Repeated roots in differential equations query

[abskebabs]

This has been bugging me so I would very much appreciate it if sum1 could clear up my difficulty ASAP. Suppose we have a general linear, homogenous ODE with constant coefficients as follows:

[math]\sum_{i=0}^{n}Ai\frac{d^i}{dx^n}y(x) = 0[/math]

We can propose a solution to the equation to be:

[math]y=e^kx[/math]

Whereby k can be one of the solutions of the resulting polynomial:

[math]\sum_{i=0}^{n}Aik^n[/math]

I know that if I have 2 repeated roots the independent solutions should be:

[math]k=e^{lx}[/math] or [math]k=le^{lx}[/math]

 

In my notes this is justified by saying that if we have 2 roots that are very close to each other e.g. [math]k=l[/math] and [math]k=l+f[/math], then their 2 independent solutions are:

[math]k=e^{lx}[/math] and [math]k=\frac{1}{f}[e^{(l+f)x}-e^{lx}][/math]

Now I can see how the 2nd solution yields the 2nd solution at the end of the previous paragraph when the limits of f tend to zero. What I cannot understand though; is how it is the independent solution of the equation for the root [math]l+f[/math]. I was under the impression that this would be just:

[math]e^{(l+f)x}[/math]

 

Thanks in advance:-)

(Edit:[math]Ai[/math] doesn't represent a multiple of A and i. I didnt know how to get subscript on latex(I'm an amateur I know) so I just wanted to be clear that this represents n different coefficients from i=0 to i=n)

[Tom Mattson]

This has been bugging me so I would very much appreciate it if sum1 could clear up my difficulty ASAP. Suppose we have a general linear, homogenous ODE with constant coefficients as follows:

[math]\sum_{i=0}^{n}Ai\frac{d^i}{dx^n}y(x) = 0[/math]

 

You can make [math]A_i[/math] by typing A_i in math tags.

 

We can propose a solution to the equation to be:

[math]y=e^kx[/math]

 

You can put both the k and the x in the exponent by typing e^{kx} in math tags.

 

Whereby k can be one of the solutions of the resulting polynomial:

[math]\sum_{i=0}^{n}Aik^n[/math]

 

That should be [math]\sum_{i=0}^{n}A_ik^i[/math].

 

I know that if I have 2 repeated roots the independent solutions should be:

[math]k=e^{lx}[/math] or [math]k=le^{lx}[/math]

 

This doesn't make sense, because k shouldn't depend on x. Don't you mean [math]y=e^{kx}[/math] and [math]y=xe^{kx}[/math]?

[abskebabs]

This doesn't make sense, because k shouldn't depend on x. Don't you mean [math]y=e^{kx}[/math] and [math]y=xe^{kx}[/math']?

You're right it should be:

[math]y=e^{lx}[/math] and [math]xe^{lx}[/math]

But I still have my difficulty with the equation that preceded the one on the right above and how it relates to the 2nd root.

 

Or is what I have written here wrong too?

[Tom Mattson]

I'm not exactly sure of what it is you're having a problem with. Are you trying to justify why we use those two solutions for a repeated root? If so then the simple answer is that we do it because it works. It's just an ansatz that happens to get the job done.

[abskebabs]

I'm not exactly sure of what it is you're having a problem with. Are you trying to justify why we use those two solutions for a repeated root? If so then the simple answer is that we do it because it works. It's just an ansatz that happens to get the job done.

As you have now seen the rest of this thread, I think I can how summarise and clarify my difficulty. I have noticed from looking at this thread now that I have made several careless mistakes, even adter editing, so it is not surprising you may be finding difficulties if what I ave presented is so unclear!

 

Anyway, you notice that if we have 2 roots of the auxiliary equation for the differential equation(if we take the solution to be of the form [math]y=e^{kt}[/math]) as [math]k=l[/math] or [math]k=l+e[/math]:

Then I cannot understand how the independent solution to the equation for the root [math]k=l+e[/math] is [math]y=\frac{1}{f}[e^{(l+f)x}-e^{lx}][/math] (I made a typo on this originally). Once(or if) I understand this, personally I think I may feel a slight feeling of euphoria:D as I will fully understand why the solution [math]y=xe^{kx}[/math] is correct for repeated roots, rather than just relying on the notion "that it works".

 

Sometimes I feel perhaps I am more of a mathematician at heart than a physicist as I think most physicists wouldn't really care about this and just simply be glad it works. I guess looking at it from their perspective tho I may just be acting like a pain in the a## .

 

Thanks a lot for taking the time to look at this Tom:-)

[abskebabs]

It's taken me a while admittedly but scrolling through this old thread I've realised that my initial difficulties with this were not resolved. Could someone else have a look at this and try to help me out please:-) ? Thanks in advance

[shadowacct]

For a second order system with two eigenvalues l and l + f, the full solution, with two freedoms A and B can be written as

 

y(x) = A * exp(l*x) + B * exp((l+f)*x)

 

Let f--->0 and apparently, this reduces to a single solution of the form c*exp(l*x).

 

But now comes the 'trick'. As long as l and f are constants (and they are, we have a time-independent and linear differential equation), then we may choose A and B as functions of l and f, they still are constant.

 

Choose A = b - c/f, and B = c/f, with b and c constants. For any f not equal to 0, we still can obtain any solution with two eigenmodes l and l + f by proper choice of b and c.

 

For f, equal to 0, the situation seems erroneuous, but if we rearrange the terms, then you see your result:

 

y(x) = b*exp(l*x) + c/f*(exp((l+f)*x) - exp(l*x)).

 

The second term goes to the form d*x*exp(l*x) when f--->0.

 

I hope this explanation makes sense to you.

 

A similar form of reasoning is possible for triple sets of eigenvalues, or even sets of larger multipicity. You'll see that for each set of multiplicity N, the solution, corresponding to that set, will be a polynomial of degree N-1 times exp(l*x), with l the eigenvalue of multiplicity N.

 

So, what really happens is that for a set of multiplicity N, the N eigenmodes simply blend into a single eigenmode, with a polynomial factor. In the case of two equal eigenvalues, I would not call this as having two corresponding solutions exp(l*x) and x*exp(l*x), where one of these corresponds to one of the eigenvalues and the other to the other eigenvalue. No, I would say that the set of eigenvalues has a corresponding single solution of the form P*exp(l*x) + Q*x*exp(l*x), with P, Q constants.

 

So, each unique eigenvalue l in a system has a single corresponding solution of the form (A0 + A1*x + A2*x^2 + ....)*exp(l*x), where the polynomial factor has degree N-1, with N being the multiplicity of eigenvalue l.

[Country Boy]

Applying the differential operator [math]\frac{d}{dx}- k[/math] to the function [math]e^{kx}[/itex] gives you 0: Applying the differential operator [math](\frac{d}{dx}-k)^2[/itex] either of [math]e^{kx}[/math] or [math]xe^{kx}[/math] gives 0.

 

In general, applying the differential operator [math](\frac{d}{dx}-k)^n[/math] to the n functions, [math]e^{kx}[/math], [math]xe^{kx}[/math], [math]x^2e^{kx}[/math], ..., [math]x^{n-1}e^{kx}[/math] gives 0. You should be able to prove that by induction.