34. Applications to Differential Equations

Dated: 17-06-2025

Differential Equations

A differential equation is any equation which contains derivatives,¹ either ordinary derivatives¹ or partial derivatives.²

System of `Linear Differential Equations`³

A system of linear differential equations³ can be expressed as

\[ \begin{array}{ccccc} x_1^\prime &= &a_{11} x_1 &+ \ldots + &a_{1n}x_n \\ x_2^\prime &= &a_{21} x_1 &+ \ldots + &a_{2n}x_n \\ \vdots &&\vdots &&\vdots \\ x_n^\prime &= &a_{n1} x_1 &+ \ldots + &a_{nn}x_n \\ \end{array} \]

\[x^\prime = A x\]

Where

\[ x(t) = \begin{bmatrix} x_1(t) \\ \vdots \\ x_n(t) \\ \end{bmatrix} \]

\[ x^\prime(t) = \begin{bmatrix} x_1^\prime(t) \\ \vdots \\ x_n^\prime(t) \\ \end{bmatrix} \]

\[ A = \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \\ \end{bmatrix} \]

Superposition of Solutions

If \(u\) and \(v\) are solutions to \(x^\prime = Ax\) then their linear combination⁴ (i.e. \(au + bv\)) is also a solution.
This property is called superposition of solutions.

Fundamental `Set`⁵ of Solutions

If \(A\) is an \(n \times n\) matrix⁶ with \(n\) linearly independent⁷ functions⁸ then each solution to the equation \(x^\prime = Ax\) is a linear combination⁴ of those functions⁸ and set⁵ of these functions⁸ is the basis for the solution set which is also \(n\) dimensional vector space⁹ of functions.⁸

Initial Value Problem

If a vector⁹ \(\vec x_0\) is given then the initial value problem is to construct a function⁸ \(x\) such that \(x^\prime = Ax\) and \(x(0) = x_0\).

Example

Find the solution to \(x^\prime = Ax\) where

\[ A = \begin{bmatrix} 3 & 0\\ 0 & -5 \end{bmatrix} \]

\[ \begin{bmatrix} x_1^\prime(t) \\ x_2\prime(t) \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & -5 \end{bmatrix} \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} \]

\[\implies x_1^\prime(t) = 3x_1(t) \text{ and } x_2^\prime(t) = -5x_2(t)\]

\[\frac{d x_1(t)}{dt} = 3x_1\]

\[\frac {d x_1} {x_1} = 3dt\]

Integrating¹⁰ both sides

\[\int \frac{dx_1}{x_1} = 3 \int dt\]

\(\(\ln x_1 = 3t + \ln c_1\)\)

\(\(\ln x_1 - \ln c_1 = 3t\)\)

\(\(\ln\left(\frac{x_1}{c_1}\right) = 3t\)\)

Taking anti log¹¹ on both sides

\[\frac{x_1}{c_1} = e^{3t}\]

\(\(x_1(t) = c_1 e^{3t}\)\)

Similarly, for \(x^\prime_2(t) = -5x_2(t)\), solution will be

\[x_2(t) = c_2 e^{-5t}\]

\[ \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} = \begin{bmatrix} c_1 e^{3t} \\ c_2 e^{-5t} \end{bmatrix} = c_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} e^{3t} + c_2 \begin{bmatrix} 0 \\ 1 \end{bmatrix} e^{-5t} \]

Observation

General solution to \(x^\prime = Ax\) might be of the form

\[x(t) = v e^{\lambda t}\]

Multiply \(A\) on both sides

\[Ax(t) = Ave^{\lambda t}\]

where \(\lambda\) is a scalar and \(v\) is some non zero vector¹²
Differentiating¹ original equation with respect to \(t\).

\[x^\prime(t) = \lambda ve^{\lambda t}\]

\[x^\prime(t) = Ax(t) \iff \lambda v = Av\]

Eigenfunctions

Each pair of eigenvalue¹³ and its corresponding eigenvector provides a solution of the equation \(x^\prime = Ax\) which is called eigenfunctions of the differential equation.³

Decoupling a Dynamical System

Let a dynamical system be defined by \(x^\prime = Ax\) where \(A\) is diagonalizable.¹⁴
Suppose that eigen functions for \(A\) are

\[\vec v_1e^{\lambda_1 t}, \ldots, \vec v_ne^{\lambda_n t}\]

where \(\{\vec v_1, \ldots, \vec v_n\}\) are linearly independent⁷ and let \(P = \begin{bmatrix}\vec v_1 & \cdots & \vec v_n\end{bmatrix}\) and \(D\) be diagonal matrix¹⁵ with entries as \(\lambda_1, \ldots, \lambda_n\) so that \(A = PDP^{-1}\).
Now replace \(x(t) = Py(t)\) and substitute in \(x^\prime = A x\) and in the end, we will get

\[y^\prime(t) = Dy(t)\]

\[ \begin{bmatrix} y_1^\prime(t) \\ y_2^\prime(t) \\ \vdots \\ y_n^\prime(t) \\ \end{bmatrix} = \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & \vdots \\ \vdots & \vdots & \ddots & 0\\ 0 & \cdots & 0 & \lambda_n \end{bmatrix} \begin{bmatrix} y_1(t) \\ y_2(t) \\ \vdots \\ y_n(t) \\ \end{bmatrix} \]

Change of variable from \(x\) to \(y\) has decoupled the system because now the derivative¹ of each function⁸ \(y_k\) depends only on \(y_k\).

\[\because y_1^\prime = \lambda_1y_1\]

\[\implies y_1(t)= \lambda_1 e^{\lambda_1 t}\]

\[ y(t) = \begin{bmatrix} c_1 e^{\lambda_1 t} \\ \vdots \\ c_n e^{\lambda_n t} \end{bmatrix}, \text{ where } \begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix} = y(0) = P^{-1} x(0) = P^{-1} x_0 \]

To obtain the general solution \(x\) of the original system, compute

\[x(t) = Py(t) = \begin{bmatrix}v_1 & \dots & v_n\end{bmatrix}y(t) = c_1v_1 e^{\lambda_1 t} + \dots + c_nv_n e^{\lambda_n t}\]

This is the eigenfunction expansion

Complex Eigenvalues

Let \(A\) be a matrix⁶ and if \(\lambda\) is an eigen value¹³ then \(\bar \lambda\) is the 2nd eigen value.¹³
Similarly, if \(\vec v\) is an eigen vector¹³ then \(\bar{\vec v}\) is the 2nd eigen vector.¹³
The solutions for \(x^\prime = Ax\) are

\[x_1(t) = ve^{\lambda t}, \ x_2(t) = \bar v e^{\bar \lambda t}\]

\[x_2(t) = \overline{x_1(t)}\]

\[\textbf{Re }(ve^{\lambda t}) = \frac 1 2 \left(x_1(t) + \overline{x_1(t)}\right)\]

\[\textbf{Im }(ve^{\lambda t}) = \frac 1 {2i} \left(x_1(t) - \overline{x_1(t)}\right)\]

\[\because e^x = 1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} + \dots\]

\[e^{\lambda t} = 1 + \lambda t + \frac{(\lambda t)^2}{2!} + \dots + \frac{(\lambda t)^n}{n!} + \dots\]

\[\because \lambda = a + b \iota\]

\[e^{(a+b \iota)t} = e^{at} e^{\iota bt} = e^{at} (\cos bt + \iota \sin bt)\]

\[\therefore ve^{\lambda t} = (\textbf{Re } v + \iota \textbf{Im } v) \cdot e^{at} (\cos bt + \iota \sin bt)\]

After working it out, we will get

\[ve^{\lambda t} = e^{at}(\textbf{Re }v \cos(bt) - \textbf{Im }v \sin(bt)) + \iota e^{at}(\textbf{Re }v \sin(bt) - \textbf{Im }v \cos(bt))\]

Solutions of \(x^\prime= Ax\) are

\[y_1(t) = \textbf{Re } x_1(t) = [\textbf{Re }v \cos(bt) - \textbf{Im }v \sin(bt)]e^{at}\]

\[y_2(t) = \textbf{Im } x_1(t) = [\textbf{Re }v \sin(bt) + \textbf{Im }v \cos(bt)]e^{at}\]