Complex conjugate eigenvalues case

For a real constant-coefficient linear system ${\mathbf{x}}^{\prime }=A\mathbf{x}$, complex eigenvalues come in conjugate pairs: if $\lambda =\alpha +i\beta$ is an eigenvalue, so is $\overline{\lambda }=\alpha -i\beta$. The corresponding eigenvectors are also complex conjugates: $\mathbf{u}=\mathbf{a}+i\mathbf{b}$ and $\overline{\mathbf{u}}=\mathbf{a}-i\mathbf{b}$.

The general solution can be written as a real-valued linear combination using sines and cosines.

Real-valued solutions

Two linearly independent real-valued solutions for the conjugate pair $\lambda =\alpha \pm i\beta$:

${\mathbf{x}}_1(t)=e^{\alpha t}[\cos (\beta t)\mathbf{a}-\sin (\beta t)\mathbf{b}]$

${\mathbf{x}}_2(t)=e^{\alpha t}[\sin (\beta t)\mathbf{a}+\cos (\beta t)\mathbf{b}]$

where $\mathbf{u}=\mathbf{a}+i\mathbf{b}$ is the eigenvector for $\alpha +i\beta$.

The general solution combines the contributions from all eigenvalue pairs.

Why this works

The complex-valued solution $\mathbf{x}(t)=\mathbf{u}{e}^{\lambda t}=(\mathbf{a}+i\mathbf{b})e^{(\alpha +i\beta )t}$ expands using Euler’s formula:

$\mathbf{x}(t)=(\mathbf{a}+i\mathbf{b})e^{\alpha t}(\cos (\beta t)+i\sin (\beta t))$

Real part: $e^{\alpha t}[\mathbf{a}\cos (\beta t)-\mathbf{b}\sin (\beta t)]$.

Imaginary part: $e^{\alpha t}[\mathbf{a}\sin (\beta t)+\mathbf{b}\cos (\beta t)]$.

For a real-coefficient ODE, real and imaginary parts of any complex solution are themselves real solutions. So the two real-valued solutions above are real and linearly independent (their Wronskian is non-zero whenever $\beta \ne 0$).

Worked example

Solve ${\mathbf{x}}^{\prime }=A\mathbf{x}$ where $A=\left[\begin{array}{ccc}1& 2& -1\\ 0& 1& 1\\ 0& -1& 1\end{array}\right]$.

Eigenvalues (by computing $\det (A-\lambda I)=0$):

${\lambda }_1=1$ (real), ${\lambda }_2=1+i$, ${\lambda }_3=1-i$ (complex conjugate pair).

Eigenvector for ${\lambda }_1=1$:

${\mathbf{u}}_1=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$

Gives solution ${\mathbf{x}}_1(t)=e^t{\mathbf{u}}_1$.

Eigenvector for ${\lambda }_2=1+i$:

${\mathbf{u}}_2=\left[\begin{array}{c}-2\\ 0\\ 1\end{array}\right]+i\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]=\mathbf{a}+i\mathbf{b}$

So $\alpha =1$, $\beta =1$, $\mathbf{a}=(-2,0,1{)}^T$, $\mathbf{b}=(1,-1,0{)}^T$.

Two real solutions:

${\mathbf{x}}_2(t)=e^t\left[\cos t\left[\begin{array}{c}-2\\ 0\\ 1\end{array}\right]-\sin t\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]\right]$

${\mathbf{x}}_3(t)=e^t\left[\sin t\left[\begin{array}{c}-2\\ 0\\ 1\end{array}\right]+\cos t\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]\right]$

The eigenvector for ${\lambda }_3=1-i$ is ${\overline{\mathbf{u}}}_2=\mathbf{a}-i\mathbf{b}$, which would give the same two real solutions (just rearranged) — no new information. So we stop at three real solutions.

General solution:

$\mathbf{x}(t)=c_1{\mathbf{x}}_1(t)+c_2{\mathbf{x}}_2(t)+c_3{\mathbf{x}}_3(t)$

Behavior

The factor $e^{\alpha t}$ controls amplitude; $\cos (\beta t)$ and $\sin (\beta t)$ control oscillation. Three regimes:

• $\alpha <0$: amplitude decays. Trajectories spiral into origin. Asymptotically stable spiral.
• $\alpha =0$: amplitude constant. Trajectories form closed orbits. Center (stable but not asymptotically stable).
• $\alpha >0$: amplitude grows. Trajectories spiral outward. Unstable spiral.

The frequency of oscillation is $\beta$ — the imaginary part of the eigenvalue. The period is $2\pi /\beta$.

For the corresponding 2D phase portraits, see Phase plane behaviour cases 5 (spiral) and 6 (center).

For other eigenvalue scenarios, see Distinct real eigenvalues case and Repeated eigenvalues case.