Solving a System of Differential Equation by Finding Eigenvalues and Eigenvectors

Differential Equations Problems and Solutions

Problem 668

Consider the system of differential equations
\begin{align*}
\frac{\mathrm{d} x_1(t)}{\mathrm{d}t} & = 2 x_1(t) -x_2(t) -x_3(t)\\
\frac{\mathrm{d}x_2(t)}{\mathrm{d}t} & = -x_1(t)+2x_2(t) -x_3(t)\\
\frac{\mathrm{d}x_3(t)}{\mathrm{d}t} & = -x_1(t) -x_2(t) +2x_3(t)
\end{align*}

(a) Express the system in the matrix form.

(b) Find the general solution of the system.

(c) Find the solution of the system with the initial value $x_1=0, x_2=1, x_3=5$.

 
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Hint.

Use the following theorem.

Theorem.
Let $A$ be a diagonalizable $n\times n$ matrix.
Let $\{\mathbf{v}_1,\dots, \mathbf{v}_n\}$ be an eigenbasis for $A$, with associated eigenvalues $\lambda_1, \dots, \lambda_n$. Then the general solution of the linear dynamical system
\[\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =A\mathbf{x}\] is
\[\mathbf{x}(t)=c_1 e^{\lambda_1 t}\mathbf{v}_1+\cdots +c_n e^{\lambda_n t}\mathbf{v}_n,\] where $c_1, \dots, c_n$ are arbitrary complex numbers.

Solution.

(a) Express the system in the matrix form.

Writing
\[\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \text{ and } A=\begin{bmatrix}
2 & -1 & -1 \\
-1 &2 &-1 \\
-1 & -1 & 2
\end{bmatrix},\] the system of differential equations can be written in the matrix form
\[\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =A\mathbf{x}.\]

(b) Find the general solution of the system.

The eigenvalues of the matrix $A$ are $0$ and $3$. The eigenspaces are
\[E_0=\Span \left(\, \begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix} \,\right) \text{ and } E_3=\Span \left(\, \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix} \,\right).\] (See the post “Quiz 13 (Part 1) Diagonalize a Matrix” for details.)


Thus, the formula in Theorem yields the general solution
\begin{align*}
\mathbf{x}(t)&=c_1 e^{0t}\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}+c_2e^{3t}\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}+c_3e^{3t}\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}\\[6pt] &= c_1 \begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}+c_2e^{3t}\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}+c_3e^{3t}\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}, \tag{*}
\end{align*}
where $c_1, c_2, c_3$ are arbitrary constants.

Equivalently, one may write the solution as a single vector
\[ \mathbf{x}(t)=\begin{bmatrix}
c_1-e^{3t}(c_2+c_3) \\
c_1+c_2e^{3t} \\
c_1+c_3e^{3t}
\end{bmatrix}.\]

(c) Find the solution of the system with the initial value $x_1=0, x_2=1, x_3=5$.

Substituting $t=0$ in the solution (*) obtained in part (b) yields
\[\begin{bmatrix}
0 \\
1 \\
5
\end{bmatrix}=\mathbf{x}(0)=c_1 \begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}+c_2\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}+c_3\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}.\] Solving this system gives $c_1=2, c_2=-1, c_3=3$.
Thus, the solution of the system of differential equations with the given initial value is
\[\mathbf{x}(t)=2\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}-e^{3t}\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}+3e^{3t}\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}.\] Or equivalently,
\[\mathbf{x}(t)=\begin{bmatrix}
2-2e^{3t} \\
2-e^{3t} \\
2+3e^{3t}
\end{bmatrix}.\]


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