Repeated Eigenvalues

Consider the linear homogeneous system

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In order to find the eigenvalues consider the Characteristic polynomial

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In this section, we consider the case when the above quadratic equation has double real root (that is if tex2html_wrap_inline97 ) the double root (eigenvalue) is

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In this case, we know that the differential system has the straight-line solution

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where tex2html_wrap_inline103 is an eigenvector associated to the eigenvalue tex2html_wrap_inline105 . We also know that the general solution (which describes all the solutions) of the system will be

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where tex2html_wrap_inline109 is another solution of the system which is linearly independent from the straight-line solution tex2html_wrap_inline111 . Therefore, the problem in this case is to find tex2html_wrap_inline109 .

Search for a second solution.

Let us use the vector notation. The system will be written as

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where A is the matrix coefficient of the system. Write

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The idea behind finding a second solution tex2html_wrap_inline109 , linearly independent from tex2html_wrap_inline123 , is to look for it as

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where tex2html_wrap_inline127 is some vector yet to be found. Since

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and

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(where we used tex2html_wrap_inline133 ), then (because tex2html_wrap_inline109 is a solution of the system) we must have

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Simplifying, we obtain

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or

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This equation will help us find the vector tex2html_wrap_inline127 . Note that the vector tex2html_wrap_inline127 will automatically be linearly independent from tex2html_wrap_inline103 (why?). This will help establish the linear independence of tex2html_wrap_inline109 from tex2html_wrap_inline123 .

Example. Find two linearly independent solutions to the linear system

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Answer. The matrix coefficient of the system is

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In order to find the eigenvalues consider the Characteristic polynomial

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Since tex2html_wrap_inline159 , we have a repeated eigenvalue equal to 2. Let us find the associated eigenvector tex2html_wrap_inline103 . Set

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Then we must have tex2html_wrap_inline165 which translates into

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This reduces to y=0. Hence we may take

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Next we look for the second vector tex2html_wrap_inline127 . The equation giving this vector is tex2html_wrap_inline175 which translates into the algebraic system

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where

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Clearly we have y=1 and x may be chosen to be any number. So we take x=0 for example to get

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Therefore the two independent solutions are

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The general solution will then be

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Qualitative Analysis of Systems with Repeated Eigenvalues

Recall that the general solution in this case has the form

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where tex2html_wrap_inline195 is the double eigenvalue and tex2html_wrap_inline103 is the associated eigenvector. Let us focus on the behavior of the solutions when tex2html_wrap_inline199 (meaning the future). We have two cases

tex2html_wrap_inline201
If tex2html_wrap_inline203 , then clearly we have

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In this case, the equilibrium point (0,0) is a sink. On the other hand, when t is large, we have

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So the solutions tend to the equilibrium point tangent to the straight-line solution. Note that is tex2html_wrap_inline213 , then the solution is the straight-line solution which still tends to the equilibrium point.

tex2html_wrap_inline201
If tex2html_wrap_inline217 , then Y(t) tends to infinity as tex2html_wrap_inline199, except of course the constant solution. Note again that if tex2html_wrap_inline213 , then we are moving along the straight-line solution.

Another example of the repeated eigenvalue's case is given by harmonic oscillators.

If you would like more practice, click on Example.

[Differential Equations] [First Order D.E.]
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Author: Mohamed Amine Khamsi

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