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 $x_1=0, x_2=1, x_3=5$.

Hint.

Use the following theorem.

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,
\[\mathbf{x}(t)=\begin{bmatrix}
2-2e^{3t} \\
2-e^{3t} \\
2+3e^{3t}
\end{bmatrix}.\] Click here if solved 34

The post Solving a System of Differential Equation by Finding Eigenvalues and Eigenvectors first appeared on Problems in Mathematics.

(a) Find all solutions of the linear dynamical system
\[\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =\begin{bmatrix}
1 & 0\\
0& 3
\end{bmatrix}\mathbf{x},\] where $\mathbf{x}(t)=\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}$ is a function of the variable $t$.

(b) Solve the linear dynamical system
\[\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}=\begin{bmatrix}
2 & -1\\
-1& 2
\end{bmatrix}\mathbf{x}\] with the initial value $\mathbf{x}(0)=\begin{bmatrix}
1 \\
3
\end{bmatrix}$.

Solution.

(a) Find all solutions of the linear dynamical system $\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =\begin{bmatrix}
1 & 0\\
0& 3
\end{bmatrix}\mathbf{x}$

Note that the given system is
\[\begin{bmatrix}
\frac{\mathrm{d}{x_1}}{\mathrm{d} t} \\[6pt] \frac{\mathrm{d}{x_2}}
{\mathrm{d} t}
\end{bmatrix}
=\begin{bmatrix}
x_1 \\
3x_2
\end{bmatrix}.\] Thus, we have two uncoupled differential equations
\begin{align*}
\frac{\mathrm{d}{x_1}}{\mathrm{d} t} &=x_1 \\[6pt] \frac{\mathrm{d}{x_2}}{\mathrm{d} t} &=3x_2.
\end{align*}

The solutions to these differential equations are
\begin{align*}
x_1(t)&=e^t x_1(0)\\
x_2(t)&=e
^{3t} x_2(0).
\end{align*}
Thus we see that
\[\mathbf{x}(t)=\begin{bmatrix}
e^t x_1(0) \\[6pt] e
^{3t} x_2(0)
\end{bmatrix}.\]

(b) Solve the linear dynamical system $\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}=\begin{bmatrix}
2 & -1\\
-1& 2
\end{bmatrix}\mathbf{x}$

Let $A=\begin{bmatrix}
2 & -1\\
-1& 2
\end{bmatrix}$.
Then the matrix $A$ has eigenvalues $1, 3$ and corresponding eigenvectors are
\[\begin{bmatrix}
1 \\
1
\end{bmatrix} \text{ and } \begin{bmatrix}
-1 \\
1
\end{bmatrix},\] respectively.
(See the post Diagonalize a 2 by 2 Symmetric Matrix for details.)

Thus, if we put $S=\begin{bmatrix}
1 & -1\\
1& 1
\end{bmatrix}$, then $A$ is diagonalizable by $S$. That is,
\[S^{-1}AS=D,\] where
\[D=\begin{bmatrix}
1 & 0\\
0& 3
\end{bmatrix}.\]

Substituting $A=SDS^{-1}$ into the given system, we have
\begin{align*}
\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}&=(SDS^{-1})\mathbf{x}\\[6pt] \Leftrightarrow \quad S^{-1}\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}&=(DS^{-1})\mathbf{x}\\[6pt] \Leftrightarrow \quad \frac{\mathrm{d}(S^{-1} \mathbf{x})}{\mathrm{d}t}&=(DS^{-1})\mathbf{x}.
\end{align*}

Let $\mathbf{u}(t)=S^{-1}\mathbf{x}$. Then we obtain the system
\[\frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t} =\begin{bmatrix}
1 & 0\\
0& 3
\end{bmatrix}\mathbf{u}.\] We solved this system in part (a) and the general solution is given by
\[\mathbf{u}(t)=\begin{bmatrix}
e^t u_1(0) \\[6pt] e^{3t} u_2(0)
\end{bmatrix}.\]

We determine the values of $u_1(0)$ and $u_2(0)$ using the given initial value $\mathbf{x}(0)=\begin{bmatrix}
1 \\
3
\end{bmatrix}$.
We have
\[\begin{bmatrix}
u_1(0) \\
u_2(0)
\end{bmatrix}=\mathbf{u}(0)=S^{-1} \mathbf{x}(0)=\frac{1}{2}\begin{bmatrix}
1 & 1\\
-1& 1
\end{bmatrix}\begin{bmatrix}
1 \\
3
\end{bmatrix}=\begin{bmatrix}
2 \\
1
\end{bmatrix}.\] (Note that $\begin{bmatrix}
2 \\
1
\end{bmatrix}$ is the coordinate vector of $\mathbf{x}(0)=\begin{bmatrix}
1 \\
3
\end{bmatrix}$ with respect to the eigenbasis $\begin{bmatrix}
1 \\
1
\end{bmatrix}, \begin{bmatrix}
-1 \\
1
\end{bmatrix}$.)

It follows that the solution of the original system is
\begin{align*}
\mathbf{x}=S\mathbf{u}=\begin{bmatrix}
1 & -1\\
1& 1
\end{bmatrix}
\begin{bmatrix}
2e^t \\[6pt] e^{3t}
\end{bmatrix}
=2e^t\begin{bmatrix}
1 \\
1
\end{bmatrix}+ e^{3t}\begin{bmatrix}
-1 \\
1
\end{bmatrix}.
\end{align*}

Another Solution of (b)

In this solution, we use the following theorem.

As in the above solution, we know that the matrix $A=\begin{bmatrix}
2 & -1\\
-1& 2
\end{bmatrix}$ has eigenvalues $1, 3$ and corresponding eigenvectors are
\[\begin{bmatrix}
1 \\
1
\end{bmatrix} \text{ and } \begin{bmatrix}
-1 \\
1
\end{bmatrix},\] respectively.

So the formula in the theorem yields the general solution
\[\mathbf{x}(t)=c_1 e^{t}\begin{bmatrix}
1 \\
1
\end{bmatrix}+c_2 e^{3t}\begin{bmatrix}
-1 \\
1
\end{bmatrix},\] where $c_1, c_2$ are constants.

Since the initial is $\mathbf{x}(0)=\begin{bmatrix}
1 \\
3
\end{bmatrix}$, we have
\begin{align*}
\begin{bmatrix}
1 \\
3
\end{bmatrix}=c_1\begin{bmatrix}
1 \\
1
\end{bmatrix}+c_2\begin{bmatrix}
-1 \\
1
\end{bmatrix}.
\end{align*}
Solving this system, we obtain $c_1=2$ and $c_2=1$.
Thus, the solution of the linear dynamical system with the given initial value is
\[\mathbf{x}(t)=2 e^{t}\begin{bmatrix}
1 \\
1
\end{bmatrix}+e^{3t}\begin{bmatrix}
-1 \\
1
\end{bmatrix}.\] Click here if solved 18

The post Solve the Linear Dynamical System $\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =A\mathbf{x}$ by Diagonalization first appeared on Problems in Mathematics.