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5. Constant-coefficient problems

While the theory of homogeneous linear ODE systems is straightforward, the good news for the general case pretty much ends there. In most problems we cannot get a reasonably simple expression for a fundamental matrix. But there is a very important special case in which we can say a lot: when the coefficient matrix is constant.

Accordingly, we now specialize to

\[ \bfx' = \bfA \bfx, \]

where \(\bfA\) is an \(n\times n\) constant matrix. We are finally ready to see the significance of the eigenvalue condition

\[\mathbf{A} \mathbf{v} = \lambda \mathbf{v}.\]

If we seek a solution in the form \(\mathbf{x}(t)=g(t)\mathbf{v}\) for this eigenvector, then

\[ g'\mathbf{v} = \mathbf{A}(g\mathbf{v}) = g\cdot (\mathbf{A}\mathbf{v}) = (g\lambda) \mathbf{v}, \]

which can be satisfied if \(g'=\lambda g\). That is, we have a solution,

\[ \bfx(t) = e^{\lambda t} \bfv. \]

5.1. General solution

Counting algebraic multiplicities, we know that \(\mathbf{A}\) has eigenvalues \(\lambda_1,\ldots,\lambda_n\). Say that we have eigenvectors \(\mathbf{v}_1,\ldots,\mathbf{v}_n\) to go with them. Then we have \(n\) homogeneous solutions

\[\mathbf{x}_j(t) = e^{t \lambda_j} \mathbf{v}_j, \quad j=1,\ldots,n.\]

Our next move is to form

\[ \bfX(t) = \bigl[ \mathbf{x}_1(t) \; \mathbf{x}_2(t) \; \cdots \; \mathbf{x}_m(t) \bigr] \]

and determine whether this is a fundamental matrix. According to Abel’s theorem, we can ask that question at any value of \(t\), including \(t=0\). So the key issue is whether

(5.4)\[\bfV = \bigl[ \mathbf{v}_1 \; \mathbf{v}_2 \; \cdots \; \mathbf{v}_m \bigr]\]

is invertible. If so, then we have the ingredients of the general homogeneous solution.

Theorem 5.8 (Solution by eigenvectors)

Let \(\mathbf{A}\) have eigenvalues \(\lambda_1,\ldots,\lambda_n\) and corresponding eigenvectors \(\mathbf{v}_1,\ldots,\mathbf{v}_n\). If the matrix \(\bfV\) in (5.4) is invertible, then

(5.5)\[\bfX(t) = \begin{bmatrix} e^{t \lambda_1} \mathbf{v}_1 & \cdots & e^{t \lambda_n} \mathbf{v}_n \end{bmatrix}\]

is a fundamental matrix for \(\mathbf{x}'=\mathbf{A}\mathbf{x}\). Hence the general solution can be expressed as

(5.6)\[\mathbf{x}(t) = \bfX(t)\mathbf{c} = c_1 e^{t \lambda_1} \mathbf{v}_1 + \cdots + c_n e^{t \lambda_n} \mathbf{v}_n.\]

Example

Given that

\[\begin{split}\mathbf{A} = \begin{bmatrix} 1 & 1 \\ 4 & 1 \end{bmatrix}\end{split}\]

has eigenpairs

\[ \lambda_1=3,\: \mathbf{v}_1 = \twovec{1}{2}, \quad \lambda_2=-1,\: \mathbf{v}_2=\twovec{1}{-2},\]

find the general solution of \(\bfx'=\bfA\bfx\).

Solution

The determinant of

\[ \bfV = \twomat{1}{1}{2}{-2} \]

is \(-4\), so this matrix is not singular. Hence a general solution is

\[\mathbf{x}(t) = c_1 e^{3t} \twovec{1}{2} + c_2 e^{-t} \twovec{1}{-2} = \twovec{ c_1 e^{3t} + c_2 e^{-t}}{2c_1 e^{3t} - 2 c_2e^{-t}}. \]

We can often skip the singularity check in the previous example.

Theorem 5.9

If the eigenvalues of \(\bfA\) are distinct, then their corresponding eigenvectors are the columns of an invertible matrix.

We will return later to the situation of a repeated eigenvalue.

5.2. Complex eigenvalues

Example

Find the general solution of

\[\bfx' = \twomat{1}{1}{-1}{1} \bfx.\]
Solution

In the first chapter we found the eigenvalues \(1\pm i\) and corresponding eigenvectors \(\twovec{1}{\pm i}\). This leads to the solution

\[ \bfx(t) = c_1 e^{(1+i)t} \twovec{1}{i} + c_2 e^{(1-i)t} \twovec{1}{-i}. \]

If we expect a real solution (because of real initial values, say), then \(c_2 = \overline{c_1}\) and we can instead write

\[\begin{split} \begin{split} \bfx(t) &= \Re \left[ c_1 e^{(1+i)t} \twovec{1}{i} \right] \\ &= e^t \Re \left[ c_1 e^{it} \twovec{1}{i} \right]. \end{split} \end{split}\]

Heading toward a completely real form, decompose \(c_1= a_1 - i a_2\) and note that

\[ e^{it} \twovec{1}{i} = \twovec{\cos t + i \sin t}{i\cos t - \sin t} = \twovec{\cos t}{-\sin t} + i \twovec{\sin t}{\cos t}. \]

Then

\[\begin{split} \begin{split} c_1 &= (a_1 -i a_2)\left(\twovec{\cos t}{-\sin t} + i \twovec{\sin t}{\cos t} \right) \\ &= a_1 \twovec{\cos t}{-\sin t} + a_2 \twovec{\sin t}{\cos t} + \cdots, \end{split} \end{split}\]

where everything omitted is purely imaginary. Since we only want the real part, we get

\[ \bfx(t) = a_1 \twovec{\cos t}{-\sin t} + a_2 \twovec{\sin t}{\cos t} \]

for arbitrary real \(a_1\) and \(a_2\).

The essence of the example is that to convert the solution to an entirely real expression, decompose the complex eigensolution into real and imaginary parts. Suppose that

\[ \lambda = r + i\omega, \qquad \bfv = \bfu_1 + i\, \bfu_2. \]

Then

\[ e^{\lambda t} \bfv = e^{rt} (\cos(\omega t) + i\sin(\omega t) ) (\bfu_1 + i\, \bfu_2). \]

The real and imaginary parts of the above product form the basis of the real solution. Specifically,

(5.7)\[\bfx(t) = a_1 e^{rt} \Bigl( \cos(\omega t)\,\bfu_1 - \sin(\omega t) \,\bfu_2(t) \Bigr) + a_2 e^{rt} \Bigl( \cos(\omega t)\,\bfu_2 + \sin(\omega t) \,\bfu_1(t) \Bigr),\]

for arbitrary real \(a_1\) and \(a_2\).

5.3. The oscillator reloaded

In an earlier section we showed that the harmonic oscillator

\[ u'' + 2 Z \omega_0 u' + \omega_0^2 u = 0 \]

can be converted to the constant-coefficient system

\[ \bfx' = \twomat{0}{1}{-\omega_0^2}{-2Z\omega_0} \bfx. \]

via \(x_1=u\), \(x_2=u'\). The eigenvalues of the coefficient matrix are the roots of the characteristic polynomial

\[ \twodet{-\lambda}{1}{-\omega_0^2}{-2 Z\omega_0-\lambda} = \lambda^2 + 2 Z\omega_0 \lambda + \omega_0^2. \]

This is also what we called the characteristic polynomial of the oscillator ODE! That is, the characteristic values we used there are the eigenvalues here. The exponential solutions we saw before also come from the first component of the general vector solution. To repeat, all second-order problems are really first-order systems in a cheap disguise.

4. Homogeneous solutions 6. Stability and the phase plane