Mathematics 22: Lecture 10
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1 Mathematics 22: Lecture 10 Euler s Method Dan Sloughter Furman University January 22, 2008 Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
2 Euler s method Consider the initial-value problem du dt = f (t, u), u(t 0) = u 0. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
3 Euler s method Consider the initial-value problem For a small h > 0, let t 1 = t 0 + h. du dt = f (t, u), u(t 0) = u 0. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
4 Euler s method Consider the initial-value problem For a small h > 0, let t 1 = t 0 + h. du dt = f (t, u), u(t 0) = u 0. To approximate u(t 1 ), first note that implies t1 t 0 u(t 1 ) u(t 0 ) = du t1 dt dt = f (s, u)ds t 0 t1 t 0 f (s, u)ds hf (t 0, u(t 0 )). Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
5 Euler s method Consider the initial-value problem For a small h > 0, let t 1 = t 0 + h. du dt = f (t, u), u(t 0) = u 0. To approximate u(t 1 ), first note that implies Hence t1 t 0 u(t 1 ) u(t 0 ) = du t1 dt dt = f (s, u)ds t 0 t1 t 0 f (s, u)ds hf (t 0, u(t 0 )). u(t 0 + h) u 0 + f (t 0, u 0 )h. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
6 Euler s method (cont d) To approximate a solution to the initial-value problem on an interval [t 0, t 0 + T ]: du dt = f (t, u), u(t 0) = u 0, Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
7 Euler s method (cont d) To approximate a solution to the initial-value problem du dt = f (t, u), u(t 0) = u 0, on an interval [t 0, t 0 + T ]: divide the interval into N subintervals of equal length h = T N, Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
8 Euler s method (cont d) To approximate a solution to the initial-value problem du dt = f (t, u), u(t 0) = u 0, on an interval [t 0, t 0 + T ]: divide the interval into N subintervals of equal length let ti = t 0 + ih, i = 1, 2,..., N, h = T N, Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
9 Euler s method (cont d) To approximate a solution to the initial-value problem du dt = f (t, u), u(t 0) = u 0, on an interval [t 0, t 0 + T ]: divide the interval into N subintervals of equal length let ti = t 0 + ih, i = 1, 2,..., N, for i = 0, 1, 2,..., N 1, let h = T N, u i+1 = u i + hf (t i, u i ). Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
10 Euler s method (cont d) To approximate a solution to the initial-value problem du dt = f (t, u), u(t 0) = u 0, on an interval [t 0, t 0 + T ]: divide the interval into N subintervals of equal length let ti = t 0 + ih, i = 1, 2,..., N, for i = 0, 1, 2,..., N 1, let h = T N, u i+1 = u i + hf (t i, u i ). Then u i u(t i ). Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
11 Example Consider the initial-value problem du dt = u cos(t), u(0) = 1. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
12 Example Consider the initial-value problem du dt = u cos(t), u(0) = 1. To approximate u on [0, 6], we will take h = 0.1, that is, N = 60. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
13 Example Consider the initial-value problem du dt = u cos(t), u(0) = 1. To approximate u on [0, 6], we will take h = 0.1, that is, N = 60. We have u 0 = 1.0, u 1 = (0.1)(1.0) cos(0) = 1.1, u 2 = (0.1)(1.1) cos(0.1) = , u 3 = (0.1)( ) cos(0.2) = Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
14 Example Consider the initial-value problem du dt = u cos(t), u(0) = 1. To approximate u on [0, 6], we will take h = 0.1, that is, N = 60. We have u 0 = 1.0, u 1 = (0.1)(1.0) cos(0) = 1.1, u 2 = (0.1)(1.1) cos(0.1) = , u 3 = (0.1)( ) cos(0.2) = Hence, for example, u(0.3) Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
15 Example (cont d) Note: the exact solution of the equation is u(t) = e sin(t), so u(0.3) = e sin(0.3) = Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
16 Using Octave Solution in Octave: octave:1> t=[0:0.1:6]; octave:2> u(1)=1; octave:3> for i = 1:60 > u(i+1) = u(i) + 0.1*u(i)*cos(t(i)); > endfor Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
17 Using Octave Solution in Octave: octave:1> t=[0:0.1:6]; octave:2> u(1)=1; octave:3> for i = 1:60 > u(i+1) = u(i) + 0.1*u(i)*cos(t(i)); > endfor Note: indices begin with 1 in Octave, so u starts with u(1) and ends with u(61). The same comment holds for t. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
18 Using Octave Solution in Octave: octave:1> t=[0:0.1:6]; octave:2> u(1)=1; octave:3> for i = 1:60 > u(i+1) = u(i) + 0.1*u(i)*cos(t(i)); > endfor Note: indices begin with 1 in Octave, so u starts with u(1) and ends with u(61). The same comment holds for t. In particular, u 60 is u(61) in Octave. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
19 Using Octave Solution in Octave: octave:1> t=[0:0.1:6]; octave:2> u(1)=1; octave:3> for i = 1:60 > u(i+1) = u(i) + 0.1*u(i)*cos(t(i)); > endfor Note: indices begin with 1 in Octave, so u starts with u(1) and ends with u(61). The same comment holds for t. In particular, u 60 is u(61) in Octave. For example, we now see that u(6) u 60 = Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
20 Using Octave Solution in Octave: octave:1> t=[0:0.1:6]; octave:2> u(1)=1; octave:3> for i = 1:60 > u(i+1) = u(i) + 0.1*u(i)*cos(t(i)); > endfor Note: indices begin with 1 in Octave, so u starts with u(1) and ends with u(61). The same comment holds for t. In particular, u 60 is u(61) in Octave. For example, we now see that u(6) u 60 = Exact value: u(6) = Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
21 Octave output The command octave:4> [t,u ] will print out the values of t and u in side-by-side columns. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
22 Octave output The command octave:4> [t,u ] will print out the values of t and u in side-by-side columns. Note: We need t, and u, because t and u are row vectors. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
23 Octave output The command octave:4> [t,u ] will print out the values of t and u in side-by-side columns. Note: We need t, and u, because t and u are row vectors. Note: In some versions of Octave, u is not a row vector. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
24 Octave output The command octave:4> [t,u ] will print out the values of t and u in side-by-side columns. Note: We need t, and u, because t and u are row vectors. Note: In some versions of Octave, u is not a row vector. If we define octave:5> z=exp(sin(t)); then octave:6> [t,u,z ] will print out the values of t, u, and z (the exact solution) in side-by-side columns. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
25 Octave output (cont d) The command octave:7> plot(t,u) will plot the approximate solution, and the command octave:8> plot(t,u,t,z) will plot a comparison of the approximate solution with the exact solution. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
26 Octave output (cont d) Comparison of exact (green) and approximate (red) solutions: Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
27 Modified Euler s method Again consider the initial-value problem du dt = f (t, u), u(t 0) = u 0. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
28 Modified Euler s method Again consider the initial-value problem du dt = f (t, u), u(t 0) = u 0. Let T, h, N, and t i, i = 0, 1, 2,..., N be as before. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
29 Modified Euler s method Again consider the initial-value problem du dt = f (t, u), u(t 0) = u 0. Let T, h, N, and t i, i = 0, 1, 2,..., N be as before. Assuming we have computed u 1, u 2,..., u i, let be the predicted value for u(t i+1 ). ũ i+1 = u i + hf (t i, u i ). Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
30 Modified Euler s method Again consider the initial-value problem du dt = f (t, u), u(t 0) = u 0. Let T, h, N, and t i, i = 0, 1, 2,..., N be as before. Assuming we have computed u 1, u 2,..., u i, let be the predicted value for u(t i+1 ). Now let ũ i+1 = u i + hf (t i, u i ). u i+1 = u i + f (t i, u i ) + f (t i+1, ũ i+1 ) h. 2 Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
31 Modified Euler s method Again consider the initial-value problem du dt = f (t, u), u(t 0) = u 0. Let T, h, N, and t i, i = 0, 1, 2,..., N be as before. Assuming we have computed u 1, u 2,..., u i, let be the predicted value for u(t i+1 ). Now let ũ i+1 = u i + hf (t i, u i ). u i+1 = u i + f (t i, u i ) + f (t i+1, ũ i+1 ) h. 2 This is an example of a predictor-corrector method. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
32 Example For our previous example, we would have ũ 1 = (0.1)(1.0) cos(0) = 1.1, and so u 1 = cos(0) cos(0.1) (0.1) = Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
33 Example For our previous example, we would have ũ 1 = (0.1)(1.0) cos(0) = 1.1, and so Next u 1 = cos(0) cos(0.1) (0.1) = and ũ 2 = (0.1)( ) cos(0.1) = u 2 = = cos(0.1) cos(0.2) (0.1) 2 Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
34 Using Octave Note: octave:9> clear u will clear the old values of u. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
35 Using Octave Note: octave:9> clear u will clear the old values of u. To define f (t, u) in Octave: octave:10> function w = f(t, u) > w = u*cos(t); > endfunction Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
36 Using Octave Note: octave:9> clear u will clear the old values of u. To define f (t, u) in Octave: octave:10> function w = f(t, u) > w = u*cos(t); > endfunction Modified Euler s method in Octave: octave:11> t = [0:0.1:6]; octave:12> u(1) = 1.0; octave:13> for i=1:60 > p = u(i) + 0.1*f(t(i),u(i)); > u(i+1) = u(i) + 0.1*(f(t(i),u(i)) + f(t(i+1),p))/2; > endfor Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
37 Using Octave (cont d) This time we have u(6) u 60 = Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
38 Using Octave (cont d) This time we have u(6) u 60 = Note: again, this is u(61) in Octave. Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
39 Using Octave (cont d) This time we have u(6) u 60 = Note: again, this is u(61) in Octave. Recall: exact value is u(6) = Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
40 Using Octave (cont d) This time we have u(6) u 60 = Note: again, this is u(61) in Octave. Recall: exact value is u(6) = Recall: the approximation from Euler s method was Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
41 Using Octave (cont d) Comparison of exact (green) and approximate (red) solutions: Dan Sloughter (Furman University) Mathematics 22: Lecture 10 January 22, / 14
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