From 2 1st order ODEs to 1 2nd order ODE?

[pauperrimo]

Hi,
it is well known that a second order ODe can be transformed into a system of two ODEs through the transformation u=y', v= y.
Is the other way round possible? I mean, I have a system of 2 ODEs and want to transform it into a sucession on higher order problems that can be solved one after the other. Is this possible?
thanx a lot to anyone who can provide some insight!

 

[Defennder]

Why would you want to do that? How would you solve the higher-order ODE then?

 

[HallsofIvy]

Actually, Defennder, that was how I first learned to solve systems of equations!

For example, if we have the system x'= x- y, y'= x+ 2y, we can differentiate the first equation again to get x"= x'- y'= x'+ x+ 2y. That still has a "y" in it but from the first equation y= x- x' so we have x"= x'+ x+ 2(x- x')= 3x- x' or x"+ x'- 3x= 0.That's a second order, linear equation with constant coefficients so it can be easily solved. We can then determine y from y= x- x'. You will note that the "characteristic" equation, r²+ r- 3= 0 is exactly the same as the equation for the eigenvalues for the system of equations.

 

[Defennder]

That's odd. I thought the reason why 2nd order homogeneous linear ODEs could be solved was because they could be reduced to a system of first order ODEs. At least that's how my lecturer explained why that technique works for 2nd order ODEs.

 

[HallsofIvy]

Certainly the theory, existence and uniqueness of solutions for initial value problems, can be simplest done in terms of systems of first order equations. But the standard way of deriving the solution for a higher order linear, homogeneous ODE with constant coefficients, at least when I was learning it (in years "BC"- "before calculators") was to try y= e^rx as a solution and derive the characteristic equation from that. Of course it can be done either way. I would think that changing to a system of equations and using the matrix form would be just a little more "sophisticated".

 

[pauperrimo]

Actually, Defennder, that was how I first learned to solve systems of equations!

For example, if we have the system x'= x- y, y'= x+ 2y, we can differentiate the first equation again to get x"= x'- y'= x'+ x+ 2y. That still has a "y" in it but from the first equation y= x- x' so we have x"= x'+ x+ 2(x- x')= 3x- x' or x"+ x'- 3x= 0.That's a second order, linear equation with constant coefficients so it can be easily solved. We can then determine y from y= x- x'. You will note that the "characteristic" equation, r²+ r- 3= 0 is exactly the same as the equation for the eigenvalues for the system of equations.

Thanks for the input! However, what if y and y' cannot be isolated? Is there any general way to get rid of one of them by implicitly deriving or something? I'm afraid that this substitution technique is very problem-dependent.

The motivation is that I have a numerical method based on spline collocation that works very well with derivatives of any order, but can hardly cope with systems. Therefore, an approach seems to go for a single higher order ODE.

 

[HallsofIvy]

Yes, it is very problem dependent. For example, if we have the non-linear equations
x'= x- y²
y'= x+ y²,

then, differentiating the first equation, x"= x'- 2yy' and y'= x+ y so we have
x"= x'- 2y(x+ y)= x'- 2xy+ y²

Clearly y²= x- x', but for the xy term we will have to solve for y: y²= x- x' so $y= \pm\sqrt{x- x'}$ and you will have to do the problem in two parts, using the plus and minus signs. There is no way to avoid that with non-linear equations.