To understand we can solve the system \({\mathbf x}' = A {\mathbf x}\) by finding eigenvalues and eigenvectors for \(A\text{,}\) where \(A\) is an \(n \times n\) matrix.
To understand that the Principle of Superposition holds for higher-order systems.
To understand that the geometry of a system in \({\mathbb R}^3\) is characterized by stable lines and stable planes.
Suppose that we have two masses on a table, \(m_1\) and \(m_2\text{,}\) connected by three springs with the outside springs connected to two walls (Figure5.8.1), and the masses are free to move horizontally. We will assume that the springs are uniform and all have the same spring constant \(k\text{.}\) The horizontal displacements of the springs are denoted by \(x_1(t)\) and \(x_2(t)\) for the masses \(m_1\) and \(m_2\text{,}\) respectively. Assuming that there is no damping, the only forces acting on mass \(m_1\) at time \(t\) are those of left and middle springs. The force from the left spring will be \(-kx_1\) while the force from middle spring will be \(k(x_2 - x_1)\text{.}\) By Newton's Second Law of motion, we have
The strategy for finding solutions to the system \({\mathbf x}' = A {\mathbf x}\) is the same as for systems of two equations. If \(\lambda\) is an eigenvalue of \(A\) with eigenvector \({\mathbf v}\text{,}\)
Thus, the eigenvalues of \(A\) are \(\lambda = 2\text{,}\) \(\lambda = 1\text{,}\) and \(\lambda = -1\text{.}\) To find an eigenvector for \(\lambda = 2\text{,}\) we must find a nontrivial solution for system of equations \((A - 2I){\mathbf x} = {\mathbf 0}\text{,}\)
It is easy to check that \((2, -3, 5)\) is a solution. Similarly, we can determine that \((5, -7, 13)\) is an eigenvector for \(\lambda = 1\) and \((1, -1, 2)\) is an eigenvector for \(\lambda = -1\text{.}\) Thus, we have found three solutions for the system \({\mathbf x}' = A {\mathbf x}\text{,}\)
The Principle of Superposition also holds for higher-order systems. If \({\mathbf x}_1(t)\) and \({\mathbf x}_2(t)\) are solutions for \({\mathbf x}' = A {\mathbf x}\text{,}\) then
is a solution for our system. This is, in fact, the general solution for the system.
Although we shall not cover the notions of linear independence, canonical matrices, and change of coordinates for \({\mathbb R}^n\text{,}\) the same ideas that we used for systems of first-order linear differential equations in \({\mathbb R}^2\) carry over to \({\mathbb R}^n\text{.}\) The necessary linear algebra is covered in any good linear algebra course. Also, see Chapters 5 and 6 in [12]. In addition, finding eigenvalues for matrices greater than \(2 \times 2\text{,}\) we will need to find the roots of a characteristic polynomial of degree greater than two, which can be difficult. A good course in linear algebra will cover techniques of finding eigenvalues for larger matrices.
the solution to the system \(\mathbf x' A \mathbf x\text{.}\)
Subsection5.8.2The Geometry of Solutions
In Section5.6, we classified all of the geometry of the solutions for planar systems using the trace-determinant plane. The geometry for linear systems in three variables is a bit more complicated. For a system
\begin{align*}
x' \amp = a_1 x + a_1 y + a_3 z\\
y' \amp = b_1 x + b_2 y + b_3 z\\
z' \amp = c_1 x + c_2 y + c_3 z,
\end{align*}
our solution curves live in \(\mathbb R^3\text{,}\) and there is simply a lot more room to move around in three dimensions than in two dimensions. The origin is still an equilibrium solution for a system of linear differential equations in three variables. The origin is a stable equilibrium solution if any solution \(\mathbf x(t)\) approaches \(\mathbf 0 = (0, 0, 0)\) as \(t \to \infty\text{;}\) otherwise, \(\mathbf 0\) is an unstable equilibrium solution. In the case of planar systems, an unstable solution is a nodal saddle, a nodal source, a spiral source, or a source with a single unstable line. In the case of \(\mathbb R^3\text{,}\) we could have a stable line of solutions and an unstable plane of solutions. In this case, all solutions of the system with initial condition lying on the stable line would approach the origin as \(t \to \infty\text{,}\) but all solutions with initial conditions that are a nonzero point on the unstable plane would move away from the origin.
The straight line through the origin and the point \((1, -1, 2)\) is a stable line. That is, for any initial condition \({\mathbf x}(0) = (x_0, y_0, z_0)\) lying on this line, our solution will tend toward the origin as \(t \to \infty\text{.}\) On the other hand, the plane spanned by \((2, -3, 5)\) and \((5, -7, 13)\) is unstable plane. Solutions on this plane move away from the origin as \(t \to \infty\text{.}\) Of course, \((0, 0, 0)\) is an equlibrium solution for our system. We say that the origin is a saddle in this example (Figure5.8.5).
we have a very different type of unstable equilibrium solution. The eigenvalues of this matrix are \(\lambda = \pm i\) and \(\lambda = -1\text{.}\) Thus, a solution satisfying the initial condition \({\mathbf x}(0) = (x_0, y_0, z_0)\) is given by
\begin{equation*}
{\mathbf x}(t)
=
x_0
\begin{pmatrix}
\cos t \\ - \sin t \\ 0
\end{pmatrix}
+
y_0
\begin{pmatrix}
\sin t \\ \cos t \\ 0
\end{pmatrix}
+ z_0
e^{-t}
\begin{pmatrix}
0 \\ 0 \\ 1
\end{pmatrix}.
\end{equation*}
If \(z_0 = 0\text{,}\) then our initial condition is in the \(xy\)-plane and all of the solutions lie on circles centered at the origin. On the other hand, if \(x_0 = 0\) and \(y_0 = 0\text{,}\) we have a stable line of solutions lying along the \(z\)-axis. In fact, each solution that does not lie on the stable line lies on a cylinder in \({\mathbb R}^3\) given by \(x^2 + y^2 = r^2\) for some constant \(r \gt 0\text{.}\) These solutions spiral towards the circular solution of radius \(r\) in the \(xy\)-plane if \(z_0 \neq 0\) (Figure5.8.7).
we will let \({\mathbf v}_1 = (1, 0, 0)\) and \({\mathbf v}_2 = (0, 1, 0)\text{.}\) Since \({\mathbf v}_3 = (0, 0, 1)\) is an eigenvector for \(\lambda = 1\text{,}\) our system has solution
\begin{equation*}
{\mathbf x}(t)
=
x_0 e^{-t}
\begin{pmatrix}
\cos t \\ - \sin t \\ 0
\end{pmatrix}
+
y_0 e^{-t}
\begin{pmatrix}
\sin t \\ \cos t \\ 0
\end{pmatrix}
+
z_0 e^t
\begin{pmatrix}
0 \\ 0 \\ 1
\end{pmatrix},
\end{equation*}
where \(\mathbf x(0) = (x_0, y_0, z_0)\text{.}\) If \(z_0 = 0\text{,}\) our initial condition lies in the \(xy\)-plane and solution curves spiral in towards the origin. Thus, we have a stable plane. On the other hand, if \(x_0 = 0\) and \(y_0 = 0\) but \(z_0 \neq 0\text{,}\) then our solution approaches \(\pm \infty\) as \(t \to \infty\text{.}\) In this case, the \(z\)-axis is an unstable line (Figure5.8.9).
For an example of a stable equilibrium solution at the origin, consider the system
where \(\lambda_3 \lt \lambda_2 \lt \lambda_1 \lt 0\text{.}\) For an initial condition \((x_0, y_0, z_0)\) with at least one coordinate nonzero, the corresponding solution tends towards the origin tangentially to the \(x\)-axis as \(t \to \infty\) (Figure5.8.10).
Changing the system that in ExampleExample5.8.8 to be
Thus, the eigenvalues of \(A\) are \(\lambda = \pm i\) and \(\lambda = \pm i \sqrt{3}\text{.}\) We can find eigenvectors
\begin{equation*}
{\mathbf v}_1
=
\begin{pmatrix}
i / \sqrt{3} \\ - i / \sqrt{3} \\ -1 \\ 1
\end{pmatrix},
{\mathbf v}_2
=
\begin{pmatrix}
-i / \sqrt{3} \\ i / \sqrt{3} \\ -1 \\ 1
\end{pmatrix},
{\mathbf v}_3
=
\begin{pmatrix}
-i \\ - i \\ 1 \\ 1
\end{pmatrix},
{\mathbf v}_4
=
\begin{pmatrix}
i \\ i \\ 1 \\ 1
\end{pmatrix},
\end{equation*}
corresponding to the eigenvalues \(\lambda_1 = i \sqrt{3}\text{,}\) \(\lambda_2 = - i \sqrt{3}\text{,}\) \(\lambda_3 = i\text{,}\) \(\lambda_4 = - i\text{,}\) respectively. Consequently, the general solution to our system is
however, this form of the solution is not very useful. By examining real and imaginary parts of \(e^{ i \sqrt{3} t} {\mathbf v}_1\) and \(c_1 e^{ i t} {\mathbf v}_3\text{,}\) we can rewrite the solution as
\begin{equation*}
{\mathbf x}(t)
=
c_1
\begin{pmatrix}
\cos \sqrt{3} t \\ - \cos \sqrt{3}t \\ - \sqrt{3} \sin \sqrt{3}t \\ \sqrt{3} \sin \sqrt{3}t
\end{pmatrix}
+
c_2
\begin{pmatrix}
- \sin \sqrt{3} t \\ \sin \sqrt{3}t \\ -\sqrt{3} \cos \sqrt{3} t \\ \sqrt{3} \cos \sqrt{3} t
\end{pmatrix}
+
c_3
\begin{pmatrix}
\cos t \\ \cos t \\ -\sin t \\ - \sin t
\end{pmatrix}
+
c_4
\begin{pmatrix}
\sin t \\ \sin t \\ \cos t \\ \cos t
\end{pmatrix}.
\end{equation*}
As in the case of \({\mathbb R}^2\text{,}\) we can solve the system \({\mathbf x}' = A {\mathbf x}\) by finding eigenvalues and eigenvectors for \(A\text{.}\)
The Principle of Superposition holds for higher-order systems. If \({\mathbf x}_1(t)\) and \({\mathbf x}_2(t)\) are solutions for \({\mathbf x}' = A {\mathbf x}\text{,}\) then
The geometry for a system in \({\mathbb R}^3\) is more complicated than the planar case. However, the solutions are usually characterized by stable lines or stable planes.
Exercise5.8.11
Is it possible for a \(3 \times 3\) to have a line of stable solutions and a plane of unstable solutions? Explain.