \right] \right)\right) \\ &= \left[\begin{array}{c} y_{0,1} \\ y_{0,2} \end{array} \right] + h f\left(t_0 +\frac{1}{2} h , \left[\begin{array}{c} y_{0,1} \\ y_{0,2} \end{array} \right] + \frac{1}{2} h \left[\begin{array}{c} y_{0,2} \\ {-y_{0,1}} \end{array} \right] \right) \\ &= \left[\begin{array}{c} 1 \\ 0 \end{array} \right] + h f\left(0 +\frac{1}{2} h , \left[\begin{array}{c} 1 \\ 0 \end{array} \right] + \frac{1}{2} h \left[\begin{array}{c} 0 \\ {-1} \end{array} \right] \right) \\ &= \left[\begin{array}{c} 1 \\ 0 \end{array} \right] + h \left[\begin{array}{c} {-\frac{1}{2} h} \\ {-1} \end{array} \right] \\ &= \left[\begin{array}{c} { 1 -\frac{1}{2} h^2}\\{-h} \end{array} \right]\,. \end{align}

Plugging in $\ h=\frac{\pi}{8}$ , then we have

\left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array}\right] = \left[\begin{array}{c} 0.922893716 \\ {-0.392699082} \end{array} \right]\,.

2. Second apply to get $y_2$

y_2 = y_1 + h f\left(t_1 +\frac{1}{2} h , y_1 + \frac{1}{2} h f\left(t_1, y_1\right)\right).

Now we convert $\ y_1$ as a vector form, where $\ t_1=t_0+h = 0+h=h ,h=\frac{\pi}{8}$ and $\left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array}\right] = \left[\begin{array}{c} {1-\frac{1}{2}h^2} \\ {-h} \end{array} \right]$ , then we have

\begin{align} \left[\begin{array}{c} y_{2,1} \\ y_{2,2} \end{array}\right] &= \left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array} \right] + h f\left(t_1 +\frac{1}{2} h , \left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array} \right] + \frac{1}{2} h f\left(t_1, \left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array} \right] \right)\right) \\ &= \left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array} \right] + h f\left(h +\frac{1}{2} h , \left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array} \right] + \frac{1}{2} h \left[\begin{array}{c} y_{1,2} \\ {-y_{1,1}} \end{array} \right] \right) \\ &= \left[\begin{array}{c} {1-\frac{1}{2}h^2} \\ {-h} \end{array} \right] + h f\left(\frac{3}{2} h , \left[\begin{array}{c} {1-\frac{1}{2}h^2} \\ {-h} \end{array} \right] + \frac{1}{2} h \left[\begin{array}{c} {-h} \\ {-1+\frac{1}{2}h^2} \end{array} \right] \right) \\ &= \left[\begin{array}{c} {1-\frac{1}{2}h^2} \\ {-h} \end{array} \right] + h f \left( \frac{3}{2} h , \left[\begin{array}{c} {1- \frac{1}{2} h^2 - \frac{1}{2} h^2} \\ {-h- \frac{1}{2}h + \frac{1}{4}h^3} \end{array} \right] \right) \\ &= \left[\begin{array}{c} {1-\frac{1}{2}h^2} \\ {-h} \end{array} \right] + h f\left( \frac{3}{2} h , \left[\begin{array}{c} {1- h^2} \\ {- \frac{2}{3}h+ \frac{1}{4} h^3} \end{array} \right] \right) \\ &= \left[\begin{array}{c} {1- \frac{1}{2} h^2} \\ {-h} \end{array} \right] + h \left[\begin{array}{c} {- \frac{2}{3} h+ \frac{1}{4} h^3} \\ {-1+ h^2} \end{array} \right] \\ &= \left[\begin{array}{c} {1-\frac{1}{2}h^2} \\ {-h} \end{array} \right] + \left[\begin{array}{c} {-\frac{2}{3}h^2+\frac{1}{4}h^4} \\ {-h+h^3} \end{array} \right] \\ &= \left[\begin{array}{c} {1- \frac{1}{2}h^2- \frac{2}{3}h^2+ \frac{1}{4}h^4} \\ {-h-h+h^3} \end{array} \right] \\ &= \left[\begin{array}{c} {1- 2h^2+ \frac{1}{4}h^4} \\ {-2h+h^3} \end{array} \right]\,. \end{align}

Plugging in $\ h=\frac{\pi}{8}$ , then we have

\left[\begin{array}{c} y_{2,1} \\ y_{2,2} \end{array}\right] = \left[\begin{array}{c} 0.694547552011673 \\ {-0.7248390292562377} \end{array} \right]\,.

Exercise 5: Using the values from the Midpoint method at $\ t=h$ in exercise3, apply the Two-step Adams-Bashforth method once.

Solution:

By the Midpoint method, we found

\left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array}\right] =\left[\begin{array}{c} {1- \frac{1}{2}h^2} \\ {-h} \end{array} \right]\,.

Now apply the Two-step Adams-Bashforth method when $\ n=0$ , then by formula

\ y_2 = y_1 + \frac{3}{2}h f\left(t_1 , y_1\right) - \frac{1}{2}h f\left(t_0 , y_0\right).

Vector form is following by

\begin{align} \left[\begin{array}{c} y_{2,1} \\ y_{2,2} \end{array}\right] & = \left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array} \right] + \frac{3}{2}h f \left(t_1, \left[\begin{array}{c} y_{1,1} \\ y_{1,2} \end{array}\right]\right) - \frac{1}{2}h f \left( t_0, \left[\begin{array}{c} y_{0,1} \\ y_{0,2} \end{array}\right]\right) \\ & = \left[\begin{array}{c} {1- \frac{1}{2}h^2 }\\ {-h} \end{array} \right] + \frac{3}{2}h f \left(t_1, \left[\begin{array}{c} {1- \frac{1}{2}h^2 }\\ {-h} \end{array}\right]\right) - \frac{1}{2}h f \left( t_0, \left[\begin{array}{c} 1 \\ 0 \end{array}\right]\right) \\ & = \left[\begin{array}{c} {1- \frac{1}{2}h^2} \\ {-h} \end{array} \right] + \frac{3}{2} h \left[\begin{array}{c} {-h} \\ {-1+ \frac{1}{2}h^2} \end{array}\right] - \frac{1}{2}h \left[\begin{array}{c} 0 \\ {-1} \end{array}\right] \\ & = \left[\begin{array}{c} {1-2h^2 } \\ {-2h+ \frac{3}{4}h^3 } \end{array} \right]\,. \end{align}

Plugging in $h=\frac{\pi}{8},$

\left[\begin{array}{c} y_{2,1} \\ y_{2,2} \end{array}\right] = \left[\begin{array}{c} 0.691574862 \\ {-0.739978812} \end{array} \right]\,.

Reference

http://en.wikipedia.org/wiki/Ordinary_differential_equation

http://www.math.ohiou.edu/courses/math3600/lecture29.pdf

http://www.ohio.edu/people/mohlenka/20131/4600-5600/hw7.pdf

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Numerical Analysis/ODE in vector form Exercises

Reduction to a first order system

Exercise

Exercise 1: Convert this second order differential equation to a system of first order equations.

Exercise 2: Apply the Euler method twice.

I. First apply to get

2. Second apply to get

Exercise 3: Apply the Backward Euler method twice.

I. First apply to get

2. Second apply to get

Exercise 4: Apply the Midpoint method twice.

I. First apply to get

2. Second apply to get

Exercise 5: Using the values from the Midpoint method at in exercise3, apply the Two-step Adams-Bashforth method once.

Reference

I. First apply to get $y_1$

2. Second apply to get $\ y_2$

I. First apply to get $\ y_1$

2. Second apply to get $\ y_2$

I. First apply to get $y_1$

2. Second apply to get $y_2$

Exercise 5: Using the values from the Midpoint method at $\ t=h$ in exercise3, apply the Two-step Adams-Bashforth method once.