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ENGG2013 Unit 22 Modeling by Differential Equations

ENGG2013 Unit 22 Modeling by Differential Equations. Apr, 2011. FREE FALLING BODY. Height, velocity and acceleration. Parabola. y = –5t 2 +15. v = –10t. a = –10. Newton’s law of motion. Assume no air friction. F = ma Force = mass  acceleration a = y’’(t) F = mg

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ENGG2013 Unit 22 Modeling by Differential Equations

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  1. ENGG2013 Unit 22Modeling byDifferential Equations Apr, 2011.


  3. Height, velocity and acceleration • Parabola y = –5t2+15 v = –10t a = –10 kshum

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

  5. Differential equation dx/dt = x + 2t d2y/dt2 = t2 + y2 kshum • 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.

  6. Initial conditions y(t) = –5t2+10t+20 • y(0)=0 • y’(0)=10 y(t) = –10t+10 y’’(t) = –10 kshum

  7. Initial conditions y(t) = –5(t+1)2 • y(0)= –5 • y’(0)= –10 y(t) = –10t –10 y’’(t) = –10 kshum

  8. Variables and parameters kshum • 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.

  9. Solutions to a differential equation y(t) = –5t2+15 is a solution to y’’(t) = –10 because after differentiating –5t2+15 twice, we get –10. y(t) = 4t2 is not a solution to y’’(t) = –10 because after differentiating 4t2 twice, we get 8, not –10. 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. kshum

  10. General solution • If every solution to a differential equation can be obtained from a family of solutions f(t,c1,c2,c3,…,cn) by choosing the constants c1, c2, c3,…, cn appropriately, then we say that f(t,c1,c2,c3,…,cn) is a general solution. kshum

  11. General solution to y’’(t) = -10 • For this simple example, just integrate two times. • Integrate both sides of y’’(t) = – 10  y’(t) = –10t+c1 • Integrate both sides of y’(t) = – 10+c1  y(t) = – 5t2+c1t+c2 (general solution) • The constants c1 and c2 can be obtained from the initial conditions. kshum


  13. Brief review of derivatives • Derivative is the slope of tangent line. • Tangent line is a line touching a curve at a point Slope of the tangent line at (x,x2) equals 2x. y=x2. Derivative of x2. kshum

  14. Slope of tangent line • Derivative is the instantaneous rate of change. y=x3+x y’ =3x2+1 y’ evaluated at x=-1 is 3(-1) 2+1=4. Slope = 4 at (-1,-2). kshum

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

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

  17. Zoom in at (1, 6.1548) Line segment with slope -1+e1=7.1548. kshum

  18. Zoom in at (2, 19.1672) Line segment with slope –1 + 3e2=21.1672 kshum

  19. 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. kshum

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

  21. Newton’s law of cooling dT/dt = – k (T – T0) (k>0) kshum • 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 T0 of the environment. • Rate of change in temperature is directly proportional to (T – T0). • k is a positive constant. • T > T0, T decreases with rate k (T – T0). • T < T0, T increases with rate k (T0 – T).

  22. Rate of change in temperature dT/dt = – 0.2 (T – 23) T  T  T0

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

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

  25. 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. T0 T  T  dT/dt = – k (T– T0) (k>0) Phase line Critical point at T0 Stable equilibrium

  26. Direction field for x’=2x(1-x) Slopesare zeroon thesetwo criticallines. The pattern is the same on every vertical line. kshum

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

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

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