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Math 308: Week-in-Review 11 — Solutions

Sections 7.5 – 7.6: Linear Systems and Complex Eigenvalues

Directions: Full worked solutions for Week in Review 11 (Sections 7.5 – 7.6: Linear Systems and Complex Eigenvalues). Click Solution under any problem to reveal the worked answer. Use the Contents panel or the Prev / Up / Next buttons to move between sections, and Search to find terms on this page.

Section 7.5 — Homogeneous Linear Systems with Constant Coefficients

Section 7.6 — Complex Eigenvalues

Problems

1. Section 7.5

Find the general solution, sketch the phase portrait, and determine the stability of the equilibrium point at the origin.

System equation: bold x prime equals the 2-by-2 matrix A times bold x, where A has first row 1 comma 2, and second row negative 1 comma 4.

\[ \mathbf{x}' = \begin{pmatrix} 1 & 2 \\ -1 & 4 \end{pmatrix}\mathbf{x} \]

Step 1 — Characteristic polynomial

For \(A=\begin{pmatrix}1&2\\-1&4\end{pmatrix}\) we have \(\operatorname{tr}(A)=5\) and \(\det(A)=4-(-2)=6\), giving

\[\det(A-\lambda I)=\lambda^2-5\lambda+6=(\lambda-2)(\lambda-3)=0 \quad\Longrightarrow\quad \lambda_1=2,\;\lambda_2=3.\]

Both eigenvalues are positive \(\Rightarrow\) the origin is an unstable node.

Step 2 — Eigenvectors

\(\lambda_1=2\). \(A-2I=\begin{pmatrix}-1&2\\-1&2\end{pmatrix}\). Row 1: \(-v_1+2v_2=0 \Rightarrow v_1=2v_2\). Take \(v_2=1\): \(\mathbf{v}_1=(2,1)^T\). Check: \(A\mathbf{v}_1=(2+2,-2+4)^T=(4,2)^T=2(2,1)^T.\;\checkmark\)

\(\lambda_2=3\). \(A-3I=\begin{pmatrix}-2&2\\-1&1\end{pmatrix}\). Row 1: \(-2v_1+2v_2=0 \Rightarrow v_1=v_2\). Take \(v_2=1\): \(\mathbf{v}_2=(1,1)^T\). Check: \(A\mathbf{v}_2=(1+2,-1+4)^T=(3,3)^T=3(1,1)^T.\;\checkmark\)

Step 3 — General solution and phase portrait

\[\boxed{\mathbf{x}(t)=c_1\begin{pmatrix}2\\1\end{pmatrix}e^{2t} +c_2\begin{pmatrix}1\\1\end{pmatrix}e^{3t}}\]

Portrait: unstable node. Near the origin trajectories are tangent to the slower eigenvector \(\mathbf{v}_1=(2,1)^T\) (\(e^{2t}\) decays more slowly as \(t\to-\infty\)); far from the origin they become tangent to \(\mathbf{v}_2=(1,1)^T\) (\(e^{3t}\) dominates as \(t\to+\infty\)). Stability: unstable.

-3 -2 -1 1 2 3 -3 -2 -1 1 2 3 x₁ x₂ O v₁ v₂

Unstable node: \(\lambda_1=2,\,\lambda_2=3\). All trajectories exit; near origin tangent to \(\mathbf{v}_1=(2,1)^T\).

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2. Section 7.5 (IVP from Question 1)

Using the system from Question 1, solve the initial value problem with initial condition x of zero equals the column vector with entries 2 and 1. \(\mathbf{x}(0) = \begin{pmatrix}2\\1\end{pmatrix}\). Draw the solution curve on the phase portrait below, and sketch the component graphs \(x_1(t)\) and \(x_2(t)\).

Step 1 — Apply the initial condition

Set \(t=0\) in \(\mathbf{x}(t)=c_1(2,1)^T e^{2t}+c_2(1,1)^T e^{3t}\):

\[c_1\begin{pmatrix}2\\1\end{pmatrix} +c_2\begin{pmatrix}1\\1\end{pmatrix} =\begin{pmatrix}2\\1\end{pmatrix} \quad\Longrightarrow\quad \begin{cases}2c_1+c_2=2\\c_1+c_2=1\end{cases}.\]

Subtracting: \(c_1=1\), then \(c_2=0\).

Step 2 — IVP solution and component graphs

\[\boxed{\mathbf{x}(t)=\begin{pmatrix}2e^{2t}\\e^{2t}\end{pmatrix}, \qquad x_1(t)=2e^{2t},\quad x_2(t)=e^{2t}.}\]

Since \(c_2=0\) the trajectory lies on the \(\mathbf{v}_1=(2,1)^T\) eigenvector ray (the half-line \(x_2=\tfrac{1}{2}x_1,\; x_1>0\)), moving away from the origin. Both component graphs are exponentially growing.

-3 -2 -1 1 2 3 -3 -2 -1 1 2 3 x₁ x₂ O x(0)=(2,1)

IVP solution \(\mathbf{x}(t)=(2,1)^Te^{2t}\): amber ray along \(\mathbf{v}_1\).

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3. Section 7.5

Find the general solution, sketch the phase portrait, and determine the stability of the equilibrium point at the origin.

System: x-sub-1 prime equals negative 5 x-sub-1 plus 4 x-sub-2; x-sub-2 prime equals three-halves x-sub-1 minus 4 x-sub-2.

\[ \begin{aligned} x_1' &= -5\,x_1 + 4\,x_2 \\[4pt] x_2' &= \tfrac{3}{2}\,x_1 - 4\,x_2 \end{aligned} \]

Step 1 — Matrix form and eigenvalues

\(A=\begin{pmatrix}-5&4\\\frac{3}{2}&-4\end{pmatrix}\), \(\operatorname{tr}(A)=-9\), \(\det(A)=20-6=14\).

\[\lambda^2+9\lambda+14=(\lambda+2)(\lambda+7)=0 \quad\Longrightarrow\quad \lambda_1=-2,\;\lambda_2=-7.\]

Both eigenvalues are negative \(\Rightarrow\) stable node.

Step 2 — Eigenvectors

\(\lambda_1=-2\). \(A+2I=\begin{pmatrix}-3&4\\\frac{3}{2}&-2\end{pmatrix}\). \(-3v_1+4v_2=0 \Rightarrow v_1=\tfrac{4}{3}v_2\). Take \(v_2=3\): \(\mathbf{v}_1=(4,3)^T\). Check: \(A\mathbf{v}_1=(-20+12,6-12)^T=(-8,-6)^T=-2(4,3)^T.\;\checkmark\)

\(\lambda_2=-7\). \(A+7I=\begin{pmatrix}2&4\\\frac{3}{2}&3\end{pmatrix}\). \(2v_1+4v_2=0 \Rightarrow v_1=-2v_2\). Take \(v_2=1\): \(\mathbf{v}_2=(-2,1)^T\). Check: \(A\mathbf{v}_2=(10+4,-3-4)^T=(14,-7)^T=-7(-2,1)^T.\;\checkmark\)

Step 3 — General solution and phase portrait

\[\boxed{\mathbf{x}(t) =c_1\begin{pmatrix}4\\3\end{pmatrix}e^{-2t} +c_2\begin{pmatrix}-2\\1\end{pmatrix}e^{-7t}}\]

Portrait: stable node. As \(t\to\infty\), \(e^{-7t}\) decays faster, so all trajectories approach origin tangent to the slower eigenvector \(\mathbf{v}_1=(4,3)^T\). Stability: asymptotically stable.

-3 -2 -1 1 2 3 -3 -2 -1 1 2 3 x₁ x₂ O v₁ v₂

Stable node: \(\lambda_1=-2,\,\lambda_2=-7\). All trajectories approach origin tangent to \(\mathbf{v}_1=(4,3)^T\).

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4. Section 7.6

Find the general solution, sketch the phase portrait, and determine the stability of the equilibrium point at the origin.

System: x prime equals x plus 2y; y prime equals negative 2x plus 5y.

\[ \begin{aligned} x' &= x + 2y \\[4pt] y' &= -2x + 5y \end{aligned} \]

Note: this problem has a repeated eigenvalue with only one eigenvector (defective matrix). The second independent solution uses a generalized eigenvector from Section 7.8.

Step 1 — Characteristic polynomial

\(A=\begin{pmatrix}1&2\\-2&5\end{pmatrix}\), \(\operatorname{tr}(A)=6\), \(\det(A)=5+4=9\).

\[\lambda^2-6\lambda+9=(\lambda-3)^2=0 \quad\Longrightarrow\quad \lambda=3\text{ (repeated)}.\]

Step 2 — Eigenvector

\(A-3I=\begin{pmatrix}-2&2\\-2&2\end{pmatrix}\), rank 1 (defective). \(-2v_1+2v_2=0 \Rightarrow v_1=v_2\). Take \(\mathbf{v}=(1,1)^T\). First solution: \(\mathbf{w}_1(t)=e^{3t}(1,1)^T\).

Step 3 — Generalized eigenvector \(\mathbf{u}\) satisfying \((A-3I)\mathbf{u}=\mathbf{v}\)

\[\begin{pmatrix}-2&2\\-2&2\end{pmatrix} \begin{pmatrix}u_1\\u_2\end{pmatrix} =\begin{pmatrix}1\\1\end{pmatrix} \quad\Longrightarrow\quad -2u_1+2u_2=1.\]

Set \(u_1=0 \Rightarrow u_2=\tfrac{1}{2}\). So \(\mathbf{u}=(0,\tfrac{1}{2})^T\). Second solution: \(\mathbf{w}_2(t)=e^{3t}(t\,\mathbf{v}+\mathbf{u})\).

Step 4 — General solution and phase portrait

\[\boxed{\mathbf{x}(t)=c_1\begin{pmatrix}1\\1\end{pmatrix}e^{3t} +c_2\,e^{3t}\!\left(t\begin{pmatrix}1\\1\end{pmatrix} +\begin{pmatrix}0\\1/2\end{pmatrix}\right)}\]

Portrait: unstable deficient (improper) node (\(\lambda=3>0\), one eigenvector). All trajectories become asymptotically tangent to \(\mathbf{v}=(1,1)^T\) and flee to infinity. Stability: unstable.

-3 -2 -1 1 2 3 -3 -2 -1 1 2 3 x y O v

Unstable deficient node: \(\lambda=3\) repeated. One eigenvector ray; all trajectories asymptotically tangent to \(\mathbf{v}=(1,1)^T\).

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5. Section 7.6

Find the general solution, sketch the phase portrait, and determine the stability of the equilibrium point at the origin.

System: x prime equals 3x plus y; y prime equals negative 2x plus y.

\[ \begin{aligned} x' &= 3x + y \\[4pt] y' &= -2x + y \end{aligned} \]

Step 1 — Characteristic polynomial

\(A=\begin{pmatrix}3&1\\-2&1\end{pmatrix}\), \(\operatorname{tr}(A)=4\), \(\det(A)=3+2=5\).

\[\lambda^2-4\lambda+5=0 \quad\Longrightarrow\quad \lambda=\frac{4\pm\sqrt{16-20}}{2}=2\pm i.\]

Here \(\alpha=2>0\), \(\beta=1\) \(\Rightarrow\) unstable spiral.

Step 2 — Complex eigenvector for \(\lambda_1=2+i\)

\(A-(2+i)I=\begin{pmatrix}1-i&1\\-2&-1-i\end{pmatrix}\). Row 1: \((1-i)v_1+v_2=0 \Rightarrow v_2=(i-1)v_1\). Take \(v_1=1\), \(v_2=i-1\):

\[\mathbf{v}=\begin{pmatrix}1\\i-1\end{pmatrix} =\begin{pmatrix}1\\-1\end{pmatrix}+i\begin{pmatrix}0\\1\end{pmatrix} =\mathbf{a}+i\mathbf{b}.\]

Step 3 — Real fundamental solutions

\[\mathbf{w}_1(t)=e^{2t}(\mathbf{a}\cos t-\mathbf{b}\sin t) =e^{2t}\begin{pmatrix}\cos t\\-\cos t-\sin t\end{pmatrix},\]
\[\mathbf{w}_2(t)=e^{2t}(\mathbf{a}\sin t+\mathbf{b}\cos t) =e^{2t}\begin{pmatrix}\sin t\\-\sin t+\cos t\end{pmatrix}.\]

Step 4 — General solution and phase portrait

\[\boxed{\mathbf{x}(t) =c_1\,e^{2t}\begin{pmatrix}\cos t\\-\cos t-\sin t\end{pmatrix} +c_2\,e^{2t}\begin{pmatrix}\sin t\\-\sin t+\cos t\end{pmatrix}}\]

Rotation direction: \(A(1,0)^T=(3,-2)^T\); second component \(-2<0 \Rightarrow\) clockwise. Stability: unstable.

-3 -2 -1 1 2 3 -3 -2 -1 1 2 3 x y O

Unstable spiral: \(\lambda=2\pm i\), \(\alpha=2>0\). Clockwise, outward.

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6. Section 7.6

Find the general solution, sketch the phase portrait, and determine the stability of the equilibrium point at the origin.

System: bold x prime equals A bold x where A has first row 0 comma negative 2 and second row 2 comma 0.

\[ \mathbf{x}' = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}\mathbf{x} \]

Step 1 — Characteristic polynomial

\(A=\begin{pmatrix}0&-2\\2&0\end{pmatrix}\), \(\operatorname{tr}(A)=0\), \(\det(A)=4\).

\[\lambda^2+4=0\quad\Longrightarrow\quad\lambda=\pm 2i.\]

Purely imaginary (\(\alpha=0\), \(\beta=2\)) \(\Rightarrow\) center.

Step 2 — Complex eigenvector for \(\lambda_1=2i\)

\(A-2iI=\begin{pmatrix}-2i&-2\\2&-2i\end{pmatrix}\). Row 2: \(2v_1-2iv_2=0 \Rightarrow v_1=iv_2\). Take \(v_2=1\): \(\mathbf{v}=(i,1)^T = \mathbf{a}+i\mathbf{b}\) with \(\mathbf{a}=(0,1)^T\), \(\mathbf{b}=(1,0)^T\).

Step 3 — Real fundamental solutions

\[\mathbf{w}_1(t)=\mathbf{a}\cos(2t)-\mathbf{b}\sin(2t) =\begin{pmatrix}-\sin(2t)\\\cos(2t)\end{pmatrix},\qquad \mathbf{w}_2(t)=\begin{pmatrix}\cos(2t)\\\sin(2t)\end{pmatrix}.\]

Step 4 — General solution and phase portrait

\[\boxed{\mathbf{x}(t) =c_1\begin{pmatrix}-\sin(2t)\\\cos(2t)\end{pmatrix} +c_2\begin{pmatrix}\cos(2t)\\\sin(2t)\end{pmatrix}}\]

Every trajectory is a circle centred at the origin with radius \(\sqrt{c_1^2+c_2^2}\) (since \(A\) is a pure rotation-dilation: \(\|\mathbf{x}(t)\|^2=c_1^2+c_2^2=\text{const}\)). Period \(T=2\pi/\beta=\pi\). Rotation direction: \(A(1,0)^T=(0,2)^T\); second component \(2>0 \Rightarrow\) counterclockwise. Stability: stable but not asymptotically stable.

-3 -2 -1 1 2 3 -3 -2 -1 1 2 3 x₁ x₂ O

Center: \(\lambda=\pm 2i\). Closed circles, counterclockwise.

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7. Section 7.6

Find the general solution, sketch the phase portrait, and determine the stability of the equilibrium point at the origin. Then solve the initial value problem with initial condition x of 0 equals the column vector with entries negative 1 and 2. \(\mathbf{x}(0) = \begin{pmatrix}-1\\2\end{pmatrix}\).

System: bold x prime equals A bold x where A has first row 1 comma negative 8 and second row 1 comma negative 3.

\[ \mathbf{x}' = \begin{pmatrix} 1 & -8 \\ 1 & -3 \end{pmatrix}\mathbf{x} \]

Step 1 — Characteristic polynomial

\(A=\begin{pmatrix}1&-8\\1&-3\end{pmatrix}\), \(\operatorname{tr}(A)=-2\), \(\det(A)=-3+8=5\).

\[\lambda^2+2\lambda+5=0 \quad\Longrightarrow\quad \lambda=\frac{-2\pm\sqrt{4-20}}{2}=-1\pm 2i.\]

\(\alpha=-1<0\), \(\beta=2\) \(\Rightarrow\) stable spiral.

Step 2 — Complex eigenvector for \(\lambda_1=-1+2i\)

\(A-(-1+2i)I=\begin{pmatrix}2-2i&-8\\1&-2-2i\end{pmatrix}\). Row 2: \(v_1+(-2-2i)v_2=0 \Rightarrow v_1=(2+2i)v_2\). Take \(v_2=1\): \(\mathbf{v}=(2+2i,1)^T = \mathbf{a}+i\mathbf{b}\) with \(\mathbf{a}=(2,1)^T\), \(\mathbf{b}=(2,0)^T\).

Step 3 — Real fundamental solutions

\[\mathbf{w}_1(t)=e^{-t}(\mathbf{a}\cos 2t-\mathbf{b}\sin 2t) =e^{-t}\begin{pmatrix}2\cos 2t-2\sin 2t\\\cos 2t\end{pmatrix},\]
\[\mathbf{w}_2(t)=e^{-t}(\mathbf{a}\sin 2t+\mathbf{b}\cos 2t) =e^{-t}\begin{pmatrix}2\sin 2t+2\cos 2t\\\sin 2t\end{pmatrix}.\]

Rotation direction: \(A(1,0)^T=(1,1)^T\); second component \(1>0 \Rightarrow\) counterclockwise. Stability: asymptotically stable.

Step 4 — Solve the IVP \(\mathbf{x}(0)=(-1,2)^T\)

\(\mathbf{w}_1(0)=(2,1)^T\) and \(\mathbf{w}_2(0)=(2,0)^T\):

\[c_1\begin{pmatrix}2\\1\end{pmatrix} +c_2\begin{pmatrix}2\\0\end{pmatrix} =\begin{pmatrix}-1\\2\end{pmatrix} \quad\Longrightarrow\quad \begin{cases}2c_1+2c_2=-1\\c_1=2\end{cases} \quad\Longrightarrow\quad c_1=2,\;c_2=-\tfrac{5}{2}.\]
\[x_1(t)=e^{-t}\!\left[2(2\cos 2t-2\sin 2t)+\bigl(-\tfrac{5}{2}\bigr)(2\sin 2t+2\cos 2t)\right] =e^{-t}(-\cos 2t-9\sin 2t),\]
\[x_2(t)=e^{-t}\!\left[2\cos 2t+\bigl(-\tfrac{5}{2}\bigr)\sin 2t\right] =e^{-t}\!\left(2\cos 2t-\tfrac{5}{2}\sin 2t\right).\]
\[\boxed{\mathbf{x}(t)=e^{-t}\begin{pmatrix}-\cos 2t-9\sin 2t\\2\cos 2t-\frac{5}{2}\sin 2t\end{pmatrix}}\]

Check: \(x_1(0)=-1\;\checkmark\), \(x_2(0)=2\;\checkmark\).

-3 -2 -1 1 2 3 -3 -2 -1 1 2 3 x₁ x₂ O x(0)

Stable spiral: \(\lambda=-1\pm 2i\), \(\alpha=-1<0\). Counterclockwise inward. IVP (amber) starts at \((-1,2)\).

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8. Sections 7.5–7.6 (Parameter Analysis)

Classify the type and stability of the equilibrium point at the origin for all values of the parameter \(\alpha\). Identify the critical values of \(\alpha\) at which the portrait type changes.

The system is: bold x prime equals A of alpha bold x, where A of alpha has first row alpha minus 1 comma alpha plus 1, and second row negative two-thirds comma one-third.

\[ \mathbf{x}' = \begin{pmatrix} \alpha - 1 & \alpha + 1 \\[4pt] -\dfrac{2}{3} & \dfrac{1}{3} \end{pmatrix}\mathbf{x} \]
Hint: compute the trace \(\tau(\alpha) = \operatorname{tr}(A)\) and determinant \(\Delta(\alpha) = \det(A)\) as functions of \(\alpha\). Use the discriminant \(\tau^2 - 4\Delta\) to distinguish nodes, saddles, spirals, and centers.

Step 1 — Compute \(\tau(\alpha)\) and \(\Delta(\alpha)\)

\[\tau(\alpha)=\operatorname{tr}(A)=(\alpha-1)+\tfrac{1}{3}=\alpha-\tfrac{2}{3},\]
\[\Delta(\alpha)=\det(A) =(\alpha-1)\cdot\tfrac{1}{3}-(\alpha+1)\cdot\!\left(-\tfrac{2}{3}\right) =\frac{(\alpha-1)+2(\alpha+1)}{3}=\alpha+\tfrac{1}{3}.\]

Key simplification: \(\Delta = \tau + 1\)

Since \(\Delta-\tau=(\alpha+\tfrac{1}{3})-(\alpha-\tfrac{2}{3})=1\), we have \(\Delta=\tau+1\). This lets us express everything in terms of \(\tau\) alone.

Step 2 — Critical values (in terms of \(\tau\))

  • Saddle/node boundary \((\Delta=0)\): \(\tau+1=0 \Longrightarrow \tau=-1\)  \((\alpha=-\tfrac{1}{3})\).
  • Stability boundary \((\tau=0)\): \(\alpha=\tfrac{2}{3}\), \(\Delta=1>0\) \(\Rightarrow\) center.
  • Node/spiral boundary \((\tau^2-4\Delta=0)\): substituting \(\Delta=\tau+1\) gives \[\tau^2-4(\tau+1)=0 \Longrightarrow \tau^2-4\tau-4=0 \Longrightarrow \tau=2\pm 2\sqrt{2}.\] Corresponding \(\alpha\) values: \(\alpha=\tau+\tfrac{2}{3}=\dfrac{8\pm 6\sqrt{2}}{3}\).

Step 3 — Complete classification table

Classification of the equilibrium at the origin as a function of \(\alpha\)
Range of \(\alpha\) \(\operatorname{sgn}(\tau)\), \(\operatorname{sgn}(\Delta)\) Eigenvalue type Portrait Stability
\(\alpha < -\tfrac{1}{3}\)\(\Delta < 0\)Real, opposite signsSaddleUnstable
\(-\tfrac{1}{3}<\alpha<\dfrac{8-6\sqrt{2}}{3}\)\(\tau<0,\,\Delta>0,\,\tau^2>4\Delta\)Real distinct, both \(<0\)Stable nodeAsymp. stable
\(\dfrac{8-6\sqrt{2}}{3}<\alpha<\dfrac{2}{3}\)\(\tau<0,\,\Delta>0,\,\tau^2<4\Delta\)Complex, \(\alpha<0\)Stable spiralAsymp. stable
\(\alpha=\dfrac{2}{3}\)\(\tau=0,\,\Delta=1>0\)Purely imaginaryCenterStable (not asymp.)
\(\dfrac{2}{3}<\alpha<\dfrac{8+6\sqrt{2}}{3}\)\(\tau>0,\,\Delta>0,\,\tau^2<4\Delta\)Complex, \(\alpha>0\)Unstable spiralUnstable
\(\alpha>\dfrac{8+6\sqrt{2}}{3}\)\(\tau>0,\,\Delta>0,\,\tau^2>4\Delta\)Real distinct, both \(>0\)Unstable nodeUnstable

At \(\alpha=\tfrac{8\pm 6\sqrt{2}}{3}\): repeated eigenvalue, node–spiral transition.

Δ-τ stability diagram — Q8 intersection points The line Delta=tau+1 (maroon) crosses four labelled points on the parabola and axes. See the reference table below for coordinates and significance of each point. τ Δ −2 2 4 6 8 0 2 4 6 8 −1 Δ = τ²/4 Δ = τ+1 Saddle stable node stable spiral unstable spiral unstable node center 1 2 3 4
# \(\tau\) \(\Delta = \tau+1\) Corresponding \(\alpha\) Significance
1 \(\tau=-1\) \(\Delta=0\) \(\alpha=-\tfrac{1}{3}\) Line crosses \(\tau\)-axis (\(\Delta=0\)): saddle → stable node boundary
2 \(\tau=2-2\sqrt{2}\approx-0.83\) \(\Delta=3-2\sqrt{2}\approx0.17\) \(\alpha=\dfrac{8-6\sqrt{2}}{3}\) Line meets parabola (\(\Delta=\tau^2/4\)): stable node → stable spiral
3 \(\tau=0\) \(\Delta=1\) \(\alpha=\tfrac{2}{3}\) Line crosses \(\Delta\)-axis (\(\tau=0\)): eigenvalues purely imaginary — center
4 \(\tau=2+2\sqrt{2}\approx4.83\) \(\Delta=3+2\sqrt{2}\approx5.83\) \(\alpha=\dfrac{8+6\sqrt{2}}{3}\) Line meets parabola (\(\Delta=\tau^2/4\)): unstable spiral → unstable node
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