Conversion of System of Eq's to 2nd Order Diff Eq

[bizoid]

My question is in regards to converting a system of differential equations into a higher order differential equation. I am an undergrad taking diff eq and have just learned the wonders of Euler's method of solving 2nd order differential equations with constant coefficients. It is significantly faster then using eigenvalues and eigenvectors.

My question is:

Can I always convert a system of (2) 1st order differential equations into (1) 2nd order differential equation?

If not, what attributes of the system allow me to identify that it cannot be converted?

Can I always convert a system of n-order differential equations into a single n-order differential equation? (I do not know why I would want to do this since Euler's method would not help in this instance, but you may as well generalize your answer if possible.)

Thank you in advance for any feedback.
Bizoid

 

[lurflurf]

the following

1) system of n first order differential equation
2) nth order differential equation
3) system of n differential equations of any order

This is even true without assuming constant coefficents

Furthermore euler's equation and its generalizations and variants (enhanced euler/Runge Kutta/adams) can be used on any of those even though for various reasons the system of n first order is the most common approach.

As far as faster than eigenvector methods that depends of what you mean by faster/ exact method used/ particular equation. Also eigen vector approach gives one a better idea of the solutions behavior.

To specifics
we will only consider constant coefficients
let p be a polynomial D differentiations
lowercase scalar uppercase tensor (vector or matrix)
p(D)y=f(x) nth order equation
DY=F(X)
P(D)Y=F(X)
DY=F(x)

say we have
(D^2+1)y=0
ie
y''+y=0
let u=y v=y'
(D^2+1)y=0
becomes
v'=-u
u'=v
the argument is reversable in particular
v'=-u
u'=v
implys a matrix form
AY=DY
from matrix theory we know
p(A)=0 for some polynomial p
thus
p(D)u=0
p(D)v=0
or
p(D)Y=0
can be solved but the soultions must be checked in
AY=DY
because that approach produces extraneous solutions
that is if
p(D)Y=0
then
p(D)BY=0
but
ABY is only zero for some B's

similar facts hold for more complicated systems like
(D^2-3D+2)u+(D^2-8D^2)v=e^11t
(D^3-6D+7)u+(D^4+7D^3)v=cos(4t)

 

[bizoid]

Thank you for the reply.

In regards to your comment, "Also eigen vector approach gives one a better idea of the solutions behavior."

Please correct me if I am missing something, but I found the following (on 2x2 systems) to be easier then using eigen value / vector approach. Especially for repeated, complex, and zero eigen valules. Yet I still have the same qualitative analysis capabilities.

1) Convert to 2nd order DE
2) Use Euler's to find eigen values
3) differentiate to identify the eigen vector (for qualitative analysis) or solve the IVP.

After writing the aforementioned, I see that I am in fact finding the eigen values / vectors. However, I am not using the Linear Algebra approach. My preferred method may be due to my weakness in Linear Algebra.

Any thoughts?

Thank you again for your comments.

 

[lurflurf]

Please describe the Euler method of which you speak, I am not sure we are speaking of the same one.
say as before we have
u'=v
v'=-u
eigenvalues are -i,i
instead of solving this we see
A^2+I=0
u,v each satisfy
(D^2+1)y
y=C1*cos(t)+C2*sin(t)
while u and v are of this form we have 4 constants instead of 2 and must eliminate 2
u=C1*cos(t)+C2*sin(t)
v=C3*cos(t)+C4*sin(t)

It seems to me work equivalent to finding eigenvectors needs to be done to eliminate extra constants

 

[HallsofIvy]

Yes, solving the "characteristic equation" so you can just plug the numbers into the general solutions is faster than solving for eigenvalues (which is the SAME as solving the characteristic equation) and then solving for the eigenvectors. The crucial point is that finding the eigenvectors gives you MORE information than just finding the solution to the higher order equation. You would still need to differentiate to find the OTHER functions in the system of equations.

Also, as lurflurf said, determining the eigenvalues and eigenvectors gives you a better understanding of the general behavior of solutions. That becomes VERY important in dealing with nonlinear equations which cannot, in general, be solved. I rather suspect that the next subject in your d.e. course will be applying those ideas to non-linear systems of equations.