Question on ODE's with roots of multiplicity.

[Skrew]

I was wondering how the general formula for the solutions of an nth order linear homogeneous ODE that had a characteristic equation which could be factored to (x-a)^n was derived(IE a set of solutions consisting e^(mx), x*e^(mx), ...x^(n-1)e^(mx)))?

For example the ODE,

y^(3)- 3y'' + 3y' - y = 0,

with characteristic equation,

m^3 - 3m^2 + 3m - 1 = 0

can be factored to

(m-1)^3,

where m = 1

and e^(x), x*e^(x) and x^2*e^(x) are all solutions.

For a second order ODE this can be found using a reduction of order technique but for higher order ODE's it gets very difficult to do so I am wondering what proof/explanation exists to show that we know such solutions exist?

I have looked around online and all the books/articles just say that's the case but don't provide an explanation.

 

[jasonRF]

You actually have a nonhomogeneous ODE here. The homogeneous portion is
[tex] y^{(3)} - 3 y^{''} + 3y = 0 [/tex]
The characteristic equation is
[tex] m^3 - 3m^2 + 3 = 0 [/tex]

This does not have repeated roots. I had to use Matlab to find them numerically.

You should be able to find the particular solution by inspection.

However, if you needed it, for third and higher order ODEs you can use reduction of order / variation of parameters just like you do for second order. It is a powerful technique, although it of course can get messy!

jason

 

[Skrew]

You actually have a nonhomogeneous ODE here. The homogeneous portion is
[tex] y^{(3)} - 3 y^{''} + 3y = 0 [/tex]
The characteristic equation is
[tex] m^3 - 3m^2 + 3 = 0 [/tex]

This does not have repeated roots. I had to use Matlab to find them numerically.

You should be able to find the particular solution by inspection.

However, if you needed it, for third and higher order ODEs you can use reduction of order / variation of parameters just like you do for second order. It is a powerful technique, although it of course can get messy!

jason

Ah I see I messed up when writing the post, it was late.

I will edit it, but my question still exists.

Is reduction of order/variation of parameters the only way to show that the e^(mx), x*e^(mx),...x^(n-1)e^(mx) pattern exists?

 

[elibj123]

Let's denote L{} as the differential operator of a constant coefficient ODE.
So the solutions satisfy:

L(y)=0

Now this operator has the property that for functions [tex]y(x)=e^{mx}[/tex] it satisfies:

[tex]L(e^{mx})=P(m)e^{mx}[/tex]

With P() being the characteristic polynomial. Now let's assume some m0 is a multiple root of P().

Let's take the latter equation and differentiate it with respect to m. Since L is an operator wrt x, it will commute with the m-differentiation so:

[tex]\frac{d}{dm}L(e^{mx})=L(\frac{d}{dm}e^{mx})=L(xe^{mx})[/tex]

On the other side:

[tex]\frac{d}{dm}(P(m)e^{mx})=P'(m)e^{mx}+xP(m)e^{mx}[/tex]

Since m0 is at least of multiplicity 2, it is also a root of the derivative of P. Therefore when we evaluate the latter at m0 we get a zero, which means:

[tex]L(xe^{m_{0}x})=P'(m_{0})e^{m_{0}x}+xP(m_{0})e^{m_{0}x}=0[/tex]

So

[tex]y(x)=xe^{m_{0}x}[/tex]

Is another solution. Repeating this (assuming m0 has bigger multiplicity) you'll get higher powers of x.

 

[Skrew]

Let's denote L{} as the differential operator of a constant coefficient ODE.
So the solutions satisfy:

L(y)=0

Now this operator has the property that for functions [tex]y(x)=e^{mx}[/tex] it satisfies:

[tex]L(e^{mx})=P(m)e^{mx}[/tex]

With P() being the characteristic polynomial. Now let's assume some m0 is a multiple root of P().

Let's take the latter equation and differentiate it with respect to m. Since L is an operator wrt x, it will commute with the m-differentiation so:

[tex]\frac{d}{dm}L(e^{mx})=L(\frac{d}{dm}e^{mx})=L(xe^{mx})[/tex]

On the other side:

[tex]\frac{d}{dm}(P(m)e^{mx})=P'(m)e^{mx}+xP(m)e^{mx}[/tex]

Since m0 is at least of multiplicity 2, it is also a root of the derivative of P. Therefore when we evaluate the latter at m0 we get a zero, which means:

[tex]L(xe^{m_{0}x})=P'(m_{0})e^{m_{0}x}+xP(m_{0})e^{m_{0}x}=0[/tex]

So

[tex]y(x)=xe^{m_{0}x}[/tex]

Is another solution. Repeating this (assuming m0 has bigger multiplicity) you'll get higher powers of x.

Thank you, I understand it now.

Does something similar exist to describe repeated roots on CauchyEuler equations? I'm not seeing an obvious way that it extends to it.

 

[HallsofIvy]

Another way to approach this is to use the "reduction of order" method.

Seeing that the characteristic equation, [itex](m-1)^3= 0[/itex] has m= 1 as solution, we know that one solution to the d.e. is [itex]e^x[/itex]. Now, seek a solution of the form [itex]y(x)= u(x)e^x[/itex]. By the product rule, [itex]y'= u'e^x+ ue^x[/itex], [itex]y'= u''e^x+ 2u'e^x+ ue^x[/itex], and [itex]y'''= u'''e^x+ 3u''e^x+ 3u'e^x+ ue^x[/itex]. Putting those into the differential equation, we have [itex]u''e^x+ 3u''e^x+ 3u'e^x+ ue^x-[/itex][itex] 3u''e^x- 6u'e^x- 3ue^x+ 3u'e^x+ 3ue^x- ue^x= 0[/itex] which reduces to [itex]u'''e^x= 0[/itex]. That is the same as u"'= 0 and, integrating three times, [itex]u(x)= Ax^2+ Bx+ C[/itex] which gives [itex]y(x)= Ax^2e^x+ Bxe^x+ Cx[/itex] as the general solution.

As for CauchyEuler equations, also called "equi-potential" equations since the coefficient of each derivative has x to a power equal to the order of the derivative, the change of variable, x= ln(t), changes a CauchyEuler equation, in variable t, to an equation with constant coefficients, in variable x. Multiplying by powers of x, in the constant coefficients equation, then, leads to multiplying by powers of ln(t) in the CauchEuler equation.

That is, if the characteristic equation for a CauchyEuler equation is [itex](m- a)^3[/itex], the three independent solutions to the equation are [itex]t^a[/itex], [itex]t^a ln(t)[/itex], and [itex]t^a (ln(t))^2[/itex].