Reduction of Higher order Equations

When we are solving initial value problems, we usually need to write these as sets of first order equations, because most of the program packages require this.

Example: \( y''(x)+y(x)=0,\ y(0)=a_0,\ y'(0)=b_0 \)

We may for instance write this equation in a system as follows, $$ \begin{align*} y'(x)=&g(x)\\ g'(x)=&-y(x)\\ y(0)=&a_0,\ g(0)=b_0 \end{align*} $$

Another example: $$ \begin{align*} &y'''(x)+2y''(x)-(y'(x))^2+2y(x)=x^2\\ &y(0)=a_0,\ y'(0)=b_0,\ y''(0)=c_0 \end{align*} $$ We set \( y'(x)=g(x) \) and \( y''(x)=g'(x)=f(x) \), and the system may be written as $$ \begin{align*} y'(x)=&g(x)\\ g'(x)=&f(x)\\ f'(x)=&-2f(x)+(g(x))^2-2y(x)+x^2 \end{align*} $$ with initial values \( y(0)=a_0,\ g(0)=b_0,\ f(0)=c_0 \).

This is fair enough for hand calculations, men when we use program packages a more systematic procedure is needed. Let's use the equation above as an example.

We start by renaming \( y \) to \( y_1 \). We then get the following procedure: $$ \begin{align*} y&'=y'_1=y_2\\ y&''=y''_1=y'_2=y_3 \end{align*} $$ The system may then be written as $$ \begin{align*} y'_1(x)=&y_2(x)\\ y'_2(x)=&y_3(x)\\ y'_3(x)=&-2y_3(x)+(y_2(x))^2-2y_1(x)+x^2 \end{align*} $$ with initial conditions \( y_1(0)=a_0,\ y_2(0)=b_0,\ y_3(0)=c_0 \).

The general procedure to reduce a higher order ODE to a system of first order ODEs becomes the following:

Given the equation $$ \begin{align} \tag{10} &y^{(m)}=f(x,y,y',y'',\dots,y^{(m-1)})\\ &y(x_0)=a_1,\ y'(x_0)=a_2,\ \dots,y^{(m-1)}(x_0)=a_m \nonumber\\ \end{align} $$ where $$ \begin{align*} y^{(m)}\equiv \frac{d^my}{dx^m} \end{align*} $$ with \( y=y_1 \), we set $$ \begin{align} &y'_1=y_2 \nonumber \\ \nonumber &y'_2=y_3\\ \nonumber &.\\ \nonumber &.\\ \nonumber &y'_{m-1}=y_m\\\nonumber\\\nonumber &y_1(x_0)=a_1,y_2(x_0)=a_2,\dots,y_m(x_0)=a_m \tag{11} \end{align} $$

Example 1: Reduction of higher order systems

Write the following ODE as a system of first order ODEs: $$ \begin{align*} y'''-y'y''-(y')^2+2y=x^3 \\ y(0)=a,\ y'(0)=b,\ y''(0)=c \end{align*} $$

First we write \( y'''=y'y''+(y')^2-2y+x^3 \).

By use of (11) we get $$ \begin{align*} &y_1'=y_2\\ &y_2'=y_3\\ &y_3'=y_2y_3+(y_2)^2-2y_1+x^3\\ &y_1(0)=a,\ y_2(0)=b,\ y_3=c \end{align*} $$