2.9 KiB
Systems of linear ordinary differential equations
Homogeneous systems of linear ODEs with constant coefficients
Let \mathbb{K} = \mathbb{R} \lor \mathbb{C}, n \in \mathbb{N} and A \in \mathbb{R}^{n \times n}. Seek differentiable functions y:\mathbb{R} \to \mathbb{K}^n such that
\mathbf{\dot y}(t) = A \mathbf{y}(t), \qquad t \in \mathbb{R}
The solutions from a linear space, therefore the general solutions can be written as,
\mathbf{y}(t) = \sum_{k=1}^n c_k \mathbf{y}_k(t), \qquad c_k \in \mathbb{K}
where \{\mathbf{y_1}, \dots, \mathbf{y_n}\} is a linear independent set of solutions, i.e. the basis of the solutions space.
Assume now that A is diagonalizable, and let \{\mathbf{v_1}, \dots, \mathbf{v_n}\} be a basis of \mathbb{K}^n consisting of eigenvectors of A.
AV = VD, \qquad \text{with } D = \begin{pmatrix} \lambda_1 & & \ & \ddots & \ & & \lambda_n \end{pmatrix}
then A = VDV^{-1}, let \mathbf{z}(t) = V^{-1} \mathbf{y}(t)
\begin{array}{ll}
&\mathbf{\dot z} = V^{-1} \mathbf{\dot y} = V^{-1} A \mathbf{y} = V^{-1} V D V^{-1} = D \mathbf{z}, \
& \mathbf{\dot z} = D \mathbf{z} \implies \mathbf{z}(t) = \mathbf{c} e^{\lambda t}.
\end{array}
Obtaining the general solution
$$\mathbf{y}(t) = V \mathbf{z}(t) = \sum_{k=1}^n c_k \mathbf{v_k} e^{\lambda_k t}.
Inhomogeneous systems of linear ODEs with constant coefficients
Let I \subseteq \mathbb{R} be an interval, \mathbf{f}: I \to \mathbb{R} continuous. Find functions \mathbf{y}: I \to \mathbb{R}^n such that
\mathbf{\dot y}(t) = A \mathbf{y}(t) + \mathbf{f}(t), \qquad t \in I. \qquad (*)
Theorem: let \mathbf{y}_p: I \to \mathbb{R}^n a particular solution for (*) and \mathbf{y}_h the general solution to the homegeneous system. Then the general solutions of the inhomogeneous system (*) is given by
\mathbf{y}(t) = \mathbf{y}_p(t) + \mathbf{y}_h(t), \qquad t \in I
Proof:
similar to 1d case, will be added later.
Method of variation of parameters
Let \{\mathbf{y_1}, \dotsc, \mathbf{y_n}\} be a basis for the solution space of the homogeneous system. Ansatz:
\mathbf{y}p(t) = \sum{k=1}^n c_k(t) \mathbf{y}_k(t) = (\mathbf{y}_1, \dots, \mathbf{y}_n) \begin{pmatrix} c_1(t) \ \vdots \ c_n(t) \end{pmatrix} = Y(t) \mathbf{c}(t),
where c_1(t), \dots, c_n(t): I \to \mathbb{R} are to be determined.
Then:
\begin{align*}
\mathbf{\dot y}p &= \sum{k=1}^n \dot c_k(t) \mathbf{y}k(t) + \sum{k=1}^n c_k(t) \mathbf{\dot y}k(t), \
&= \sum{k=1}^n \dot c_k(t) \mathbf{y}k(t) + A \sum{k=1}^n c_k(t) \mathbf{y}_k(t), \
&= Y(t) \mathbf{\dot c}(t) + A \mathbf{y}_p(t).
\end{align*}
Demanding that: Y(t) \mathbf{\dot c}(t) = \mathbf{f}(t) is the Wronskian. Then \mathbf{\dot c}(t) = Y^{-1}(t) \mathbf{f}(t) \iff Y(t) is nonsingular. Then solve for \mathbf{c}(t).