System of three, 2nd order diff. equations

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While trying to solve a problem in physics I got this system of 3, second order differential equations.
Solution should be equation for linear harmonic oscillator.

m\ddot{η}1+kη1-kη2=0
mn\ddot{η}2-kη1+2kη2-kη3=0
m\ddot{η}3-kη2+kη3=0


My attempts at the solution produced 6th order equation that I only manage to reduce to 4th order but after that I got completely lost.
Here is what I did if you wish to see: part1 part2
 
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You should know that the solution of a system of constant-coefficient linear equations is of the form ηi=Aieλt. In case of an oscillatory system , it is more convenient to use ηi=Aieiωt.

Substitute for the η-s and you get a third-order equation for the unknown ω2, and also relations between the coefficients Ai.

ehild
 
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Thanks, though I already know the solution. I forgot to mention that in the first post.
The problem is I don't know how to come to that solution.
I want to solve it once so that I can have an idea how to come to that.
 
I would form a matrix. You should get a 3x1 on the left side equaled to a 3x3 times some constants that you need to solve for. Solve for the eigenvalues and eigenvectors.
 
_alexis_ said:
Thanks, though I already know the solution. I forgot to mention that in the first post.
The problem is I don't know how to come to that solution.
I want to solve it once so that I can have an idea how to come to that.

All such equations are solved that way: replace the unknown functions with the exponential form. Performing the derivations, you can simplify the exponential factor and you get 3 homogeneous equations for the amplitudes. The system of homogeneous equations has non-trivial (not zero) solutions only if the determinant of the coefficients is equal to zero. That provides an equation for the eigenvalue ω2. (It is easy to solve in your problem.) Solve, and substitute the roots back into the system of equations. Solve for Ai for each eigenvalue. (One of the amplitudes is arbitrary in each case) These Ai-s, multiplied with the proper exponential factor, give the eigenvectors, or normal modes of the oscillatory system. The general solution is linear combination of these normal modes.

ehild
 
_alexis_ said:
While trying to solve a problem in physics I got this system of 3, second order differential equations.
Solution should be equation for linear harmonic oscillator.

m\ddot{η}1+kη1-kη2=0
mn\ddot{η}2-kη1+2kη2-kη3=0
m\ddot{η}3-kη2+kη3=0


My attempts at the solution produced 6th order equation that I only manage to reduce to 4th order but after that I got completely lost.
Here is what I did if you wish to see: part1 part2
That's odd. Yes, reducing to a single function will give you a 6th order equation but since only even derivatives will occur, it should be easily reducible to a third order equation.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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