1 Ordinary Differential Equations
- ODE Fundamentals
1.1 Fundamentals of ODEs
A differential equation is an equation in which the unknowns are functions of one or more variables and which contains both these functions and their derivatives.
An ordinary differential equation (ODE) to be differential equations that involve functions of a single independent variable with general form \[ F(x,y,y',\dots,y^{(n)})=0, \] where \(x\) is the independent variable, \(y\) is the function (or dependent variable) and \(y^{(n)}\) denotes the \(n\)-th derivative with respect to \(x\).
The solution of an ODE is the function \(y\) which we aim to determine the closed form expression for. Typically, if solvable, an ODE will have an infinite number of solutions parameterized by arbitrary constants, therefore our aim is to instead define the solution space of the ODE.
Often we will also be given some boundary condition (e.g. \(y(0)=0\)) from which we will be able to determine the unique solution to the ODE given that initial condition.
A boundary condition (also known as an initial condition when considering equations with respect to time) is the statement that \[ y(x_{0})=y_{0};~~\text{for some }(x_{0},y_{0})\in\mathscr{D}, \] where \(\mathscr{D}\) is the function domain.
When more than one independent variable is involved, the equation is known as a partial differential equation which are reserved for their own note. Further, there can be more than one function involved \(y_{1}(x), y_{2}(x), \dots\) in which case there will be multiple equations relating them allowing us to solve the system of equations (i.e. determine the solution space for functions \(y_{j}\) for all \(j\)).
The order of an ODE is the order of the highest derivative that it contains
When one cannot analytically solve an ODE (and most cannot be solved analytically) one has four options:
- Introduce a new function (known as special functions) defined as the solution of the previously unsolved ODE and study it in some way (e.g. power series, plot it, tabulate it);
- Find approximate (or asymptotic solutions);
- Find numerical solutions; and
- Study solutions qualitatively.
1.1.1 Notation Conventions
Here we make a quick note about notation conventions. By convention we will use \[ \frac{dy}{dt}=\dot{y}\quad\&\quad \frac{dy}{dx}=y'. \] The fractional notation both specifies the independent variable and reminds us that (with care) derivatives behave similarly to fractions.
Furthermore we note the difference between \(\partial\) and \(d\) notation.
- \(\frac{{\partial f}}{\partial x}\) is the partial derivative meaning it describes the change of \(f\) with respect to \(x\) assuming all other variables are constants with respect to \(x\); - \(\frac{df}{dx}\) is the total derivative meaning it describes the change of \(f\) with respect to \(x\) making no assumptions about other variables, they are all considered variables in \(x\).
1.2 Linear ODEs
Example: One of the most important examples of an ODE is \[ \dot{y}=\frac{dy}{dt}=ay, \] i.e. the rate of growth of a quantity is proportional to the quantity itself (i.e. a monogamous population). This example is both first-order and linear. It can also be resolved with respect to derivative, i.e. it is of the general form \[ \dot{y}=f(t,y). \]
An ODE is linear if it can be written in the form \[ a_{n}(x)y^{(n)}+a_{n-1}(x)y^{(n-1)}+\dots+a_{1}(x)y'+a_{0}(x)y=f(x)), \] or to use Einstein’s convention of implied summation over repeated indices (here mean to run from \(0\) to \(n\)) \[ a_{i}(x)y^{(i)}=f(x). \]
When \(f(x)=0\) the linear ODE is said to be homogeneous, otherwise it is inhomogeneous. Homogeneous linear ODEs have the important property that they are invariant under arbitrary rescaling \(y \to \lambda y,~\forall \lambda \neq 0\). Further, linear ODEs have many nice properties making them much easier to solve.
1.3 First Order ODEs
A first-order ODE can be written in symmetric form \[ P(x,y) dx+Q(x,y)dy=0, \] if it can be expressed implicitly in terms of \(x\) and \(y\).
This includes: (1) first-order ODEs that are explicitly or implicitly solvable; (2) separable equations; (3) exact equations; and (4) integrating factor cases; all of which we will consider in this note.
1.3.1 Equations in Full Differentials
Consider a first-order ODE in symmetric form \[ P(x,y)dx+Q(x,y)dy=0. \] It is easy to integrate if the LHS is a full differential of some function, i.e. if \(\exists \Phi(x,y)\) such that \[ \forall(x,y)\in\mathscr{D},~~P(x,y)=\frac{{\partial \Phi}}{\partial x}~~\&~~Q(x,y)=\frac{{\partial \Phi}}{\partial y}. \] Then the ODE is said to be an equation in full differentials and becomes \[ d\Phi(x,y)=0\implies \Phi(x,y)=C, \] where \(X\) is an arbitrary constant of integration.
The necessary condition for this method is that differentiating the first of them with respect to \(y\) and the second with respect to \(x\) we find that these derivatives must be equal \[ \forall(x,y)\in\mathscr{D},~~ \frac{{\partial P}}{\partial y}=\frac{{\partial Q}}{\partial x}. \] Example (Equations in Full Differentials): Consider the ODE \[ 2xydx+(x^2-y^2)dy=0. \] Since \(P(x,y)=2xy\) and \(Q(x,y)=(x^2-y^2)\) the necessary condition is satisfied. It must be that \[ \frac{{\partial \Phi}}{\partial x}=2xy\implies \Phi=x^2y+\phi(y), \] where \(\phi(y)\) is the integration constant in \(x\) but a function of \(y\). This is given by \[ \frac{{\partial \Phi}}{\partial y}=x^2-y^2\implies x^2+\phi'(y)=x^2-y^2\implies \phi(y)=-\frac{y^3}{3}+c. \] Thus the solution is \[ \Phi(x,y)=x^2y-\frac{y^3}{3}=C. \] ### Integrating Factor
Supposing that the necessary condition is not met we can instead sometimes use the integrating factor method. The integrating factor is a function \(\Lambda(x,y)\neq 0\) such that \[ \Lambda(x,y)\left[ P(x,y)dx+Q(x,y)dy \right] =0, \] is an equation in full differentials. i.e. \[ \frac{{\partial \Lambda P}}{\partial y}=\frac{{\partial \Lambda Q}}{\partial x}\quad\text{or}\quad Q \frac{{\partial \Lambda}}{\partial x}-P \frac{{\partial \Lambda}}{\partial y}=\left( \frac{{\partial P}}{\partial y} - \frac{{\partial Q}}{\partial x} \right)\Lambda. \] Generally speaking this is a PDE (first-order) and is not necessarily any easier to solve than the original ODE but it can allow us to more easily find a particular solution in some cases.
Example (Integrating Factor): Consider the equation \[ (x^2+y^2+y)dx-xdy=0. \] We spot that \(ydx-xdy=-x^2d({y/x})\) thus we have \[ (x^2+y^2)dx-x^2d(y/x)=0. \] The integrating factor is \(\Lambda=1/(x^2+y^2)\). Multiplying by the integrating factor we get \[ dx- \frac{d(y/x)}{1+(y/x)^2}=0\implies x - \arctan(y/x)=C. \] ### Separable Equations
A particularly simple and very useful special case of an ODE in full differentials is a separable ODE which takes the form \[ P(x)dx+Q(y)dy=0. \] In this case we can integrate instantly \[ \int_{}^{}{}~d{x}P(x)+\int_{}^{}{}~d{y}Q(y)=C. \] Example (Separable Equations): Consider the ODE \[ x \frac{dy}{dx}+y=y^2\implies \frac{1}{y(y-1)}dy=\frac{1}{x}dx. \] Thus we can integrate \[ \begin{align} \int_{}^{}{\frac{1}{y(y-1)}}~d{y} & =\int_{}^{}{\frac{1}{x}}~d{x} \\ \implies \log \lvert x \rvert +C & =\int_{}^{}{\frac{1}{y-1}}~d{y}-\int_{}^{}{\frac{1}{y}}~d{y}=\log \left\lvert \frac{{y-1}}{y} \right\rvert \\ \implies \frac{{y-1}}{y} & =Cx \\ \implies y & =\frac{1}{1+Cx}. \end{align} \] ### Homogeneous Equations
- Reduction to homogeneous form by linear change of variables
1.3.2 Linear Equations
- Solution via integrating factor
- Solution via Variation of Constant
- CF + PI