Higher order Linear ODEs#

Wronskian#

We define a Wronskian of two functions as the determinant

\[\begin{split} W[f,g](x) = \left| \begin{array}{cc} f(x) & g(x) \\ f'(x) & g'(x) \end{array} \right| \end{split}\]

Now consider the Wronskian of two solutions of a homogeneous ODE $y'' + p(x)y' + q(x)y = 0$:

\[ W[y_1, y_2] = y_1 y'_2 - y_2 y'_1 \]

Differentiating it

\[\begin{split} \begin{aligned} \frac{dW}{dx} &= y'_1 y'_2 + y_1 y''_2 - y'_2 y'_1 - y_2 y''_1 \\ &= y_1y''_2 - y_2 y''_1 \\ &= y_1(-p y'_2 - q u_2) - u_2 (-pu'_1 -qu_1) \\ &= -pW \end{aligned} \end{split}\]

In other words, the Wronskian satisfies the first-order linear ODE

\[ \frac{dW}{dx} + pW = 0 \]

whose solution is the Abel’s identity

\[ W(x) = W(x_0) e^{- \int_{x_0}^x p(s) ds} \]

In general, the Wronskian of a set of functions $\{f_1(x), f_2(x), \dots, f_n(x) \}$ is defined by

\[\begin{split} W[f_1, \dots, f_n] (x) := \left| \begin{array}{cccc} f_1 & f_2 & \cdots & f_n \\ f'_1 & f'_2 & \cdots & f'_n \\ \vdots & \vdots & & \vdots \\ f^{(n-1)}_1 & f^{(n-1)}_2 & \cdots & f^{(n-1)}_n \end{array} \right| \end{split}\]

Second-order linear ODE#

(119)#\[ y'' + p(x)y' + q(x)y = r(x) \]

Reduction of Order - homogeneous#

Often we can find one solution of an ODE from inspection, which we can then use to find the other linearly independent solution.

\[ \frac{d}{dx} \left( \frac{y_2}{y_1} \right) = \frac{y_1 y'_2 - y_2 y'_1}{y_1^2} = \frac{W[y_1, y_2]}{y_1^2} \]

And integrating

\[ y_2(x) = y_1 (x) \left( C + \int_{x_0}^{x} \frac{W(s)}{y_1^2 (s)} ds \right) \]

Can set C = 0 since we write the solution as a linear combination of u1 u2 anyway

Variation of parameters - inhomogeneous#

Let the general solution of the corresponding homogeneous equation be known

(120)#\[ y_h(x) = A y_1(x) + B y_2(x) \]

As we have done before, we use variation of parameters to set the ansatz for the particular solution:

(121)#\[ y_p(x) = C_1(x)y_1(x) + C_2(x)y_2(x) \]

where we have “promoted” the constants $C_1, C_2$ to functions $A(x), B(x)$. Differentiating it once:

\[ y' = C'_1y_1 + C_1y'_1 + C'_2y_2 + C_2y'_2 \]

We also require that

(122)#\[ C'_1 y_1 + C'_2 y_2 = 0 \]

so eq. (156) becomes

(123)#\[ y' = C_1 y'_1 + C_2 y'_2. \]

Differentiating again

(124)#\[ y'' = C'_1y'_1 + C_1y''_1 + C'_2y'_2 + C_2y''_2 \]

Now we substitute (159) and (160) into eq. (155):

\[\begin{split} C'_1y'_1 + C_1y''_1 + C'_2y'_2 + C_2y''_2 + p(C_1 y'_1 + C_2 y'_2) + q(C_1y_1 + C_2y_2) = r \\ (C'_1y'_1 + C'_2y'_2) + \underbrace{C_1(y''_1 + py'_1 + qy_1) + C_2(y''_2 + py'_2 + qy_2)}_{= \ 0 \quad eq. (20)} = r \end{split}\]

We get

(125)#\[ C'_1 y'_1 + C'_2 y'_2 = r \]

The solution of the system of equations (157) and (161) is given by

\[ C'_1 = - \frac{y_2}{W[y_1, y_2]} r, \quad C'_2 = \frac{y_1}{W[y_1, y_2]} r \]

and after integrating we get

\[ \begin{align}\begin{aligned} C_1(x) = - \int^x \frac{y_2(s)}{W(s)} r(s) \ ds , \quad C_2(x) = \int^x \frac{y_1(s)}{W(s)} r(s) \ ds $$ {eq}`eq27`\\where the choice of the lower bounds of integration is irrelevant since it only changes the coefficients $A$ and $B$ (see below). Therefore, the solution of the inhomogeneous second-order ODE {eq}`eq19` is\\\begin{split}$$ \begin{aligned} y(x) &= y_h(x) + y_p(x) \\ &= A y_1(x) + B y_2(x) + C_1(x)y_1(x) + C_2(x)y_2(x) \\ &= [A + C_1(x)]y_1(x) + [B + C_2(x)]y_2(x) \end{aligned} \end{split}\end{aligned}\end{align} \]

General $n$-th order linear ODE#

We can directly extend the method of variation of parameters we used to solve second-order linear ODE to $n$-th order ODEs, which are of the form

(126)#\[ \mathcal{L}_x[y] = y^{(n)} + p_n(x) y^{(n-1)} + \cdots + p_1(x)y' + p_0(x)y = r(x) \]

Let $\{y_1, y_2, \dots, y_n\}$ be solutions of the homogeneous ODE, i.e. $\mathcal{L}[y_i] = 0$. We seek solutions of the inhomogeneous ODE in the form of ansatz

(127)#\[ y_p(x) = \sum_{i=1}^n C_i(x)y_i(x) \]

where $C_i(x)$ are functions that we need to find. As before, we differentiate and define conditions:

\[\begin{split} \begin{aligned} y' = \sum_{i=1}^n C'_iy_i + C_iy'_i, \quad & \text{assume:} \sum_{i=1}^n C'_iy_i = 0 \\ y'' = \sum_{i=1}^n C'_iy'_i + C_iy''_i, \quad & \text{assume:} \sum_{i=1}^n C'_iy'_i = 0 \\ \vdots \\ y^{(n-1)} = \sum_{i=1}^n C'_iy_i^{(n-2)} + C_iy_i^{(n-1)}, \quad & \text{assume:} \sum_{i=1}^n C'_iy_i^{(n-2)} = 0 \end{aligned} \end{split}\]

\[ y^{(n)} = \sum_{i=1}^n C'_i y_i^{(n-1)} + C_i y_i^{(n)} \]

These $n$ equations define a system of linear equations with unknowns $C_1, \dots, C_n$:

(128)#\[\begin{split} \begin{bmatrix} y_1 & y_2 & \cdots & y_n \\ y'_1 & y'_2 & \cdots & y'_n \\ \vdots & \vdots & & \vdots \\ y_1^{(n-1)} & y_2^{(n-1)} & \cdots & y_n^{(n-1)} \end{bmatrix} \begin{bmatrix} C'_1 \\ C'_2 \\ \vdots \\ C'_n \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ r \end{bmatrix} \end{split}\]

A sufficient condition for the existence of a solution of the system of equations (128) is that the determinant of coefficients is nonzero for each value of $x$. This is equal to the Wronskian $W[y_1, \dots, y_n]$ and we know it is nonzero for all $x$ because the set $\{y_1, \dots, y_n \}$ is linearly independent. We determine $C'_1, \dots, C'_n$ using Cramer’s rule:

\[ C'_1(x) = \frac{W_i (x)}{W(x)}r(x) \]

where $W_i$ is the determinant obtained from $W$ by replacing $i$-th column with the RHS column $(0, 0, \dots, r)^T$. We integrate this and substitute the result in eq. (127) to get the particular solution of (186)

\[ y_p = \sum_{i=1}^n y_i(x) \int_{x_0}^x \frac{W_i (s)}{W(s)}r(s) \ ds \]

where $x_0$ is arbitrary.

Linear ODEs with Constant Coefficients#

In this section we focus on second-order ODEs where functions $p(x), q(x)$ and $r(x)$ are now constants. We therefore write Eq. inhmg2ndode as

(129)#\[ y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_1y' + a_0y = 0 \]

Homogeneous case - method of characteristic polynomial#

We use the method of characteristic polynomial which can be used for any such ODE of any order. A homogeneous linear ODE of first order with constant coefficient $a_0$

\[ y' + a_0 y = 0 \]

is separable and its integral is $y(x) = Ce^{\lambda x}$, where $\lambda$ is an unknown constant. We find it by plugging $y$ back into the ODE:

\[\begin{split} \lambda Ce^{\lambda x} + a_0 Ce^{\lambda x} = 0 \\ \lambda + a_0 = 0\end{split}\]

This is the characteristic equation of the ODE. We find $\lambda = -a_0$.

Consider now a second-order ODE with constant coefficients $a_1, a_0$

\[ y'' + a_1 y' + a_0 y = 0 \]

Inspired by the solution of the first-order equation, we assume the solution of the form $y(x) = Ce^{\lambda x}$. We plug it back into the ODE and get the characteristic equation

\[ \lambda^2 + a_1 \lambda + a_0 = 0 \]

The roots of this quadratic equation are

\[ \lambda_{1,2} = \frac{-a_1 \pm \sqrt{D}}{2}, \qquad D = a_1^2 - 4a_0 \]

so we have three cases depending whether $D > 0, D = 0$ or $D < 0$, keeping in mind that we need linearly independent solutions to form the basis:

$ \lambda_{1,2} \in \mathbb{R}, \lambda_1 \neq \lambda_2: \quad y(x) = C_1 e^{\lambda_1 x} + C_2e^{\lambda_2 x} $
$ \lambda_{1,2} \in \mathbb{R}, \lambda_1 = \lambda_2: \quad y(x) = (C_1 + C_2 x)e^{\lambda_1 x} $
$ \lambda_{1,2} \in \mathbb{C}, \lambda_2 = \lambda_1^: \quad y(x) = C_1 e^{\lambda_1 x} + C_2e^{\lambda_1^ x} $

Hint: Diagonalisation

The method of characteristic polynomial might remind us of the process of finding eigenvalues. Indeed, we can think of finding solutions in the form of an exponential function as a type of diagonalisation.

\[ \frac{d}{dx} e^{\lambda x} = \lambda e^{\lambda x} \]

The exponential function can be thought of as an eigenvector of the derivative operator and $\lambda$ as the eigenvalue. If $\lambda_1, \lambda_2$ are roots of the characteristic equation, we can write

\[ \left( \frac{d}{dx} - \lambda_1 \right) \left( \frac{d}{dx} - \lambda_2 \right) y(x) = 0 \]

Let us now generalise this to an $n$-th order ODE. The characteristic equation of eq. (129) is again obtained by substituting $y = e^{\lambda x}$:

\[ \lambda^n + a_{n-1}\lambda^{n-1} + \cdots + a_1 \lambda + a_0 = 0 \]

For any distinct real root $\lambda$, one solution is $y = e^{\lambda x} $
For any complex root $\lambda = \gamma + i \omega$, its conjugate $\lambda^* = \gamma - i \omega$ is also a root:

\[ y_1 = e^{\gamma x} \cos \omega x, \quad y_2 = e^{\gamma x} \sin \omega x \]

Multiple real roots: if a real root is repeated $m$ times, the $m$ corresponding linearly independent solutions are

\[ e^{\lambda x}, \ xe^{\lambda x}, \ \dots \ , x^{m-1}e^{\lambda x} \]

Multiple complex roots: if a complex root $\lambda = \gamma + i \omega$ is repeated, the corresponding linearly independent solutions are

\[ e^{\gamma x} \cos \omega x, \quad e^{\gamma x} \sin \omega x, \quad x e^{\gamma x} \cos \omega x, \quad xe^{\gamma x} \sin \omega x, \ \dots \]

ESE Jupyter Material

Higher order Linear ODEs

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