# Finite difference discretization for systems of higher ODEs

• Ojo Saheed
If you're trying to solve this numerically, one way it is often done is to convert to a set of first order ODEs. In this particular case, there would be 12 first order ODEs to integrate. Define,zj=yj (j = 1,3)zj+3=y'j (j = 1,3)zj+6=y''j (j=1,3)zj+9=y'''j (j=1,3)Thenz1' = z4z4' = z7z7'f

#### Ojo Saheed

How can I use finite difference to discretize a system of fourth order differential equations?
for example:
y(4)+5y(3)-2y''+3y'-y=0

This doesn't look like a system of 4th order ODEs. It looks like a single 4th order ODE.

Chet

This doesn't look like a system of 4th order ODEs. It looks like a single 4th order ODE.

Chet
Thanks for the reply Chet. Actually, I had thought that given an example of one the system of equations might suffice for a demonstration. In any case, a better representation of the system goes thus:
y1(4)=a1y1''+a2y2''+a3y3''+a4y1+a5y2+a6y3+a7
y2(4)=b1y1''+b2y2''+b3y3''+b4y1+b5y2+b6y3+b7
y3(4)=c1y1''+c2y2''+c3y3''+c4y1+c5y2+c6y3+c7
where a, b and c terms are constant coefficients

Thanks for the reply Chet. Actually, I had thought that given an example of one the system of equations might suffice for a demonstration. In any case, a better representation of the system goes thus:
y1(4)=a1y1''+a2y2''+a3y3''+a4y1+a5y2+a6y3+a7
y2(4)=b1y1''+b2y2''+b3y3''+b4y1+b5y2+b6y3+b7
y3(4)=c1y1''+c2y2''+c3y3''+c4y1+c5y2+c6y3+c7
where a, b and c terms are constant coefficients
If you're trying to solve this numerically, one way it is often done is to convert to a set of first order ODEs. In this particular case, there would be 12 first order ODEs to integrate.

Define,
zj=yj (j = 1,3)
zj+3=y'j (j = 1,3)
zj+6=y''j (j=1,3)
zj+9=y'''j (j=1,3)
Then
z1' = z4
z4' = z7
z7' = z10
z10' = a1z7+a2z8+a3z9+a4z1+a5z2+a6z3+a7
etc.

You end up with 12 coupled first order linear ODEs explicit in the derivatives. You can use an automatic integrator to solve them or use your own coding of Runge Kutta or forward Euler, or backward Euler, or whatever.

Chet