This note covers pairs of linked differential equations, called coupled systems, and second-order equations. You will solve linear systems using eigenvalues, classify the behaviour near the origin, and use Euler's method for both coupled and second-order equations. These models describe interacting populations, mixing tanks and vibrating systems.
Key ideas
Coupled systems. Two quantities \(x(t)\) and \(y(t)\) are coupled when the rate of change of each depends on both. A linear system has the form \[\frac{dx}{dt}=ax+by,\qquad \frac{dy}{dt}=cx+dy,\] which is written in matrix form as \(\dfrac{d}{dt}\begin{pmatrix}x\\y\end{pmatrix}=M\begin{pmatrix}x\\y\end{pmatrix}\) with \(M=\begin{pmatrix}a&b\\c&d\end{pmatrix}\). The origin \((0,0)\) is an equilibrium point, because both rates are zero there.
Euler's method for a system. Use the old values of both variables to update both at each step: \[x_{n+1}=x_n+h\,f(x_n,y_n),\qquad y_{n+1}=y_n+h\,g(x_n,y_n),\] where \(\dfrac{dx}{dt}=f\) and \(\dfrac{dy}{dt}=g\). Do not use the new \(x\) when finding the new \(y\).
Solving by eigenvalues. Find the eigenvalues \(\lambda\) of \(M\) from \(\det(M-\lambda I)=0\), then an eigenvector \(\mathbf v\) for each from \((M-\lambda I)\mathbf v=\mathbf 0\). You may use the GDC to check these. If the eigenvalues \(\lambda_1\ne\lambda_2\) are real, the general solution is \[\begin{pmatrix}x\\y\end{pmatrix}=Ae^{\lambda_1t}\mathbf v_1+Be^{\lambda_2t}\mathbf v_2,\] where \(A\) and \(B\) are found from the starting values \(x(0)\) and \(y(0)\).
Behaviour near the origin. The eigenvalues tell you what trajectories in the \((x,y)\) plane do:
- Both real and negative: stable node. Trajectories move towards the origin.
- Both real and positive: unstable node. Trajectories move away.
- Real with opposite signs: saddle. Trajectories approach along one eigenvector direction, then leave along the other. This is unstable.
- Complex with negative real part: stable spiral. Complex with positive real part: unstable spiral.
- Purely imaginary: a centre, with closed loops around the origin.
Second-order equations. An equation with \(\dfrac{d^2x}{dt^2}\) can be turned into a coupled pair. Let \(y=\dfrac{dx}{dt}\). Then \(\dfrac{dx}{dt}=y\), and \(\dfrac{dy}{dt}\) is the rest of the equation. For example, \(\dfrac{d^2x}{dt^2}=-4x-0.5\dfrac{dx}{dt}\) becomes \(\dfrac{dx}{dt}=y,\ \dfrac{dy}{dt}=-4x-0.5y\). You can then use Euler's method or the eigenvalues of the matrix, with \(x\) as the position and \(y\) as the velocity.
Phase portraits. A phase portrait plots \(y\) against \(x\), with arrows showing the direction of motion as \(t\) increases. For real eigenvalues, the lines through the origin in the directions of the eigenvectors are special trajectories. A trajectory that starts on one of these lines stays on it. For a saddle, the line with the negative eigenvalue brings points in, while the line with the positive eigenvalue sends them out. Every other trajectory curves between these lines and eventually follows the direction of the larger eigenvalue.
Reading a system in context. The signs of the coefficients tell a story. In Example 1, the \(-0.3x\) term says amount \(x\) drains out of lake 1, and the \(+0.3x\) term in the second equation says it flows into lake 2. The \(-0.2y\) term includes both flow back to lake 1 and a loss from the system, which is why the totals fall to zero.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\dfrac{d\mathbf X}{dt}=M\mathbf X\) | Matrix form of a linear coupled system. |
| \(\det(M-\lambda I)=0\) | Equation for the eigenvalues of \(M\). |
| \((M-\lambda I)\mathbf v=\mathbf 0\) | Gives the eigenvector for a given eigenvalue. |
| \(\mathbf X=Ae^{\lambda_1t}\mathbf v_1+Be^{\lambda_2t}\mathbf v_2\) | General solution for distinct real eigenvalues. |
| \(x_{n+1}=x_n+hf,\ y_{n+1}=y_n+hg\) | Euler step for a system. |
| \(y=\dfrac{dx}{dt}\) | Substitution that turns a second-order equation into a pair. |
Worked examples
Example 1 Two lakes exchange a pollutant. Their amounts \(x\) and \(y\) (tonnes) satisfy \(\dfrac{dx}{dt}=-0.3x+0.1y\) and \(\dfrac{dy}{dt}=0.3x-0.2y\), with \(x(0)=100\), \(y(0)=0\). Use Euler's method with \(h=1\) year to estimate \(x(2)\) and \(y(2)\). Then classify the equilibrium.
Step 1 (from \(t=0\)): \(\dfrac{dx}{dt}=-30+0=-30\) and \(\dfrac{dy}{dt}=30-0=30\). So \(x_1=70\), \(y_1=30\).
Step 2 (from \(t=1\)): \(\dfrac{dx}{dt}=-21+3=-18\) and \(\dfrac{dy}{dt}=21-6=15\). So \(x_2=52\), \(y_2=45\).
For the type, \(M=\begin{pmatrix}-0.3&0.1\\0.3&-0.2\end{pmatrix}\) has trace \(-0.5\) and determinant \(0.06-0.03=0.03\).
The eigenvalue equation is \(\lambda^2+0.5\lambda+0.03=0\), so \(\lambda=\dfrac{-0.5\pm\sqrt{0.13}}{2}\approx-0.070\) or \(-0.430\).
Both eigenvalues are real and negative, so the origin is a stable node: both amounts decay towards zero.
Answer: \(x(2)\approx52\), \(y(2)\approx45\); stable node
Example 2 Solve \(\dfrac{dx}{dt}=x+2y\), \(\dfrac{dy}{dt}=3x+2y\) with \(x(0)=3\), \(y(0)=2\), and classify the equilibrium.
\(M=\begin{pmatrix}1&2\\3&2\end{pmatrix}\). Then \(\det(M-\lambda I)=(1-\lambda)(2-\lambda)-6=\lambda^2-3\lambda-4=(\lambda-4)(\lambda+1)\).
So \(\lambda=4\) or \(\lambda=-1\).
For \(\lambda=4\): \(\begin{pmatrix}-3&2\\3&-2\end{pmatrix}\mathbf v=\mathbf0\) gives \(\mathbf v_1=\begin{pmatrix}2\\3\end{pmatrix}\).
For \(\lambda=-1\): \(\begin{pmatrix}2&2\\3&3\end{pmatrix}\mathbf v=\mathbf0\) gives \(\mathbf v_2=\begin{pmatrix}1\\-1\end{pmatrix}\).
General solution: \(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{4t}\begin{pmatrix}2\\3\end{pmatrix}+Be^{-t}\begin{pmatrix}1\\-1\end{pmatrix}\).
At \(t=0\): \(2A+B=3\) and \(3A-B=2\). Adding gives \(5A=5\), so \(A=1\) and \(B=1\).
So \(x=2e^{4t}+e^{-t}\) and \(y=3e^{4t}-e^{-t}\). Check: \(\dfrac{dx}{dt}=8e^{4t}-e^{-t}\), and \(x+2y=8e^{4t}-e^{-t}\), as required.
The eigenvalues are real with opposite signs, so the origin is a saddle (unstable). For example, at \(t=0.5\): \(x\approx15.4\), \(y\approx21.6\).
Answer: \(x=2e^{4t}+e^{-t}\), \(y=3e^{4t}-e^{-t}\); saddle
Example 3 A damped spring has position \(x\) satisfying \(\dfrac{d^2x}{dt^2}=-4x-0.5\dfrac{dx}{dt}\), with \(x(0)=1\) and \(\dfrac{dx}{dt}(0)=0\). Use Euler's method with \(h=0.1\) to estimate \(x(0.2)\), and describe the long-term behaviour.
Let \(y=\dfrac{dx}{dt}\). Then \(\dfrac{dx}{dt}=y\) and \(\dfrac{dy}{dt}=-4x-0.5y\). Start with \(x_0=1\), \(y_0=0\).
Step 1: \(\dfrac{dx}{dt}=0\), \(\dfrac{dy}{dt}=-4\). So \(x_1=1+0.1(0)=1\) and \(y_1=0+0.1(-4)=-0.4\).
Step 2: \(\dfrac{dx}{dt}=-0.4\), \(\dfrac{dy}{dt}=-4(1)-0.5(-0.4)=-3.8\). So \(x_2=1+0.1(-0.4)=0.96\) and \(y_2=-0.4+0.1(-3.8)=-0.78\).
For long-term behaviour, \(M=\begin{pmatrix}0&1\\-4&-0.5\end{pmatrix}\) gives \(\lambda^2+0.5\lambda+4=0\). The discriminant is \(0.25-16\lt0\), so \(\lambda=-0.25\pm1.98i\).
The real part is negative, so the origin is a stable spiral: the spring oscillates with decreasing size.
Answer: \(x(0.2)\approx0.96\); stable spiral (damped oscillation)
Common errors
- Updating \(x\) first and then using the new \(x\) to update \(y\) in Euler's method. Both updates use the old values.
- Choosing the wrong constants \(A\) and \(B\) by forgetting that each eigenvector multiplies its own exponential.
- Classifying by the sign of one eigenvalue only. Look at both, and at the real part when they are complex.
- Forgetting to define \(y=\dfrac{dx}{dt}\), so the second equation for \(\dfrac{dy}{dt}\) has no meaning.
- Giving an eigenvector with the wrong ratio. Check by multiplying \(M\mathbf v\) and comparing with \(\lambda\mathbf v\).
Exam tip: Check an eigenvalue and eigenvector with one quick multiplication: \(M\mathbf v\) must equal \(\lambda\mathbf v\). Quote the type of equilibrium together with the reason, such as "both eigenvalues real and negative".
Check yourself
- Classify the origin for \(M=\begin{pmatrix}-1&0\\0&-3\end{pmatrix}\).
Answer
The eigenvalues are \(-1\) and \(-3\), both real and negative, so it is a stable node. - Find the eigenvalues of \(M=\begin{pmatrix}0&1\\-9&0\end{pmatrix}\) and classify the origin.
Answer
\(\lambda^2+9=0\), so \(\lambda=\pm3i\). They are purely imaginary, so the origin is a centre. - For \(\dfrac{dx}{dt}=y\), \(\dfrac{dy}{dt}=-x\) with \(x_0=1\), \(y_0=0\) and \(h=0.2\), find \(x_1\) and \(y_1\).
Answer
\(x_1=1+0.2(0)=1\). \(y_1=0+0.2(-1)=-0.2\). - Write \(\dfrac{d^2x}{dt^2}=3x-2\dfrac{dx}{dt}\) as a first-order system and classify the origin.
Answer
With \(y=\dfrac{dx}{dt}\): \(\dfrac{dx}{dt}=y\), \(\dfrac{dy}{dt}=3x-2y\). Then \(M=\begin{pmatrix}0&1\\3&-2\end{pmatrix}\) gives \(\lambda^2+2\lambda-3=0\), so \(\lambda=1\) or \(-3\). They have opposite signs, so the origin is a saddle.