Queens College, CUNY, Department of Computer Science Numerical Methods CSCI 361 / 761 Spring 2018 Instructor: Dr. Sateesh Mane.
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1 Queens College, CUNY, Department of Computer Scence Numercal Methods CSCI 361 / 761 Sprng 2018 Instructor: Dr. Sateesh Mane c Sateesh R. Mane Lecture 16a May 3, 2018 Numercal soluton of systems of ordnary dfferental equatons We dsplay worked examples of ntal alue problems usng auxlary arables. 1
2 16.12 Worked example Equaton Consder the followng ordnary dfferental equaton d 2 y dy + 2x dx2 dx + (1 x2 y e x. ( It s a second order ordnary dfferental equaton. We ntroduce an auxlary arable a dy dx. ( We reexpress eq. ( as a par of coupled frst order equatons as follows: dy dx, d dx 2x (1 x2 y + e x. ( a ( b We express ths n the formal notaton as follows. To aod confuson between y and the orgnal arable y, defne y (y 1, y 2 (u,. Then u s the orgnal scalar arable y and the equatons are d dx dy f(x, y, dx ( u ( f1 (x, u, f 2 (x, u, ( 2x (1 x 2 y + e x. ( Therefore the rght hand sde functons are f 1 (x, u,, f 2 (x, u, 2x (1 x 2 u + e x. ( a ( b 2
3 C++ code In terms of C++ functon calls, we hae m 2 and nt f(nt m, double x, const std::ector<double> & y, std::ector<double> & g { // frst component "f1(x,u, " g[0] y[1]; // second component "f2(x,u, -2x -(1-x^2y + exp(x" g[1] -2.0*x*y[1] - (1.0 - x*x*y[0] + exp(x; } return 0; All the ntegraton schemes can be called wth the followng nputs: // m 2 // x x_ // h step sze // ector (array y_n (u, _ // ector (array y_out (u, _{+1} nt Euler_explct(nt m, double x, double h, std::ector<double> & y_n, std::ector<double> & y_out; nt mdpont(nt m, double x, double h, std::ector<double> & y_n, std::ector<double> & y_out; nt trapezod(nt m, double x, double h, std::ector<double> & y_n, std::ector<double> & y_out; nt RK4(nt m, double x, double h, std::ector<double> & y_n, std::ector<double> & y_out; 3
4 Soluton for 0, etc Let the ntal condtons at x 0 0 be y 0 1 and 0 y 0 1. We shall ntegrate eq. ( usng explct Euler ntegraton wth a stepsze h. The formal equatons are y +1 y + hf(x, y. ( In terms of u and, the equatons are ( ( ( u u + h 2x (1 x 2 u + exp(x Step 0: Intally u 0 1 and 0 1. Then. ( u 1 u 0 + h 0 1 h. ( Next we hae Step 1: h ( 2x 0 0 (1 x 2 0u 0 + exp(x h ( 0 (1 0 + exp(0 1. ( We hae 2. Next we hae u 2 u 1 + h 1 1 h h 1 2h. ( h ( 2x 1 1 (1 x 2 1u 1 + exp(x h ( 2h (1 h 2 (1 h + exp(h 1 + h ( 1 + 3h + h 2 h 3 + exp(h ( h + 3h 2 + h 3 h 4 + h exp(h. Step 2: 1. We hae u 3 u 2 + h 2 1 2h + h ( 1 h + 3h 2 + h 3 h 4 + h exp(h 1 3h h 2 + 3h 3 + h 4 h 5 + h 2 exp(h. ( Next we hae h ( 2x 2 2 (1 x 2 2u 2 + exp(x h ( 4h 2 (1 4h 2 u 2 + exp(2h ugh. (
5 16.13 Worked example Equaton Consder the followng ordnary dfferental equaton Ths s a smple equaton. We know the general soluton s d 2 y dx 2 + y 0. ( y(x c 1 cos(x + c 2 sn(x. ( It s a second order ordnary dfferental equaton. We ntroduce an auxlary arable a dy dx. ( We reexpress eq. ( as a par of coupled frst order equatons as follows: dy dx, d dx y. ( a ( b We express ths n the formal notaton as follows. Agan defne y (y 1, y 2 (u,. Then d dx dy f(x, y, dx ( u ( f1 (x, u, f 2 (x, u, (. u ( Therefore the rght hand sde functons are f 1 (x, u,, ( a f 2 (x, u, u. ( b 5
6 C++ code In terms of C++ functon calls, we hae m 2 and nt f(nt m, double x, const std::ector<double> & y, std::ector<double> & g { // frst component "f1(x,u, " g[0] y[1]; // second component "f2(x,u, -u" g[1] -y[0]; } return 0; All the ntegraton schemes can be called wth the followng nputs: // m 2 // x x_ // h step sze // ector (array y_n (u, _ // ector (array y_out (u, _{+1} nt Euler_explct(nt m, double x, double h, std::ector<double> & y_n, std::ector<double> & y_out; nt mdpont(nt m, double x, double h, std::ector<double> & y_n, std::ector<double> & y_out; nt trapezod(nt m, double x, double h, std::ector<double> & y_n, std::ector<double> & y_out; nt RK4(nt m, double x, double h, std::ector<double> & y_n, std::ector<double> & y_out; 6
7 Soluton for 0, etc Let the ntal condtons at x 0 0 be y 0 1 and 0 y 0 0. Then we know the exact soluton s y exact (x cos(x. ( We shall ntegrate eq. ( usng explct Euler ntegraton wth a stepsze h. The formal equatons are y +1 y + hf(x, y. ( In terms of u and, the equatons are ( u Step 0: Intally u 0 1 and 0 0 and so ( ( u + h. ( u u 1 u 0 + h ( Next we hae 1 0 hu 0 0 h h. ( Step 1: 1. We hae 2. Next we hae u 2 u 1 + h h( h 1 h 2. ( hu 1 h h 2h. ( Step 2: 1. We hae 2. Next we hae u 3 u 2 + h 2 1 h 2 + h( 2h 1 3h 2. ( hu 2 2h h(1 h 2 3h + h 3. ( Step 3: 1. We hae u 4 u 3 + h 3 1 3h 2 + h( 3h + h 3 1 6h 2 + h 4. ( Next we hae 4 3 hu 3 3h + h 3 h(1 3h 2 4h + 4h 3. ( We see that the explct Euler method s not ery accurate. 7
8 Soluton usng mdpont method Let us ntegrate eq. ( usng the mdpont method wth a stepsze h. In terms of u and, the equatons are ( u Howeer, now we requre the alues ntermedate ponts. +1 ( ( u + h. ( u 1. Frst we hae, at the step, ( ( g u g 1 g. ( u 2. Next we hae ( g u g 2 1 g1 Step 0: ( f1 (x h, u + h 2 gu 1, + h 2 g 1 f 2 (x h, u + h 2 gu 1, + h 2 g 1 ( h 2 u u h 2. ( We hae 2. Next g u 2 0 h 2 u 0 0 h 2. ( g 2 u 0 h ( Then u 1 u 0 + hg u 2 1 h2 2, hg 2 0 h h. (
9 Step 1: 1. Now g u 2 and g 2 refer to the step 1. (g u h 2 u 1 h h 2 (1 h2 3h h3 4. ( Next (g 2 1 u 1 h h2 2 + h h2. ( Then Step 2: 1. Now g u 2 and g 2 u 2 u 1 + h(g2 u 1 1 h2 ( 2 + h 3h 2 + h h 2 + h4 4, h(g 2 1 h + h( 1 + h 2 refer to the step 2. 2h + h 3. ( Next (g u h 2 u 2 2h + h 3 h 2 (1 2h 2 + h4 5h h3 h5 8. ( (g 2 2 u 2 h h 2 h4 4 + h2 h h2 3h4 4. ( Then u 3 u 2 + h(g2 u 2 1 2h 2 + h4 ( 4 + h 5h 2 + 2h3 h h h4 4 h6 8, h(g2 2 2h + h 3 + h ( 1 + 3h 2 3h4 4 3h + 4h 3 3h5 4. ( Ths s more accurate than the explct Euler method. It matches the exact soluton to O(h 2. 9
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Quees College, CUNY, Departmet of Computer Scece Numercal Methods CSCI 361 / 761 Fall 2018 Istructor: Dr. Sateesh Mae c Sateesh R. Mae 2018 16 Homework lecture 16 Please emal your soluto, as a fle attachmet,
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