4.5 KiB
Second-order ordinary differential equations
For simplicity, all definitions and statements are for complex values functions and vector spaces over \mathbb{C}.
Linear second-order ODEs with constant coefficients
Let L[y] = f be given by
L[y] = \ddot y + p \dot y + qy = f \qquad (*),
with f,p,q \in \mathbb{R}.
Definition: the set of all solutions to (*) is called the general solution.
Property: if y_1,y_2 are both solutions to the homogeneous case L[y]=0 then \forall c_1,c_2 \in \mathbb{R}, y=c_1y_1 + c_2y_2 is a solution.
L[y] = L[c_1y_1 + c_2y_2] = c_1L[y_1] + c_2L[y_2],
Then the consequence is that the general solution is a linear space.
(*) is said to have resonance if f can be split into linearly independent terms of which at least one lies in the solution space of (*).
Solving homogeneous linear second-order ODEs with constant coefficients
Therefore solving
L[y] = \ddot y + p \dot y + qy = 0.
Ansatz: let y(t) = e^{\lambda t} with \lambda \in \mathbb{C}. Then
L[y(t)] = \lambda^2 e^{\lambda t} + p \lambda e^{\lambda t} + q e^{\lambda t} = e^{\lambda t} (\lambda^2 + p \lambda + q) = 0,
obtaining the characteristic equation \chi(\lambda) = \lambda^2 + p \lambda + q = 0. If two roots \lambda_1,\lambda_2 \in \mathbb{C} are found the solution space is
y(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}, \quad c_1,c_2 \in \mathbb{C},
if instead one root \lambda_1 \in \mathbb{C} is foundt the solution space is
y(t) = (c_1 + c_2t) e^{\lambda_1 t}.
??? note "Proof:"
Will be added later.
Example
Let the homogeneous linear second-order ode be given by \ddot y + 4 \dot y + 8y = 0. Then the characteristic equation is given by \chi(\lambda) = \lambda^2 + 4\lambda + 8 = 0 with solutions \lambda_1 = -2 + 2i and \lambda_2 = -2 - 2i. Then the general solution is given by
y(t) = c_1 e^{(-2 + 2i)1 t} + c_2 e^{(-2 - 2i) t}, \quad c_1,c_2 \in \mathbb{C},
and we can write the real solution as
y(t) = e^{-2t}\big(d_1\cos 2t + d_2 \sin 2t \big), \quad d_1,d_2 \in \mathbb{R}.
Solving inhomogeneous linear second-order ODEs with constant coefficients
Theorem: let y_p be a particular solution to (*). Then the general solution to (*) is given by
y = y_h + y_p,
with y_h the solution to the homegeneous case.
??? note "Proof:"
Let $y$ be a solution to $(*)$, then $L[y - y_p] = L[y] - L[y_p] = f - f = 0$. Therefore $y = (y - y_p) + y_p = y_h + y_p$.
Method of variation of parameters
We need the general solution to the homogeneous case
y_h(t) = c_1 y_1(t) + c_2 y_2(t), \qquad c_1,c_2 \in \mathbb{C}.
Ansatz: let y_p(t) = c_1(t) y_2(t) + c_2(t) y_2(t), then taking the derivative of y_p(t)
\dot y_p(t) = \dot c_1(t) y_2(t) + \dot c_2(t) y_2(t) + c_1(t) \dot y_2(t) + c_2(t) \dot y_2(t),
we demand that \dot c_1(t) y_2(t) + \dot c_2(t) y_2(t) = 0. Then taking the second derivative of y_p(t)
\ddot y_p(t) = \dot c_1(t) \dot y_2(t) + \dot c_2(t) \dot y_2(t) + c_1(t) \ddot y_2(t) + c_2(t) \ddot y_2(t),
then we have for (*)
\ddot y_p(t) + p \dot y_p(t) + q = c_1\big(\ddot y_1 + p \dot y_1 + q y_1\big) + c_2\big(\ddot y_2 + p \dot y_2 + q y_2\big) + \dot c_1 \dot y_1 + \dot c_2 \dot y_2 = f
we demand that \dot c_1 \dot y_1 + \dot c_2 \dot y_2 = f. Then we can create a linear system of demands
\begin{pmatrix} y_1 && y_2 \ \dot y_1 && \dot y_2\end{pmatrix} \begin{pmatrix} \dot c_1 \ \dot c_2 \end{pmatrix} = \begin{pmatrix} 0 \ f \end{pmatrix},
named the Wronskian and we can solve for c_1(t) and c_2(t) by integration.
Ansatz method
Let f(t) = p(t)e^{\lambda t}, rule of thumb: y_p is of related type to inhomogeneity f. Then for A_n, B_n and P_n polynomials of degree \leq n and \alpha \in \mathbb{R}
| Inhomogeneity | Particular solution |
|---|---|
L[y] = P_n |
t^m A_n |
L[y] = P_n e^{\alpha t} |
t^m A_n e^{\alpha t} |
L[y] = P_n \cos \omega t |
t^m \big(A_n \cos \omega t + B_n \sin \omega t \big) |
L[y] = P_n \sin \omega t |
t^m \big(A_n \cos \omega t + B_n \sin \omega t \big) |
L[y] = P_n e^{\alpha t} \cos \omega t |
t^m e^{\alpha t} \big(A_n \cos \omega t + B_n \sin \omega t \big) |
L[y] = P_n e^{\alpha t} \sin \omega t |
t^m e^{\alpha t} \big(A_n \cos \omega t + B_n \sin \omega t \big) |
Choose m \in \mathbb{N} \cup \{0\} as small as possible such that no term in the ansatz solves the homogeneous equation L[y] = 0.