Solve the system of differential equations: dot x=-x+y, dot…
Solve the system of differential equations: $$ \dot x=-x+y,\qquad \dot y=-y+4z,\qquad \dot z=x-4z $$
Explanation of Method
The characteristic cubic factors as $-\lambda(\lambda+3)^2$: a zero eigenvalue (giving a genuine equilibrium line) and a repeated negative eigenvalue whose eigenspace is only one-dimensional, requiring a generalised eigenvector.
- Write the matrix $A$ and factor the characteristic cubic.
- Find the eigenvector for $\lambda=0$.
- Find the eigenvector and a generalised eigenvector for the repeated eigenvalue $\lambda=-3$.
- Assemble the general solution and identify the steady state.
Step 1 · Characteristic equation
With $A=\begin{pmatrix}-1&1&0\\0&-1&4\\1&0&-4\end{pmatrix}$, expanding gives
$$ -\lambda(\lambda+3)^2=0, $$
so $\lambda_1=0$ (simple) and $\lambda_{2,3}=-3$ (double).
Step 2 · Eigenvector for $\lambda_1=0$
$$ \mathbf v_1=\begin{pmatrix}4\\4\\1\end{pmatrix}. $$
Step 3 · Eigenvector and generalised eigenvector for $\lambda_{2,3}=-3$
$A+3I$ has rank 2, so the eigenspace is only one-dimensional. Solving $(A+3I)\mathbf v_2=0$ and $(A+3I)\mathbf v_3=\mathbf v_2$ gives
$$ \mathbf v_2=\begin{pmatrix}1\\-2\\1\end{pmatrix},\qquad \mathbf v_3=\begin{pmatrix}1\\-1\\0\end{pmatrix}. $$
Step 4 · General solution
$$ \mathbf u=C_1\mathbf v_1+C_2e^{-3t}\mathbf v_2+C_3e^{-3t}(t\mathbf v_2+\mathbf v_3). $$
Step 5 · Verification and steady state
Direct substitution confirms all three equations hold identically, and as $t\to\infty$ the $e^{-3t}$ terms vanish, leaving $(x,y,z)\to C_1(4,4,1)$ — exactly the null space of $A$. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=0$ (simple), $\lambda_{2,3}=-3$ (double, defective) |
| Equilibrium line | $(x,y,z)=C_1(4,4,1)$ |
| Jordan chain for $\lambda=-3$ | $\mathbf v_2=(1,-2,1)\to\mathbf v_3=(1,-1,0)$ |
Additional Notes
- The zero eigenvalue means the system has a genuine line of equilibria (rather than an isolated equilibrium point) — any point along $(4,4,1)$ is a steady state.
- Since the repeated eigenvalue $\lambda=-3$ is defective (geometric multiplicity 1), a generalised eigenvector is required, contributing a $t\,e^{-3t}$ term to the solution.
- Verified by direct substitution into all three equations and by confirming the long-time limit lands exactly on the null space of $A$.
Result
The general solution of the system is: