Decoupling Linear Systems with Constant Coefficients 2

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Decouple the following system

Let’s solve the system of differential equations:

$\begin{cases} x' = 2x + y \\\\ y' = x + 2y \end{cases}$

Notice that in its current state, the system is coupled —the rate of change of x depends on y, and the rate of change of y depends on x. Our objective is to transform this system into a decoupled form where each new variable evolves according to a single, independent first-order ODE. This “decoupling” allows us to solve the system much more easily.

Step 1. Writing the System in Matrix Form

We can express the system in matrix form as: $\vec{x'} = A\vec{x}$ where A = $(\begin{smallmatrix}2 & 1\\\ 1 & 2\end{smallmatrix})$, $\vec{x} = (\begin{smallmatrix}x\\\ y\end{smallmatrix})$

$(\begin{smallmatrix}x'\\\ y'\end{smallmatrix}) = (\begin{smallmatrix}2 & 1\\\ 1 & 2\end{smallmatrix})(\begin{smallmatrix}x\\\ y\end{smallmatrix})$

Step 2. Finding Eigenvalues of A

To decouple the system, we need to find a transformation that diagonalizes A. The first step in diagonalization is finding the eigenvalues by solving the characteristic equation: $\det(A - \lambda I) = 0$, where $I$ is the id

Compute $A - \lambda I = (\begin{smallmatrix}2-λ & 1\\\ 1 & 2-λ\end{smallmatrix})$

Calculate the determinant and set it to zero: $\det(A - \lambda I) = (2-\lambda)(2-\lambda) - 1 = 0 \iff \lambda^2 - 4\lambda + 4 - 1 = 0 \iff \lambda^2 - 4\lambda + 3 = 0$

Using the quadratic formula or by factoring $(\lambda - 3)(\lambda - 1) = 0$, we find: $\lambda = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm 2}{2}$

So the eigenvalues are $\lambda_1 = 3$ and $\lambda_2 = 1$.

Step 3. Finding Eigenvectors

For each eigenvalue $\lambda_i$, we find a corresponding eigenvector $\vec{α_i}$ by solving $(A - \lambda_iI)\vec{α_i} = 0$

For $\lambda_1 = 3$, solve $(A - 3I)\vec{α_1} = 0$ ⇒$(\begin{smallmatrix}-1 & 1\\\ 1 & -1\end{smallmatrix})(\begin{smallmatrix}a_1\\\ b_1\end{smallmatrix}) = (\begin{smallmatrix}0\\\ 0\end{smallmatrix})$.

This yields the system: $\begin{cases} -a_1 + b_1 = 0 \\\ a_1 -b_1 = 0 \end{cases}$

So, $a_1 = b_1$. Choosing $a_1 = 1$ for simplicity, an eigenvector is $\vec{α_1}=(\begin{smallmatrix}1\\\ 1\end{smallmatrix})$

For $\lambda_1 = 1$, solve $(A - I)\vec{α_2} = 0$ ⇒$(\begin{smallmatrix}1 & 1\\\ 1 & 1\end{smallmatrix})(\begin{smallmatrix}a_2\\\ b_2\end{smallmatrix}) = (\begin{smallmatrix}0\\\ 0\end{smallmatrix})$.

This yields the system: $\begin{cases} a_2 + b_2 = 0 \\\ a_2 +b_2 = 0 \end{cases}$

So, $a_2 + b_2 = 0 ⇒ a_2 = -b_2.$ Choosing $b_2 = -1$ for simplicity, an eigenvector is $\vec{α_2}=(\begin{smallmatrix}1\\\ -1\end{smallmatrix})$

Step 4. Constructing the Transformation Matrix E

The matrix E is formed by placing the eigenvectors as columns: E = $(\vec{α_1}\;\vec{α_2}) = (\begin{smallmatrix}1 & 1\\\ 1 & -1\end{smallmatrix})$

Verifying That A is Diagonalizable

Since we have two distinct eigenvalues and thus two independent eigenvectors, A is diagonalizable.

This means $AE = ED \implies A = EDE^{-1}$, where $D$ is the diagonal matrix of eigenvalues: $D = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}$

Compute $E^{-1}$

det(E) = $det(\begin{smallmatrix}1 & 1\\\ 1 & -1\end{smallmatrix}) = 1·(-1)−1·1 = -2.$

The inverse of a 2×2 matrix $(\begin{smallmatrix}a & b\\\ c & d\end{smallmatrix})$ is $\frac{1}{ad-bc}(\begin{smallmatrix}d & -b\\\ -c & a\end{smallmatrix})$

Then, $E^{-1} = \frac{1}{det(E)}·(\begin{smallmatrix}e_{22} & -e_{12}\\\ -e_{21} & e_{11}\end{smallmatrix})= \frac{1}{-2}(\begin{smallmatrix}-1 & -1\\\ -1 & 1\end{smallmatrix}) = \frac{1}{2}(\begin{smallmatrix}1 & 1\\\ 1 & -1\end{smallmatrix})$

Step 5. Changing Variables

We introduce the new decoupled variables u defined by: $\vec{u} = E^{-1}\vec{x} = \frac{1}{2}(\begin{smallmatrix}1 & 1\\\ 1 & -1\end{smallmatrix})(\begin{smallmatrix}x\\\ y\end{smallmatrix}) = (\begin{smallmatrix}u\\\ v\end{smallmatrix})$

Thus, the specific transformation is: $u = \frac{1}{2}(x+y)$, v = $\frac{1}{2}(x-y)$. They are chosen so that in the (u, v)-coordinates, the system of ODEs becomes decoupled.

Conversely, expressing $\vec{x}$ in Terms of $\vec{u}$ gives us $\vec{x} = E\vec{u} = (\begin{smallmatrix}1 & 1\\\ 1 & -1\end{smallmatrix})(\begin{smallmatrix}u\\\ v\end{smallmatrix})$

This results in: x = u + v, y = u - v

Step 6. Transforming the Original System using the Change of Variables

Substitute $\vec{x} = E\vec{u}$ and $\vec{x'} = E\vec{u'}$ into our original differential equation $\vec{x'} = A\vec{x'}: E\vec{u'} = AE\vec{u}$.

Since AE = ED, we can substitute the right side: $E\vec{u'} = ED\vec{u}$

Multiply both sides on the left by $E^{-1}$ to isolate $\vec{u'}$: $E^{-1}E\vec{u'} = E^{-1}ED\vec{u} ⇒\vec{u'} = D\vec{u} = (\begin{smallmatrix}3 & 0\\\ 0 & 1\end{smallmatrix})\vec{u}↭ (\begin{smallmatrix}u'\\\ v'\end{smallmatrix}) = (\begin{smallmatrix}3 & 0\\\ 0 & 1\end{smallmatrix})(\begin{smallmatrix}u\\\ v\end{smallmatrix})$.

This results in two perfectly decoupled equations that we can solve independently:

$\begin{cases} u' = 3u \\\ v' = v \end{cases}$

Now, solve each equation independently.

1. For u’ = 3u, $u(t) = c_1e^{3t}$ where $c_1$ is a constant determined by initial conditions.
2. For v’ = v, $v(t) = c_2e^{t}$ where $c_2$ is another constant determined by initial conditions.

Step 7. Writing the General Solution

Finally, express $x(t)$ and $y(t)$ in terms of our solved $u(t)$ and $v(t)$ using the relations $x = u + v$ and $y = u - v$: $x(t) = c_1 e^{3t} + c_2 e^{t}$ and $y(t) = c_1 e^{3t} - c_2 e^{t}$

Written cleanly in vector format, the general solution to the system is: $\boxed{\vec{x}(t) = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = c_1 e^{3t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 e^{t} \begin{pmatrix} 1 \\ -1 \end{pmatrix}}$

Interpretation of the Solution

Because both eigenvalues are positive, the system exhibits exponential growth away from the origin (an unstable node).

However, because $\lambda_1 = 3$ is greater than $\lambda_2 = 1$, the $e^{3t}$ term grows much faster than the $e^{t}$ term. This means that as time goes on, the trajectory of the system will be overwhelmingly dominated by the $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ eigenvector. No matter the starting point, the values of $x$ and $y$ will eventually grow at roughly the same rate, pushing the system along the line $y = x$.

Summary of the Method

1. Find the Eigenvalues and Eigenvectors of A: Solve $\det(A - \lambda I) = 0$ to find eigenvalues $\lambda_1, \lambda_2$. For each eigenvector $\lambda_i$, find the corresponding eigenvector $\vec{α_i}$ by solving $(A - \lambda_i I)\vec{\alpha_i} = \vec{0}$.
2. Form the Matrix E: Place the calculated eigenvectors as columns to build $E = (\vec{α_1}\;\vec{α_2})$
3. Compute $E^{-1}$. Find the inverse of the transformation matrix.
4. Change Variables: Define $\vec{x} = E\vec{u}$ and $\vec{u} = E^{-1}\vec{x}$.
5. Transform the Original System: Substitute $\vec{x} = E\vec{u}$ into the original differential equation $\vec{x'} = A\vec{x}$ to get $\vec{u'} = D\vec{u}$ (the decoupled state).
6. Solve the Decoupled Equations: Solve the simple independent equations $u' = \lambda_1 u$ and $v' = \lambda_2 v$.
7. Transform Back to Original Variables. Use $\vec{x} = E\vec{u}$ to find the final expressions for $x(t)$ and $y(t)$.

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare [18.03 Differential Equations, Spring 2006], YouTube by MIT OpenCourseWare.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.