Differential Equations Lecture 1

Convert bulk of handwritten notes from DE lecture one to markdown.

src/differential_equations_1.md

+---
+title: Differential Equations
+---
+
+# Differential equations
+
+A differential equation  or DE is any equation which involves both a function and some 
+derivative of that function. In this course we will be focusing on 
+*Ordinary Differential Equations*, meaning that our equations will involve 
+functions of one independent variable and hence any derivatives will be full 
+derivatives. Equations which involve a function of several independent variables
+and their partial derivatives are handled in courses on 
+*Partial Differential Equations*. 
+
+We consider functions $x(t)$ and define $\dot{x}(t)=\frac{dx}{dt}$, 
+$x^{(n)}(t)=\frac{d^{n}x}{dt^{n}}$. An $n$*-th* order differential equation is 
+an equation of the form 
+
+$$x^{(n)}(t) = f(x^{(n-1)}(t), \cdots, x(t), t).$$ 
+
+Typically, $n \leq 2$. Such an equation will usually be presented with a set of 
+initial conditions,
+
+$$x^{(n-1)}(t_{0}) = x^{(n-1)}_{0}, \cdots, x(t_0)=x_0. $$
+
+This is because to fully specify the solution to an $n$*-th* order differential 
+equation, $n-1$ initial conditions are necessary. To understand why we need 
+initial conditions, look at the following example.
+
+!!! check "Example: Initial conditions"
+    Consider the following calculus problem,
+    
+    $$\dot{f}(x)=x. $$ 
+
+    By integrating, one finds that the solution to this equation is 
+
+    $$\frac{1}{2}x^2 + c,$$
+
+    where $c$ is an integration constant. In order to specify the integration 
+    constant, an initial condition is needed. For instance, if we know that when 
+    $x=2$ then $f(2)=4$, we can plug this into the equation to get 
+
+    $$\frac{1}{2}*4 + c = 4, $$
+
+    which implies that $c=2$. 
+    
+Essentially initial conditions are needed when solving differential equations so
+that unknowns resulting from integration may be determined.
+
+!!! info Terminology for Differential Equations
+    1. If a differential equation does not explicitly contain the 
+        independent variable $t$, it is called an *autonomous equation*.
+    2. If the largest derivative in a differential equation is of first order, 
+        i.e. $n=1$, then the equation is called a first order differential 
+        equation.
+    3. Often you will see differential equation presented using $y(x)$ 
+        instead of $x(t)$. This is just a different nomenclature. 
+            
+In this course we will be focusing on *Linear Differential Equations*, meaning 
+that we consider differential equations $x^{(n)}(t) = f(x^{(n-1)}(t), \cdots, x(t), t)$
+where the function $f$ is a linear ploynomial function of the unknown function
+$x(t)$. A simple way to spot a non-linear differential euation is to look for 
+non-linear terms, such as $x(t)*\dot{x}(t)$ or $x^{(n)}(t)*x^{(2)}(t)$. 
+
+Often, we will be dealing with several coupled differential equations. In this 
+situation we can write the entire system of differential equations as a vector 
+equation, involving a linear operator. For a system of $m$ equations, denote 
+
+$$**x(t)** = \begin{bmatrix}
+x_1(t) \\
+\vdots \\
+x_{m}(t) \\
+\end{bmatrix}.$$
+
+A system of first order linear equations is then written as 
+
+$$\dot{**x(t)**} = **f**(**x(t)**,t) $$
+
+with initial condition $**x(t_0)** = **x_0**$.
+
+# Basic examples and strategies
+
+The simplest type of differential equation is the type learned about in the 
+integration portion of a calculus course. Such equations have the form,
+
+$$\dot{x}(t) = f(t). $$
+
+When $F(t)$ is an anti-derivative of $f(t)$ i.e. $\dot{F}=f$, then the solutions
+to this type of equation are 
+
+$$x(t) = F(t) + c. $$
+
+!!! check "Example: First order linear differential equation with constant coefficients"
+    Given the equation
+    
+    $$\dot{x}(t)=t, $$
+    
+    one finds by integrating that the solution is $\frac{1}{2}t^2 + c$. 
+    
+For first order linear differential equations, it is possible to use the 
+concept of an anti-derivative from calculus to write a general solution, in 
+terms of the independent varaible.
+    
+$$\dot{x}(t)=f(x(t)).$$ 
+    
+This implies that $\frac{\dot{x}(t)}{f(x)} = 1$. Let $F(x)$ be the 
+anti-derivative of $\frac{1}{f(x)}$. Then, making use of the chain rule
+    
+$$\frac{dot{x}(t)}{f(x(t))} = \frac{dx}{dt}} \cdot \frac{dF}{dx}} = \frac{dF}{dt} = 1$$
+    
+$$\Leftrightarrow F(x(t)) = t + c.$$
+    
+From this we notice that if we can solve for $x(t)$ then we have the 
+solution! Having a specific form for the function $f(x)$ can often makes it 
+possible to solve either implicitly or explicity for the function $x(t)$.
+
+!!! check "Example: Autonomous first order linear differential equation with constant coefficients"
+    Given the equation
+    
+    $$\dot{x} = \lambda x, $$
+    
+    we re-write the equation to be in the form 
+    
+    $$\frac{\dot{x}}{\lambda x} = 1.$$
+    
+    Now, applying the same process worked through above, let $f(x)=\lambda x$ 
+    and $F(x)$ be the anti-derivative of the $\frac{1}{f(x)}$. Integrating 
+    allows us to find the form of this anti-derivative. 
+    
+    $$F(x):= \int \frac{dx}{\lambda x} = \frac{1}{\lambda}log{\lambda x} $$
+    
+    Now, making use of the general solution we also have that $F(x(t)) =t+c$. 
+    These two equations can be combined to form an equation for $x(t)$,
+    
+    $$Log(\lambda x)  = \lambda t + c$$
+    $$x(t) = \frac{1}{\lambda} e^c e^{\lambda t} $$
+    $$x(t) = c_0 e^{\lambda t}$$
+    
+    where in the last line we defined a new constant $c_0 =\frac{1}{\lambda}e^c$.
+    Given an initial condition, we could immediately determine this constant $c_0$.
+    
+So far we have considered only DE's with constant coefficients, but it is very 
+common to encounter equations such as the following,
+
+$$\dot{x}(t)=g(t)f(x(t)).$$
+
+This type of differential equation is called a first order differential equation 
+with non-constant coefficients. If $f(x(t))$ is linear in $x$ then it is also 
+said to be a linear equation.  
+
+This equation can be re-written to isolate the coefficient function, g(t)
+
+$$\frac{dot{x}(t)}{f(x(t))} = g(t). $$
+
+Now, define $F(x)$ to be the anti-derivative of $\frac{1}{f(x)}$, and $G(t)$ to
+be the anti-derivative of $g(t)$. Without showing again the use of chain rule on
+the left side of the equation, we have
+
+$$\frac{d}{dt} F(x(t)) = g(t) $$
+$$\Rightarrow F(x(t)) = G(t) + c $$
+
+Given this form of general solution, knowledge of specific functions $f, g$ would
+make it possible to solve for $x(t)$. 
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