Numerical Lab 3: Numerical solutions of ordinary differential equations. Submission by hard copy and via turnitin by 20th October 2016. 5:00 pm Numerical Lab 3: Numerical solutions of ordinary differential equations. CHEE 2825-2019 1 This assignment is due on Wed., 23rd October 2019, 5:00 PM for the Wed. group and Friday 25th October for the Friday group. Hard copy submitted via EB box and electronic submission via Turnitin are required Numerical solutions of ordinary differential equations CHEE2825 Chemical Engineering Laboratories I, 2019 Numerical Lab 3: Numerical solutions of ordinary differential equations. CHEE 2825-2019 2 This assignment is due on Wed., 23rd October 2019, 5:00 PM for the Wed. group and Friday 25th October for the Friday group. Hard copy submitted via EB box and electronic submission via Turnitin are required Contents 1. Theoretical Background .................................................................................................... 3 1.1 Notation ....................................................................................................................... 3 1.2 Euler method ............................................................................................................... 3 1.3 Runge Kutta methods .................................................................................................. 5 2. Numerical lab: “Solving differential equations using the Euler and 4th order Runge Kutta methods” ............................................................................................................................... 6 3. Report writing .................................................................................................................... 6 3.1 Important instructions .................................................................................................. 6 3.2 Report sections............................................................................................................ 7 Numerical Lab 3: Numerical solutions of ordinary differential equations. CHEE 2825-2019 3 This assignment is due on Wed., 23rd October 2019, 5:00 PM for the Wed. group and Friday 25th October for the Friday group. Hard copy submitted via EB box and electronic submission via Turnitin are required 1. Theoretical Background 1.1 Notation The term f(x) refers to any function in a generic manner. For instance, ?(?) = ?2 + ?? When the notation specifies f(x+h), we should write ?(? + ℎ) = (? + ℎ)2 + ?(?+ℎ) We can extend the above extension to a function of two variables. For instance, ?(?, ?) = ??2 + ????(?) Therefore, if we have f(x+h/2,y+k/2), we should write x+h/2 instead of z and y+k/2 instead of y: ? (? + ℎ 2 , ? + ? 2 ) = (? + ℎ 2 ) (? + ? 2 ) 2 + (? + ? 2 ) ??? (? + ℎ 2 ) 1.2 Euler method Let’s consider the first order differential equation given by ?? ?? = ?(?, ?) (1) where x and y are the independent and dependent variables respectively. Therefore, we wish to find the values of y for any given value of x. Please note that an analytical solution to this problem may not exist. This problem is subject to the condition that when x takes the value x0, y takes the value y0. We will write this condition as (x0, y0). The solution to this problem can be found by using the Euler method which expresses the value of y at the point n+1 as a function of its derivative, which is given by Eq. 1, and a previously known value of the dependent variable yn corresponding to the value of x equal to xn. ??+1 = ?? + ℎ ?? ?? | ? (2) Where h is the integration step and therefore, ??+1 = ?? + ℎ (3) Please note that if the independent variable is time, the integration step is usually called time step. Numerical Lab 3: Numerical solutions of ordinary differential equations. CHEE 2825-2019 4 This assignment is due on Wed., 23rd October 2019, 5:00 PM for the Wed. group and Friday 25th October for the Friday group. Hard copy submitted via EB box and electronic submission via Turnitin are required If we substitute Eq. 1 into Eq. 2 we obtain. ??+1 = ?? + ℎ?(??, ??) (4) Please note that Eq. 2 corresponds to the truncation of a Taylor series in the second term. If we want to retain higher order terms we will have a 3rd, 4th and so in, Taylor method. It is important to note that the precision of the solution depends on the integration step, h. Therefore, if we know the first point (x0, y0) we can obtain the value of the function at a point 1 in the form ?1 = ?0 + ℎ (5a) ?1 = ?0 + ℎ?(?0, ?0) (5b) and thus, the value of the function at point 2 ?2 = ?1 + ℎ (6a) ?2 = ?1 + ℎ?(?1, ?1) (6b) Graphically, we can represent the numerical solution as an approximation to the real solution by considering that between point n and point n+1 the real curve is approximated by a straight line with a slope given by the derivative of the function (Eq. 1). y 2 =y 1 +h f(x 1 ,y 1 ) h f(x 1 ,y 1 ) Solution to our problem y=f(x) f(x 1 ,y 1 ) h x 0 ,y 0 y 1 =y 0 +h f(x 0 ,y 0 ) h f(x 0 ,y 0 ) h f(x 0 ,y 0 ) Independent coordinate, x y Figure 1. Graphical representation of the Euler method and comparison with the exact solution. Numerical Lab 3: Numerical solutions of ordinary differential equations. CHEE 2825-2019 5 This assignment is