Section 5.6. Solve a system of homogeneous differential equations using the coefficient matrix

Objective:

1. Solve a system of homogeneous differential equations using the coefficient matrix with repeated eigenvalues

2. Understand connection between the phase plan and the solution

In the previous sections, we only have the case that the matrix [latex]A[/latex] has two distinct eigenvalues and they each give different eigenvectors. One can check that those will give linearly independent eigenfunctions, hence general solutions. How about the case when the matrix only has one eigenvalue? This case has two situations, if this eigenvalue has two linearly independent eigenvectors, then we still has two linearly independent eigenfunctions. 

Example 1: Given [latex]\left[\begin{array}{c} x_{1}'\\ x_{2}' \end{array}\right]=\left[\begin{array}{cc} -1 & 0\\ 0 & -1 \end{array}\right]\left[\begin{array}{c} x_{1}\\ x_{2} \end{array}\right]=A\overrightarrow{x}[/latex], find the general solution of the system of equations. Sketch its phase plane.

Exercise 1: Given [latex]\left[\begin{array}{c} x_{1}'\\ x_{2}' \end{array}\right]=\left[\begin{array}{cc} -2 & 0\\ 0 & -2 \end{array}\right]\left[\begin{array}{c} x_{1}\\ x_{2} \end{array}\right]=A\overrightarrow{x}[/latex], find the general solution of the system of equations. Sketch its phase plane.

For the case when the matrix has one eigenvalue and one eigenvector, we need to use what we did before for the second order ODE, [latex]y''+2y'+y=0[/latex], then the general solution of this ODE is [latex]y=c_{1}e^{-t}+c_{2}te^{-t}[/latex]. We multiply [latex]t[/latex] to the only solution [latex]y=e^{-t}[/latex] coming from the root of the characteristics polynomial. Go back to the matrix [latex]A[/latex] that is coming from the [latex]\overrightarrow{x'}=A\overrightarrow{x}[/latex], the system of differential equation. Suppose [latex]A[/latex] has only one eigenvalue, [latex]\lambda[/latex], and one eigenvector, [latex]\overrightarrow{v}[/latex]. This gives an eigenfunction, [latex]\overrightarrow{v(t)}=e^{\lambda t}\overrightarrow{v}[/latex]. We could try [latex]\overrightarrow{w(t)}=te^{\lambda t}\overrightarrow{v}[/latex] vector function. We can check that if [latex]\overrightarrow{x}(t)=c_{1}e^{\lambda t}\overrightarrow{v}+c_{2}te^{\lambda t}\overrightarrow{v}[/latex] is a solution, then in order to satisfy [latex]\overrightarrow{x'}=A\overrightarrow{x}[/latex], we must have [latex]\overrightarrow{v}=0[/latex], a contradiction. We need to find the [latex]\overrightarrow{w}[/latex] vector such that [latex]\overrightarrow{w(t)}[/latex] is a solution of the system of differential equation and with [latex]\overrightarrow{v(t)}[/latex], they give the general solution of the differential equation. Hence, we consider [latex]\overrightarrow{w}(t)=e^{\lambda t}\overrightarrow{w}+te^{\lambda t}\overrightarrow{v}[/latex] and assume it is also a solution, i.e. [latex]\overrightarrow{x}'(t)=\lambda e^{\lambda t}\overrightarrow{w}+e^{\lambda t}\overrightarrow{v}+\lambda te^{\lambda t}\overrightarrow{v}=A(e^{\lambda t}\overrightarrow{w}+te^{\lambda t}\overrightarrow{v})[/latex].  We have  \[\begin{align*} \lambda e^{\lambda t}\overrightarrow{w}+e^{\lambda t}\overrightarrow{v}+\lambda te^{\lambda t}\overrightarrow{v} & =Ae^{\lambda t}\overrightarrow{w}+Ate^{\lambda t}\overrightarrow{v}\\ \lambda e^{\lambda t}\overrightarrow{w}+e^{\lambda t}\overrightarrow{v}-Ae^{\lambda t}\overrightarrow{w} & =Ate^{\lambda t}\overrightarrow{v}-\lambda te^{\lambda t}\overrightarrow{v}. \end{align*}\] Therefore \[\begin{align*} \lambda e^{\lambda t}\overrightarrow{w}+e^{\lambda t}\overrightarrow{v}-Ae^{\lambda t}\overrightarrow{w} & =0\\ Ate^{\lambda t}\overrightarrow{v}-\lambda te^{\lambda t}\overrightarrow{v} & =0. \end{align*}\] Hence [latex]A\overrightarrow{v}=\lambda\overrightarrow{v}[/latex] and [latex]A\overrightarrow{w}=\lambda\overrightarrow{w}+\overrightarrow{v}[/latex]. In order to find the vector [latex]\overrightarrow{w}[/latex], we find the solution set of [latex](A-\lambda I)\overrightarrow{x}=\overrightarrow{v}[/latex]. Then the general solution is [latex]\overrightarrow{x}(t)=c_{1}e^{\lambda t}\overrightarrow{v}+c_{2}(e^{\lambda t}\overrightarrow{w}+te^{\lambda t}\overrightarrow{v})[/latex].

Example 2: Given [latex]\left[\begin{array}{c} x_{1}'\\ x_{2}' \end{array}\right]=\left[\begin{array}{cc} -1 & 2\\ 0 & -1 \end{array}\right]\left[\begin{array}{c} x_{1}\\ x_{2} \end{array}\right]=A\overrightarrow{x}[/latex], find the general solution of the system of equations. Sketch its phase plane.

Exercise 2: Given [latex]\left[\begin{array}{c} x_{1}'\\ x_{2}' \end{array}\right]=\left[\begin{array}{cc} -2 & 3\\ 0 & -2 \end{array}\right]\left[\begin{array}{c} x_{1}\\ x_{2} \end{array}\right]=A\overrightarrow{x}[/latex], find the general solution of the system of equations. Sketch its phase plane.

Example 3: Given [latex]\left[\begin{array}{c} x_{1}'\\ x_{2}' \end{array}\right]=\left[\begin{array}{cc} 1 & -1\\ 1 & 3 \end{array}\right]\left[\begin{array}{c} x_{1}\\ x_{2} \end{array}\right]=A\overrightarrow{x}[/latex], find the general solution of the system of equations. Sketch its phase plane.

Exercise 3: Given [latex]\left[\begin{array}{c} x_{1}'\\ x_{2}' \end{array}\right]=\left[\begin{array}{cc} 4 & 1\\ -1 & 2 \end{array}\right]\left[\begin{array}{c} x_{1}\\ x_{2} \end{array}\right]=A\overrightarrow{x}[/latex], find the general solution of the system of equations. Sketch its phase plane.

Example 4: Given [latex]\left[\begin{array}{c} x_{1}'\\ x_{2}' \end{array}\right]=\left[\begin{array}{cc} -6 & 4\\ -1 & -2 \end{array}\right]\left[\begin{array}{c} x_{1}\\ x_{2} \end{array}\right]=A\overrightarrow{x}[/latex], find the general solution of the system of equations. Sketch its phase plane.

Exercise 4: Given [latex]\left[\begin{array}{c} x_{1}'\\ x_{2}' \end{array}\right]=\left[\begin{array}{cc} -5 & 3\\ -3 & 1 \end{array}\right]\left[\begin{array}{c} x_{1}\\ x_{2} \end{array}\right]=A\overrightarrow{x}[/latex], find the general solution of the system of equations. Sketch its phase plane.

Group Work: 

1. Solve the initial value problem and describe the behavior of the solution as [latex]t[/latex] approaches infinity.

(a) [latex]\left[\begin{array}{c} x_{1}'\\ x_{2}' \end{array}\right]=\left[\begin{array}{cc} 1 & -4\\ 4 & -7 \end{array}\right]\left[\begin{array}{c} x_{1}\\ x_{2} \end{array}\right],[/latex] [latex]\left[\begin{array}{c} x_{1}(0)\\ x_{2}(0) \end{array}\right]=\left[\begin{array}{c} 1\\ 2 \end{array}\right][/latex]. 

(b) [latex]\left[\begin{array}{c} x_{1}'\\ x_{2}' \end{array}\right]=\left[\begin{array}{cc} 3 & 9\\ -1 & -3 \end{array}\right]\left[\begin{array}{c} x_{1}\\ x_{2} \end{array}\right],[/latex] [latex]\left[\begin{array}{c} x_{1}(0)\\ x_{2}(0 \end{array}\right]=\left[\begin{array}{c} 2\\ -1 \end{array}\right][/latex]. 

(c) [latex]\left[\begin{array}{c} x_{1}'\\ x_{2}' \end{array}\right]=\left[\begin{array}{cc} -\frac{5}{2} & \frac{3}{2}\\ -\frac{3}{2} & \frac{1}{2} \end{array}\right]\left[\begin{array}{c} x_{1}\\ x_{2} \end{array}\right],[/latex] [latex]\left[\begin{array}{c} x_{1}(0)\\ x_{2}(0) \end{array}\right]=\left[\begin{array}{c} 0\\ 1 \end{array}\right][/latex].