3.5 Special Cases: Repeated and Zero Eigenvalues
Systems with zero as an eigenvalue
If λ₁=0 and λ₂≠0 for dY/dt=AY, and V₁ is an eigenvector for λ₁ and V₂ is an eigenvector for λ2, we have two real distinct eigenvalues and the general solution is: Y(t)=k₁e^(λ₁t)V₁+k₂e^(λ₂t)V₂ But since λ₁=0, Y(t)=k₁V₁+k₂e^(λ₂t)V₂ Therefore, all solutions with k₂=0 are equilibrium solutions since the first term is constant. Every point on the line of eigenvectors for the eigenvalue λ₁=0 is an equilibrium point. If λ₂<0 then the second term tends to 0 as t increases, so the solution tends to k₁V₁ along a line parallel to V₂. If λ₂>0 then the solution moves away from the line of equilibrium points as t increases.
What is the issue with repeated eigenvalues?
We only find one solution, when we need two independent solutions to obtain the general solution. To find a second solution, we can set the system of equations we used to find the first eigenvector equal to that first eigenvector, rather than equal to [0;0]. Going from there, we can determine the new eigenvalue/eigenvector, possibly using linear analysis.
When is Y(t)=e^(λt)V₀+te^(λt)V₁ a solution?
When V₁=(A-λI)V₀ and V1 is an eigenvector, unless it is zero. We can determine this by using dY/dt=AY, setting the differential of the solution equal to multiplying the solution by A, and isolating V₁. The two parts of this linear equations are not solutions, since when there are multiple eigenvalues, V₁ is determined by V₀.
Repeated eigenvalue
When instead of 2 distinct eigenvalue, there are 2 that are the same value.
Form of general solution of systems with repeated eigenvalues
Y(t)=e^(λt)V₀+te^(λt)V₁, where V₁=(A-λI)V₀ is either an eigenvector or 0
Systems for which every vector is an eigenvector
dY/dt=[a 0;0 a]Y λ=a here, and every nonzero vector is an eigenvector for λ=a. In this case finding the general solution is equivalent to finding the general solutions of the two equations dx/dt=ax, dy/dt=ay, so that the system completely decouples. Since every vector is an eigenvector, every solution curve (except the equilibrium point at the origin) is a ray that approaches/leaves the origin as t increases. if a>0, it is a source, and if a<0, it is a sink.
Graphical representation of systems with repeated eigenvalues
λ<0, both parts of linear equation tends to 0 as t increases, and therefore the equilibrium point at the origin is a sink. To determine directions, use Y(t)=e^(λt)(V₀+tV₁). The tV₁ term dominates if t is large, thus the solution tends to the origin in a direction that is tangent to the line of eigenvectors. λ>0, then all solutions (except the equilibrium solution) tend to infinity as t increases, so the origin is a source. The te^(λt) term dominates for large t if V₁ is not -.
