April 25, 2024
An ordinary differential equation is a constraint on the derivatives of a function of one variable. The theory of ODE’s show how to find all functions satisfying the constraint. Constraints on the derivatives of a function of more than one variable leads to partial differential equations and finding all solutions is a much more difficult problem.
If x(t) is a function and dx(t)/dt = ax(t), a\in\mathbf{R}, then x(t) = e^{at} is a solution with x(0) = 1. This ODE is linear since \alpha x(t) is also a solution for any constant \alpha\in\mathbf{R}. The solution of dx/dt = ax with x(0) = x_0 is x(t) = x_0e^{at}. We also write x' for dx/dt.
Exercise. Find all solutions of x' = ax with x(t_0) = x_0.
If x'(t) = ax(t) and x(t_0) = x_0 then x(t_0 + h) \approx x(t_0) + x'(t_0) h = x_0 + a x_0 h for small h. This suggests that an ODE with an initial condition (x(t_0) = x_0) should have a unique solution. We know the function at one point and the constraint on its derivative allows us to find nearby points. Note x^{(n)}(t) = d^nx(t)/dt^n = a^nx(t) so using all terms in the Taylor expansion gives x(t_0 + h) = \sum_{n \ge 0} x^{(n)}(t_0) h^n/n! = \sum_{n \ge 0} a^n x_0 h^n/n! = x_0 e^{ah} and we see the solution to the exercise is x(t) = x_0 e^{a(t - t_0)}.
This equation can be written x' - ax = 0. If u' - au = 0 and v' - av = 0 are any two solutions then a linear combination of u and v, cu + dv for c,d\in\mathbf{R}, is also a solution. This shows the set of all solutions is a vector space.
Exercise. If u and v are solutions of x' - ax = 0 and u(t_0) = v(t_0) for some t_0\in\mathbf{R} then u = v.
Hint: If x' - ax = 0 and x(t_0) = 0 for some t_0\in\mathbf{R} then x(t) = 0 for all t.
This shows the vector space of solutions is one dimensional.
The ODE x' - ax = 0 is called homogeneous. The ODE x' - ax = b, b\in\mathbf{R}, is called inhomogeneous. Clearly the constant function x(t) = -b/a is a solution of this inhomogeous ODE.
Exercise. If x' - ax = b then y = x - b/a satisfies y' - ay = 0.
Exercise. Find all solutions of x' - ax = b with x(t_0) = x_0.
If x(t) is vector-value and x'(t) = Ax(t) where A is a matrix then a similar argument shows x(t) = e^{A(t - t_0)}x(t_0) where e^{At} = \sum_{n\ge 0}A^nt^n/n!.
An ODE of the form \sum_{k = 0}^n a_k x^{(k)}(t) = 0 is a homogeneous ODE of order n with constant coefficients a_k, 0\le k \le n, a_n \not= 0. Dividing by a_n we can assume the ODE has the form x^{(n)} + \sum_{0\le k <n} a_k x^{(k)}(t) = 0. If u and v are any two solutions then a linear combination of u and v, cu + dv for c,d\in\mathbf{R}, is also a solution. This shows the set of all solutions is a vector space.
A solution of the form x(t) = e^{\omega t} must satisfy \sum_{k = 0}^n a_k \omega^k = 0 since x^{(k)}(t) = \omega^k x(t). If the polynomial has roots \omega_k then any linear combination \sum_k \alpha_k e^{\omega_k t} is a solution. If the roots are not distinct then this does not include all soltuions. For example x''(t) = 0 has solutions x(t) = 1 and also x(t) = t. The general solution is x(t) = \sum_k \alpha_k \sum_{0\le j<\mu_k} t^j e^{\omega_k t} where \mu_k is the multiplicity of the root \omega_k, but we don’t know that yet.
Differentiation is a linear operator Dx = x'. We can write the ODE as (\sum_n a_n D^n)x = 0 where \sum_n a_n D^n is also a linear operator. The solution of the ODE is the kernel of this operator, hence must be a subspace.
Letting x_k(t) = x^{(k)}(t) = D^kx(t) (and assuming a_n = 1) we can write this as \begin{aligned} x_0'(t) &= x_1(t) \\ x_1'(t) &= x_2(t) \\ &\cdots \\ x_{(n-1)}'(t) &= -a_0 x_0(t) - \cdots - a_{n-1} x_{n-1}(t)) \\ \end{aligned} or as a matrix \begin{bmatrix} x_0(t) \\ x_1(t) \\ \dots \\ x_{n-1}(t) \\ \end{bmatrix}' = \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \dots & & & & \\ -a_0/a_n & -a_1/a_n & -a_2/a_n & \cdots & -a_{n-1}/a_n \\ \end{bmatrix} \begin{bmatrix} x_0(t) \\ x_1(t) \\ \dots \\ x_{n-1}(t) \\ \end{bmatrix} For example, the ODE x''(t) + x(t) = 0 has matrix A = \begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix} so A^2 = \begin{bmatrix}-1 & 0 \\ 0 & -1 \end{bmatrix} = -I, A^3 = -A, A^4 = I, A^5 = A, . We have e^{At} = \begin{bmatrix}\cos t & \sin t \\ -\sin t & \cos t \end{bmatrix}. Given initial conditions x_0(0) = x(0) = c and x_1(0) = x'(0) = d we have the solution x(t) = c\cos t + d\sin t.
We are now in a position to solve any homogeneous ODE with constant coefficents. Recall every matrix can be reduced to Jordan normal form so all we need to do is compute e^{At} where A = \lambda I + J where J is …