ENGG2013 Unit 22 Modeling by Differential Equations Apr, 2011.

FREE FALLING BODY kshum2

Height, velocity and acceleration Parabola kshum3 y = –5t v = –10t a = –10

Newton’s law of motion F = ma – Force = mass  acceleration – a = y’’(t) F = mg – Gravitational Force is proportional to the mass, the proportionality constant g  –10 ms -2. kshum4 y’’(t) = g Assume no air friction

Differential equation kshum5 A differential equation is an equation which involves derivatives. Examples: The variable x is a function of time t. The variable y is a function of t. dx/dt = x + 2t d 2 y/dt 2 = t 2 + y 2

Initial conditions y(0)=0 y’(0)=10 kshum6 y(t) = –5t 2 +10t+20 y(t) = –10t+10 y’’(t) = –10

Initial conditions y(0)= –5 y’(0)= –10 kshum7 y(t) = –5(t+1) 2 y(t) = –10t –10 y’’(t) = –10

Variables and parameters kshum8 The dependent variable is called the system state, or the phase of the system. The independent variable is usually time. A constant which does not change with time is called a parameter. In the example Newton’s law of motion y’’(t) = g – Phase = system state = height of the mass – Independent variable = time – g is parameter.

Solutions to a differential equation A solution is a function which satisfies the given differential equation. In solving differential equation, the solutions are function of time. In general, there are many solutions to a given differential equation. We have different solutions for different initial condition. Deriving a solution is difficult, but checking whether a given function is a solution is easy. kshum9 y(t) = –5t is a solution to y’’(t) = –10 because after differentiating –5t twice, we get –10. y(t) = 4t 2 is not a solution to y’’(t) = –10 because after differentiating 4t 2 twice, we get 8, not –10.

General solution If every solution to a differential equation can be obtained from a family of solutions f(t,c 1,c 2,c 3,…,c n ) by choosing the constants c 1, c 2, c 3,…, c n appropriately, then we say that f(t,c 1,c 2,c 3,…,c n ) is a general solution. kshum10

General solution to y’’(t) = -10 For this simple example, just integrate two times. Integrate both sides of y’’(t) = – 10  y’(t) = –10t+c 1 Integrate both sides of y’(t) = – 10+c 1  y(t) = – 5t 2 +c 1 t+c 2 (general solution) The constants c 1 and c 2 can be obtained from the initial conditions. kshum11

FIRST-ORDER DIFFERENTIAL EQUATION kshum12

Brief review of derivatives Derivative is the slope of tangent line. – Tangent line is a line touching a curve at a point kshum13 y=x 2. Slope of the tangent line at (x,x 2 ) equals 2x. Derivative of x 2.

Slope of tangent line Derivative is the instantaneous rate of change. kshum14 y=x 3 +x Slope = 4 at (-1,-2). y’ =3x 2 +1 y’ evaluated at x=-1 is 3(-1) 2 +1=4.

First-order differential equation No second or higher derivative, for example First-order derivative defines slope. Example kshum15 dx/dt = a function of x and t General solution constant

An illustration If an initial condition is given, then we can solve for the constant C. Suppose that x(0) = 2.  C=3. kshum16

Zoom in at (1, ) kshum17 Line segment with slope -1+e 1 =

Zoom in at (2, ) kshum18 Line segment with slope –1 + 3e 2 =

Direction field or slope field A graphical method for solving differential equation. Systematically evaluate f(x,t) on a grid on points. On a grid point (t,x), draw a short line segment with slope f(x,t). A solution must follow the flow pattern. kshum19

Direction Field for x’=x+t kshum20 Sample solution for x(0)=2 Each grid point (t,x) is associated with a line segment with slope x+t.

Newton’s law of cooling kshum21 Imagine a can of coffee in an air-conditioned room. The rate of change of the temperature T(t) is directly proportional to the difference between T and the temperature T 0 of the environment. Rate of change in temperature is directly proportional to (T – T 0 ). – k is a positive constant. – T > T 0, T decreases with rate k (T – T 0 ). – T < T 0, T increases with rate k (T 0 – T). dT/dt = – k (T – T 0 ) (k>0)

Rate of change in temperature T  T  dT/dt = – 0.2 (T – 23) T0T0

Direction field dT/dt = – 0.2 (T – 23)

Sample solutions Some typical solution paths, corresponding initial temperature 0, 5, 10, 15, 20, 25, 30, 35, 40, are shown in the graph. dT/dt = – 0.2 (T – 23)

Autonomous DE and Phase line Autonomous DE: x’(t) = a function of x only. – no independent variable on the R. H. S. For autonomous DE, we can understand the system via the phase line. T0T0 T  T  dT/dt = – k (T– T 0 ) Phase line Stable equilibrium Critical point at T 0 (k>0)

Direction field for x’=2x(1-x) kshum26 The pattern is the same on every vertical line. Slopes are zero on these two critical lines.

Phase line for x’=2x(1-x) kshum27 1 x  Phase line Critical points at x=0 and x=1 0 x  Stable equilibriumUnstable equilibrium Without solving the differential equation explicitly, we know that the solution x(t) converges to 1 if it starts at positive x(0), but diverges to negative infinity if it starts at negative x(0).

Main concepts Independent variable, dependent variable and parameters Initial conditions General solution Direction field Autonomous differential equations. – Phase line Equilibrium – Stable and unstable