Introduction to computational modelling

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1 Introduction to computational modelling Lecture 6 : Differential equations Physics: Pendulum Algorithm: Taylor s method Programming: Code explained Instructor : Cedric Weber Course : 4CCP1000

Schedule Class/Week Chapter Topic Milestones 1 Monte Carlo UNIX system / Fortran 2 Monte Carlo Fibonacci sequence 3 Monte Carlo Random variables 4 Monte Carlo Central Limit Theorem 5 Monte Carlo

2 Schedule Class/Week Chapter Topic Milestones 1 Monte Carlo UNIX system / Fortran 2 Monte Carlo Fibonacci sequence 3 Monte Carlo Random variables 4 Monte Carlo Central Limit Theorem 5 Monte Carlo Monte Carlo integration Milestone 1 6 Differential equations The Pendulum 7 Differential equations A Quantum Particle in a box 8 Differential equations The Tacoma bridge Milestone 2 9 Linear Algebra System of equations 10 Linear Algebra Matrix operations Milestone 3 2

3 What did you learn last time? 1. Performing an integral with Monte Carlo ý 2. Using a subroutine with the call statement ý 3. Using a random generator to produce random values ý 4. Convert integers to real numbers ý 5. Generate and plot two columns files with xmgrace ý 3

20% 10% Milestone 2 : Solving differential

4 Where are we going? 100% * Lecture 1-5 : * Statistics and integrals 90% 80% 70% 60% * Lecture 6 : * The pendulum : we will combine functions and arrays 50% 40% 30% 20% 10% Milestone 2 : Solving differential equations 0% week 1 week 2 week 3 week 4 week 5 week 6 week 7 week 8 week 9 week 10 4

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6 Today s experiment Pendulum! Step 1 : Let s write down Forces R ² At time t=0, pendulum is dropped from angle q* with initial zero velocity ² Length R, gravity constant g=9.81 ² Forces: ² e n : ² e t : normal direction tangential direction 6

Equation of motion for a circular motion Step 2 : Let s deﬁne the acceleration d ² Newton : F = M a = M v(t) [a=acceleration, v=velocity, M=mass] dt ² point moving on a circle of radius R can be

7 Equation of motion for a circular motion Step 2 : Let s deﬁne the acceleration d ² Newton : F = M a = M v(t) [a=acceleration, v=velocity, M=mass] dt ² point moving on a circle of radius R can be described by polar coordinates (R,q) ² In the Cartesian basis, a point on the circle : ² Carthesian Position: OP(t) = ( R cos( q(t) ), R sin( q(t) ) ) ² velocity: d v(t) = dt OP (t) d d v (t) = OP ( ) d dt ² v(t) = ( - R sin(q) dq/dt, R cos(q) dq/dt ) ² The velocity is along the tangential direction!

8 Newton s law Ø Acceleration : a = d dt v Ø we obtained: v(t) = ( - R sin(q) dq/dt, R cos(q) dq/dt ) Ø a = d/dt ( - R sin(q) dq/dt, R cos(q) dq/dt ) = [ Problem 1 ] Ø You will find that : Ø a = R (d 2 q(t)/dt 2 ) e t - R (dq/dt) 2 e n Ø e t =tangential direction, e n =normal direction Ø Acceleration along the tangential direction: at = R (d 2 q/dt 2 ) 8

looking for the function q(t) which satisfies this equation * With initial condition :

9 Differential equation F = M a Newton law along tangential direction : * Force = - Mg sin(q(t)) * Acceleration = R (d 2 q(t) / dt 2 ) d 2 (t) dt 2 = g sin( (t)) R * We are looking for the function q(t) which satisfies this equation * With initial condition : (t = 0) = d d (t = 0) = 0 * To simplify the equation we use R=g : d 2 (t) dt 2 = sin( (t)) 9

10 How can we solve this? d 2 dx 2 = sin( ) * Let s start with small oscillations * In this limit, we can use : 0 sin( ) * Ha! And the differential equation simplifies to : * Guess for the solution? [Fill in] d 2 dx 2 = (x) =... * For the general case, we need a computer 10

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12 A Differential equation: we know the derivative of a function at each point, but we do not know the function 12

² Brook Taylor (1685-1731) born in Edmonton ² Mathematician (Taylor series, foundation of

If I start with the initial condition : y(x=a)=a, ² à what is the value b of y(x=b)?

13 ² Brook Taylor ( ) born in Edmonton ² Mathematician (Taylor series, foundation of derivation calculus) ² Differential equation: d y(x) =f(x, y) dx ² Question: Taylors s method ² If I start with the initial condition : y(x=a)=a, ² à what is the value b of y(x=b)? ² All we know is: where we are starting from (y(a)), and the derivative of y(x) at each point along the way up to x=b 13

14 * Taylor Algorithm : Taylor s method 1. Start at x=a, we know y(a) 2. Estimate y(a+h), where h is a small number 3. y(a+h) is obtained by using the slope (derivative) of the function y(x) at the point x=a 4. Do the same for x=a+h, starting from the obtained y(a+h), and do the estimation for y(a+2h), repeat until x=b. y i+1 = y i + hf(x i,y i ) * Iterative procedure (do loop) * The smaller the interval h, the smaller the error h 14

15 Discretization and error * For the program, we will use a fixed number of interval N * The larger N, the smaller the obtained error * a and b are given, this is the interval in which we want to solve the differential equation * The value y(a) is given by the initial condition h =(b a)/n x i = a + ih y i = y(a + ih) yi 0 = y 0 (a + ih) =f(x i,y i ) 15

16 Taylor s method: second order * The pendulum equation is a second order differential equation (double derivative), we need to modify the algorithm d 2 y dx 2 = sin(y) y(0) = y y 0 (0) = 0 * Simple idea: let s define a new variable: * We get the equation : Initial conditions dy dx = z dz dx = sin(y) * We obtain a system of coupled first order differential equations z(0) = 0 y(0) = y 16

17 Algorithm : F1(y,z) and F2(y,z) are given * More generally : dy dx = F 1(y, z) dz dx = F 2(y, z) y(0) = A z(0) = B 1. Start with : y(0)=a, z(0)=b 2. Compute : F1(y(0),z(0)) F2(y(0),z(0)) 3. Estimate y(h),z(h) : y(h) = y(0) + h*f1(y(0),z(0)) z(h) = z(0) + h*f2(y(0),z(0)) 4. Compute : F1(y(h),z(h)) F2(y(h),z(h)) 5. Estimate y(2h),z(2h) : y(2h) = y(h) + h*f1(y(h),z(h)) z(2h) = z(h) + h*f2(y(h),z(h)) 6... Repeat until y( N*h ) and z( N*h ) are obtained 17

18 Let s see what we get * pendulum angle as a function of time, with an initial dropping angle q i =p/4 Period When y(x) is zero the pendulum is in the down right position Periodic oscillation (no dissipation, perpetual movement). 18

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Main structure of the code MODULE module library! contains!

implicit none real(8) :: F2 real(8) :: a,b F2= [ FILL IN ] end function end module!

implicit none integer,parameter :: N=100000 Integer :: j real(8) :: h,a,b,alpha Real(8) ::

20 Main structure of the code MODULE module library! contains! function F1(a,b) implicit none real(8) :: F1 real(8) :: a,b F1= [ FILL IN ] end function function F2(a,b) implicit none real(8) :: F2 real(8) :: a,b F2= [ FILL IN ] end function end module! PROGRAM program pendulum use library! implicit none integer,parameter :: N= Integer :: j real(8) :: h,a,b,alpha Real(8) :: x(0:n),y(0:n),z(0:n) a=0.0 ; b=30.0 ; h = [ FILL IN ] write(*,*) 'please enter initial angle' [FILL IN] x(0)=a y(0)=alpha z(0)=0.0 do j = 1, N y(j) = y(j-1) + h * F1( y(j-1),z(j-1) ) z(j) = z(j-1) + h * F2( y(j-1),z(j-1) ) x(j)= j*h End do end program 20

function F1(a,b) Implicit none is an implicit none optional statement: if present, all real(8) :: F1

21 Reminder: Function Function name is F1 a and b are the arguments (inputs) The function is also defined as a variable which will contain the return value of the function module library! contains! function F1(a,b) Implicit none is an implicit none optional statement: if present, all real(8) :: F1 variables used in the real(8) :: a,b scope of the function F1= [ FILL IN ] have to be declared/ end function defined end module! 21

22 Part 1 Part 2 Part 3 22 program pendulum use library! implicit none integer, parameter :: N= integer :: j real(8) :: h,a,b,alpha real(8) :: x(0:n),y(0:n),z(0:n) a=0.0 ; b=30.0 ; h = [ FILL IN ] write(*,*) 'please enter initial angle' [FILL IN] x(0)=a y(0)=alpha z(0)=0.0 do j = 1, N y(j) = y(j-1) + h * F1( y(j-1), z(j-1) ) z(j) = z(j-1) + h * F2( y(j-1), z(j-1) ) x(j)= j*h End do end program Program structure Part 1 : use module and implicit none Part 2 : declaration of variables, the list of all objects you are going to work with (real(8) for double precision rational numbers and integer(4) for whole numbers) Part 3 : the calculations, we use variables to perform additions/ multiplications

23 program pendulum use library implicit none Integer(4),parameter :: N= Integer(4) :: j real(8) :: h,a,b,alpha real(8) :: x(0:n),y(0:n),z(0:n) Part 2 : variable declaration 1) Parameter statement (follows integer(4) or real(8) statements) : it will assign once and for all a value to this variable, which cannot be further changed. Basically a parameter variable is identical to replacing everywhere in your code N with in this example. 2) x(0:n) : the bracket after x denotes that x is an array (a vector). The general notation to declare an array is : x(i1:i2), where i1 is the first index of the vector, and i2 the last index of the vector. The dimension or size of this vector is hence i2-i1+1 The notation x(i2) is also valid, in this case the compiler assumes that i1=1 23

24 program pendulum use library implicit none integer,parameter :: N= integer :: j real(8) :: h,a,b,x(0:n),y(0:n),z(0:n),alpha a=0.0 ; b=30.0 ; h = [ FILL IN ] write(*,*) 'please enter initial angle' [FILL IN] x(0)=a y(0)=alpha z(0)=0.0 do j = 1, N y(j) = y(j-1) + h * F1( y(j-1),z(j-1) ) z(j) = z(j-1) + h * F2( y(j-1),z(j-1) ) x(j)= j*h End do end program We apply here the Taylor s method. The first thing to do is to find the small parameter h. The curve y(x) will be obtained for x=0,h,2h,3h If there are N points along y(x), and x goes from a to b, what is h? We then ask the initial angle (where we drop the pendulum from). How can you read the initial angle from the keyboard? 24

25 program pendulum use library implicit none integer,parameter :: N= integer :: j real(8) ::h,a,b,x(0:n),y(0:n),z(0:n),alpha x(0)=a y(0)=alpha z(0)=0.0 do j = 1, N y(j) = y(j-1) + h * F1( y(j-1),z(j-1) ) z(j) = z(j-1) + h * F2( y(j-1),z(j-1) ) x(j)= j*h End do end program x(0) : x was declared as an array, the indices of the array x range from 0 to N. x : array for the time, x(0)=first time t=0, x(1) second time t=h, x(2) third time t=2h y : array for the solution of the differential equation z: array for the function z(x)=dy/dx y(0)=alpha and z(0)=0 are the initial conditions of the problem 25

26 program pendulum use library implicit none integer,parameter :: N= integer :: j real(8) ::h,a,b,x(0:n),y(0:n),z(0:n),alpha x(0)=a y(0)=alpha z(0)=0.0 do j = 1, N y(j) = y(j-1) + h * F1( y(j-1),z(j-1) ) z(j) = z(j-1) + h * F2( y(j-1),z(j-1) ) x(j)= j*h End do end program We apply the algorithm (see page 15), and we compute the function y(h) and z(h) from the functions known at t=0 (y(0) and z(0)). y(h) = y(0) + h F1(y(0),z(0)) z(h) = z(0) + h F2(y(0),z(0)) This is a call to the function defined in the module. The argument a of the function F1(a,b) will take the value y(j-1), where y(j- 1) is the element j- 1 of the array y. First iteration of the do loop, j=1, so y(j- 1)=y(0), and z(j- 1)=z(0), these are the initial conditions. 26

27 program pendulum use library implicit none integer,parameter :: N= integer :: j real(8) ::h,a,b,x(0:n),y(0:n),z(0:n),alpha x(0)=a y(0)=alpha z(0)=0.0 do j = 1, N y(j) = y(j-1) + h * F1( y(j-1),z(j-1) ) z(j) = z(j-1) + h * F2( y(j-1),z(j-1) ) x(j) = j*h [FILL IN] End do end program For j=1: from the initial condition y(0) and z(0), we just found y(1) and z(1). Y(1) is the solution of the differential equation at time x=h. We also store the time positions at which we compute y(x) in the array x. Final step: we need to write the solution of the differential equation y(x) into a file, such that we can plot it with gnuplot or xmgrace. How can we do that? Fill the blank during the problem session. 27

28 Friday, problem session 1. Problem 1 : derive analytically the differential equations of the pendulum. 2. Problem 2 : calculate the solution y(x) of the differential equation for various initial condition, and plot the result. 28

Computational modeling

Computational modeling Lecture 1 : Linear algebra - Matrix operations Examination next week: How to get prepared Theory and programming: Matrix operations Instructor : Cedric Weber Course : 4CCP1 Schedule