Differential and Difference Equations¶

The Nature of Mathematical Modeling (draft)¶

$ \def\%#1{\mathbf{#1}} \def\*#1{\vec{#1}} \def\ds#1#2{\cfrac{d #1}{d #2}} \def\ps#1#2{\cfrac{\partial #1}{\partial #2}} \def\ba{\begin{eqnarray}} \def\ea{\end{eqnarray}} $

Ordinary Differential Equations¶

Order¶

Differential equations relate derivatives of functions. The order is the highest derivative that occurs.
A first-order equation can be immediately solved by integration:

$$ \ba \ds{y}{t} &=& f(t) \\ dy &=& f(t)~dt \\ y(t_1)-y(t_0) &=& \int_{t_0}^{t_1} f(t)~dt \ea $$

Ansatz¶

Differental equation solutions exist and are unique, so that if you find a solution it is the solution
Ansatz is a German word meaning approach or estimate
An ansatz can be used to solve a differental equation by guessing a form for the solution and then adjusting parameters; if it works then it is the solution

Linear Constant Coefficient¶

A linear constant coefficient differential equation is:

$$\frac{d^Ny}{dt^N}+A_1\frac{d^{N-1}y}{dt^{N-1}}+\cdots+A_{N-1}\ds{y}{y}+A_Ny = f(t)$$

These are frequently encountered, in part because they occur naturally, and in part because they are easy to solve
Try an ansatz $y = e^{kt}$ in the homogeneous problem $f(t)=0$:

$$e^{kt}\left(k^N+A_1k^{N-1}+\cdots+A_{N-1}k+A_N\right) = 0$$

That's solved by the characteristic equation:

$$k^N+A_1k^{N-1}+\cdots+A_{N-1}k+A_N = 0$$

This is an $N$th order polynomial that will have $N$ roots. The real part of the roots represent exponentially growing or decaying solutions, and the complex part oscillatory behavior. If the roots are distinct:

$$k^N+A_1k^{N-1}+\cdots+A_{N-1}k+A_N = (k-k_1)(k-k_2)\cdots (k-k_N)$$

then the general solution is:

$$y = \sum_{n=1}^N C_n e^{k_nt}$$

The $C_n$ are determined from the boundary conditions. In an initial value problem those are the value of the $y$ and its derivates; in a boundary value problem they've given at the beginning and end of an internval. If the roots are not distinct, the $M$ solutions for an $M$th order root are:

$$y = \left(C_1+C_2t+C_3t^2+\ldots C_Mt^{M-1}\right)e^{k_M}t$$

The complete solution is the sum of the general solution (from the homogeneous problem) and a particular solution to the inhomogeneous problem ($f(t) \ne 0$).

As an example, consider a simple RC circuit:

import matplotlib.pyplot as plt
import numpy as np
scale = .6
width = 7*scale
height = 3*scale
line = width/3
marker = 1.5*width
font = 3*width
fig,ax = plt.subplots(figsize=(width,height))
ax.axis([0,width,0,height])
ax.set_axis_off()
def resistor(x,y,width):
    x = scale*x
    y = scale*y
    width = scale*width
    dx = width/16
    dy = 2.5*dx
    xplot = np.linspace(x-width/2,x+width/2,17)
    yplot = y+np.array([0,0,0,dy,0,-dy,0,dy,0,-dy,0,dy,0,-dy,0,0,0])
    plt.plot(xplot,yplot,'k-',linewidth=line)
def capacitor(x,y,width):
    x = scale*x
    y = scale*y
    width = scale*width
    dy = width/2.5
    xplot = [[x-width,x-width,x,x],[x+width,x+width,x,x]]
    yplot = [[y+dy,y-dy,y+dy,y-dy],[y+dy,y-dy,y+width/2,y-width/2]]
    plt.plot(xplot,yplot,'k-',linewidth=line)    
def ground(x,y,width):
    x = scale*x
    y = scale*y
    width = scale*width
    xplot = [[x-.8*width,x-.55*width,x-.3*width,x],[x+.8*width,x+.55*width,x+.3*width,x]]
    yplot = [[y+.3*width,y,y-.3*width,y+.3*width],[y+.3*width,y,y-.3*width,y+.5*width]]
    plt.plot(xplot,yplot,'k-',linewidth=line)        
def wire(x0,y0,x1,y1):
    x0 = scale*x0
    x1 = scale*x1
    y0 = scale*y0
    y1 = scale*y1
    plt.plot([x0,x1],[y0,y1],'k-',linewidth=line)
def point(x,y):
    x = scale*x
    y = scale*y
    plt.plot(x,y,'ko',markersize=marker)
def circle(x,y):
    x = scale*x
    y = scale*y
    plt.plot(x,y,'ko',markersize=marker,markeredgewidth=width/2,
        markerfacecolor='white',markeredgecolor='black')
resistor(2,2.25,2)
wire(.75,2.25,1,2.25)
circle(.75,2.25)
wire(3,2.25,4,2.25)
circle(4,2.25)
point(3.25,2.25)
capacitor(3.25,1.25,.5)
wire(3.25,2.25,3.25,1.5)
ground(3.25,.25,.5)
wire(3.25,1,3.25,.5)
def text(x,y,text):
    x = scale*x
    y = scale*y
    ax.text(x,y,text,
        ha='center',va='center',
        math_fontfamily='cm',
        fontsize=font,color='black')
text(4,1.25,'$C$')
text(4.5,2.25,'$V_o$')
text(.3,2.25,'$V_i$')
text(2,1.6,'$R$')
plt.show()

The current through the resistor and capacitor must match:

$$C\dot{V}_o = \cfrac{V_i-V_o}{R}$$

therefore:

$$RC\dot{V}_o+V_o = V_i$$

The characteristic equation is:

$$ \ba RCr+1 &=& 0 \\ r &=& \cfrac{-1}{RC} \ea $$

therefore the undriven response is a decaying exponential:

$$V_o = Ae^{-t/RC}$$

To find the driven response, assume periodic forcing $V_i = \exp(i\omega t)$ (where ...). Trying an ansatz $V_o = A\exp(i\omega t)$ shows that:

$$ \ba RCi\omega+A &=& 1\\ A &=& \cfrac{1}{1+i\omega RC} \ea $$

At low frequencies the output is equal to the input; at high frequencies it rolls off as $1/\omega$ (it is a low-pass filter) and is out of phase by $90^\circ$. Problem 1 covers the important example of a damped, driven harmonic oscillator.

Systems of differential equations¶

Normal modes¶

diagonalization

Laplace Transforms¶

Perturbation Expansions¶

Partial Differential Equations¶

chain of oscillators
hyperbolic
parabolic
elliptic
SymPy

Differential and Difference Equations¶

The Nature of Mathematical Modeling (draft)¶

Ordinary Differential Equations¶

Order¶

Ansatz¶

Linear Constant Coefficient¶

Systems of differential equations¶

Normal modes¶

Laplace Transforms¶

Perturbation Expansions¶

Partial Differential Equations¶

Origin¶

Types¶

Separation of Variables¶

Coordinate Systems¶

Difference Equations¶

z-Transforms¶

Problems¶