1. PART 7 Ordinary Differential Equations ODEs

2. Ordinary Differential Equations Part 7
Equations which are composed of an unknown function and its derivatives are called differential equations.
Differential equations play a fundamental role in engineering because many physical phenomena are best formulated mathematically in terms of their rate of change.
v - dependent variable
t - independent variable

3. Ordinary Differential Equations
When a function involves one dependent variable, the equation is called an ordinary differential equation (ODE).
A partial differential equation (PDE) involves two or more independent variables.

4. Ordinary Differential Equations
Differential equations are also classified as to their order:
A first order equation includes a first derivative as its highest derivative.
- Linear 1st order ODE
- Non-Linear 1st order ODE

5. Ordinary Differential Equations
A second order equation includes a second derivative.
- Linear 2nd order ODE
- Non-Linear 2nd order ODE
Higher order equations can be reduced to a system of first order equations, by redefining a variable.

6. Ordinary Differential Equations

7. Runge-Kutta Methods
This chapter is devoted to solving ODE of the form:
Euler’s Method
solution

8. Runge-Kutta Methods

9. Euler’s Method: Example
Obtain a solution between x = 0 to x = 4
with a step size of 0.5 for:
Initial conditions are: x = 0 to y = 1
Solution:

10. Euler’s Method: Example
Although the computation captures the general trend solution, the error is considerable.
This error can be reduced by using a smaller step size.

11. System of Ordinary Differential Equations
The solution is finding y1, y2,y3,…yn

12. Euler’s Method: Example
Obtain a solution between x= 0 & 4 (step size of 0.5)
Initial conditions are: x = 0 ; y1= 4 & y2 = 6
Solution:
X y1 y2

13. Improvements of Euler’s method
A fundamental source of error in Euler’s method is that the derivative at the beginning of the interval is assumed to apply across the entire interval.
Simple modifications are available:
Heun’s Method
The Midpoint Method
Ralston’s Method

14. Runge-Kutta Methods
Runge-Kutta methods achieve the accuracy of a Taylor series approach without requiring the calculation of higher derivatives.
Increment function (representative slope over the interval)

15. Runge-Kutta Methods
Various types of RK methods can be devised by employing different number of terms in the increment function as specified by n.
First order RK method with n=1 is Euler’s method.
Second order RK methods:
Values of a1, a2, p1, and q11 are evaluated by setting the second order equation to Taylor series expansion to the second order term.

16. Runge-Kutta Methods
Three equations to evaluate the four unknowns constants are derived:
A value is assumed for one of the unknowns to solve for the other three.

17. Runge-Kutta Methods
We can choose an infinite number of values for a2,there are an infinite number of second-order RK methods.
Every version would yield exactly the same results if the solution to ODE were quadratic, linear, or a constant.
However, they yield different results if the solution is more complicated (typically the case).

18. Runge-Kutta Methods Three of the most commonly used methods are:
Huen Method with a Single Corrector (a2=1/2)
The Midpoint Method (a2= 1)
Raltson’s Method (a2= 2/3)

19. Runge-Kutta Methods Heun’s Method:y
f(xi,yi) xi
xi+h x
Heun’s Method:
Involves the determination of two derivatives for the interval at the initial point and the end point.
f(xi+h,yi+k1h) y ea xi
xi+h x
Slope: 0.5(k1+k2)

20. Heun’s Method - Example
Obtain a solution between x=0 & 4 (step size = 0.5) for:
Initial conditions are: x = 0 to y = 1,use Heun’s method
Chapter 25

21. Runge-Kutta Methods Midpoint Method:y
f(xi,yi) xi
xi+h/2 x
Midpoint Method:
Uses Euler’s method t predict a value of y at the midpoint of the interval:
f(xi+h/2,yi+k1h/2) y ea xi
xi+h x
Slope: k2
Chapter 25

22. Midpoint Method (Example 25.6, P705)
Obtain a solution between x=0 & 4 (step size = 0.5) for:
Initial conditions are: x = 0 to y = 1,use Midpoint method
Chapter 25

23. Runge-Kutta Methods Ralston’s Method: y f(xi,yi) xi xi+3/4h ea
Slope: (1/3k1+2/3k2)
Ralston’s Method:
f(xi+ 3/4 h, yi+3/4k1h) x
xi+h x
Chapter 25

24. Chapter 25

25. Runge-Kutta Methods
Third order RK methods
Chapter 25

26. Runge-Kutta Methods
Fourth order RK methods
Chapter 25

27. Comparison of Runge-Kutta Methods
Use first to fourth order RK methods to solve the equation from x = 0 to x = 4
Initial condition y(0) = 2, exact answer of y(4) =
Chapter 25

28. Implicit Versus Explicit Euler’s Methods
f(xi,yi) xi h
xi +h y x
Implicit Versus Explicit Euler’s Methods
Implicit Euler’s Method:
f(xi+1,yi+1)
f(xi,yi) xi h
xi +1 y x
Chapter 27

29. The Finite Difference Method
The previous central divided differences are substituted for the derivatives in the original equation.
Thus a differential equation is transformed into a set of simultaneous algebraic equations that can be solved by the methods described earlier.
Chapter 27

30. The Central divided differences
f(x)
f(x)
f(xi+1)
f(xi-1)
f(xi)
f\(xi) Dx Dx x
xi-1 xi
xi+1
Chapter 27

31. The Finite Difference Method-Example
Solve the following 2nd order ODE for initial conditions of T(0) = T1 and T(L) = T2
Where Ta and h/ are constants.
Solution:
Using the divided differences equations:
Substitute into the original DE for d2T/dx2:
Chapter 27

32. The Finite Difference Method-Example
Now collecting terms gives:
The above equation is now applied at each of the interior point.
The first point and the last point are specified by the boundary conditions.
So if there is 5 points (including the first and the last point) we will have 3 equations in 3 unknowns.
Chapter 27

33. The Finite Difference Method:Example
Solve the following 2nd order ODE for initial conditions of y(0) = 5 and y(10) = -1.5, Use Dx = 2
Using the divided differences equations:
Chapter 27

34. The Finite Difference Method: Example
Substitute into the original DE for d2y/dx2 and dy/dx:
Now collecting terms gives and substitute for Dx =2:
The above equations will be used to find the y’s at 2, 4, 6 and 8 where y(0) = 5 and y(10) = -1.5
Chapter 27

35. The Finite Difference Method: Example
Chapter 27

36. The Finite Difference Method: Example
Or in a matrix format:
Now you can use for example Gauss elimination to solve these equations.
Chapter 27

37. PART 8 Partial Differential Equations PDEs
Finite Difference Method
Finite Element Method
Chapter 29

38. Partial Differential Equations
Chapter 29

39. Partial Differential Equations
A partial differential equation (PDE) involves two or more independent variables. For example:
1. Laplace equation
2. Diffusion equation
3. Wave equation
Chapter 29

40. Partial Differential Equations 1. Finite Difference Method
Similar to the ODE, central divided differences are substituted for the partial derivatives in the original equation.
Thus a partial differential equation is transformed into a set of simultaneous algebraic equations that can be solved by the methods described earlier.
Chapter 29

41. Partial Differential Equations
The Central divided differences x y
i,j
i-1,j
i+1,j
i,j+1
i,j-1
Chapter 29

42. Finite Difference: Elliptic Equations
Because of its simplicity and general relevance to most areas of engineering, we will use a heated plate as an example for solving elliptic PDEs.
Chapter 29

43. Figure 29.1
by Lale Yurttas, Texas A&M University
Chapter 29

44. Figure 29.3
by Lale Yurttas, Texas A&M University
Chapter 29

45. The Laplacian Difference Equations/
Laplace Equation
O[D(x)2]
O[D(y)2]
Laplacian difference equation.
Holds for all interior points
Chapter 29

46. Figure 29.4
Chapter 29

47. A balance for node (1,1) is:
In addition, boundary conditions along the edges must be specified to obtain a unique solution.
The simplest case is where the temperature at the boundary is set at a fixed value, Dirichlet boundary condition.
A balance for node (1,1) is:
Similar equations can be developed for other interior points to result a set of simultaneous equations.
Chapter 29

48. The result is a set of nine simultaneous equations with nine unknowns:
Chapter 29

49. Partial Differential Equations: Example 1: Laplace Equation
Substitute with the Central divided differences and assuming that Dx = Dy = h

50. Partial Differential Equations: Example 1: Laplace Equationy
Finite Difference Grid
At i and j:
i,j+1 -4 1
i,j
i,j
i-1,j
i+1,j
i,j-1 x

51. Partial Differential Equations: Example 2: Diffusion Equation
Substitute with the Central divided differences and assuming that Dx = Dy = h:

52. Partial Differential Equations: Example 2: Diffusion Equationy
Finite Difference Grid
At i and j:
i,j+1 2 -1
h/2a
i,j
i,j
i-1,j
i+1,j
i,j-1 x