1. Consider the initial value problem: find y(t) such that. y = y 2 t, y(0) = 1.

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1 Engineering Mathematics CHEN30101 solutions to sheet 3 1. Consider the initial value problem: find y(t) such that y = y 2 t, y(0) = 1. Take a step size h = 0.1 and verify that the forward Euler approximation to y(0.1) is y 1 = 1.0 and that the approximations to subsequent points are y 2 = 1.01, y 3 = and y 4 = Compare these estimates with the exact values y(0.1), y(0.2), y(0.3) and y(0.4) obtained by solving the ODE algebraically. Running the forward Euler (fe) process the estimates y j y(t j ) are tabulated below. The results are rounded to four decimal places. h t j y j f(y j,t j ) y(t j ) Note that the exact solution is the function y(t) = 2 2 t Bytakingastepsizeofh = 0.5, computetheforward Eulerapproximation of z(5), where z(t) satisfies the initial value problem: z = 1 z, z(0) = 1. 50(1+t) Running (fe) the approximate solution z j is tabulated below. The results are rounded to four decimal places. h t j z j f(z j,t j ) z(t j ) Note that the exact solution is the function z(t) = (1+t) 1/50. 1

2 3. Consider the initial value problem: find y(t) such that y = π 1 y 2, y(0) = 0. Take a step size h = 0.1 and show that computing the forward Euler approximation to y(1) is problematic. Running the forward Euler process gives... h t j y j f(y j,t j ) If y = , y = π 1 y 2 cannot be calculated and the method stops. 4. Consider the initial value problem: find y(x) such that y = 1 2 ex y, y(0) = 0. Compute the forward Euler approximation of y(1) using (a) a single step of size h = 1, (b) two steps each of size h = 0.5, (c) four steps each of size h = 0.25, and (d) ten steps each of size h = 0.1. Compare your estimates with the exact value. (a) Running (fe) for h = 1 gives so y(1) 0.5. (b) Running (fe) for h = 0.5 gives so y(1) (c) Running (fe) for h = 0.25 gives so y(1) h x j y j f(y j,x j ) h x j y j f(y j,x j ) h x j y j f(y j,x j )

3 (d) Running (fe) for h = 0.1 gives so y(1) h x j y j f(y j,x j ) The solution is y(x) = ln ( ) 1+e x 2 so y(1) = to four decimal places. The accuracy of the forward Euler approximation improves when the step size h is reduced. 5. By taking a step size of h = 0.5, compute the modified Euler approximation of z(2), where z(t) satisfies the initial value problem: z = 1 z, z(0) = 1. 50(1+t) Running the modified Euler (me) process the approximate solution y j is tabulated below. Note that t m = t j + h 2 and z m = z j + h 2 f(z j,t j ). h t j z j f(z j,t j ) t m z m f(z m,t m ) z(t j ) These results are clearly much more accurate than those obtained for the forward Euler approximation scheme (cf. question 2). 6. Consider the initial value problem: find y(x) such that y = 17 12x 3y, y(0) = 9. Using four steps of size h = 0.25, estimate y(1) using (i) the forward Euler method, (ii) the modified Euler method. Compare your estimates with the exact value. 3

4 The exact solution is the function y(x) = 2e 3x + 7 4x. Running the forward Euler process gives... h x j y j f(y j,x j ) Running the modified Euler process gives... h x j y j f(y j,x j ) x m y m f(y m,x m ) y(x j ) Neither solution is particularly accurate but the modified Euler method gives more accurate results than the forward Euler method. 7. Heun s method for approximating the solution of an initial value problem is given by K 1 = f(x n,y n ), K 2 = f(x n +h,y n +hk 1 ) y n+1 = y n + h 2 (K 1 +K 2 ). Carry out two steps of the method (with step size h = 0.5) to compute an estimate of the solution y(x) of the problem y = y x, y(0) = 20. Running the process with x = x n +h and y = y n +hk 1 gives... h x n y n K 1 x y K 2 y(t n ) Consider the initial value problem: find y(t) such that y = 1+lny, y(0) = 1. Carry out a step of the 4th-order Runge Kutta method (with step size h = 0.1) and verify that K 1 = 1, K 2 = , K 3 = and K 4 = so that the approximation to y(0.1) is given by y 1 =

5 For the next step of the method, show that K 1 = , K 2 = , K 3 = and K 4 = so that y 2 = Continue for one more step and verify that y 3 = Running the Runge Kutta (rk4) process gives... h = 0.1 t = 0.00 y 0 = K 1 = first temporary point t = 0.05 y m = K 2 = second temporary point t = 0.05 y m = K 3 = third temporary point t = 0.10 y = K 4 = new point t = 0.10 y 1 = h = 0.1 t = 0.10 y 1 = K 1 = first temporary point t = 0.15 y m = K 2 = second temporary point t = 0.15 y m = K 3 = third temporary point t = 0.20 y = K 4 = new point t = 0.20 y 2 = h = 0.1 t = 0.20 y 2 = K 1 = first temporary point t = 0.25 y m = K 2 = second temporary point t = 0.25 y m = K 3 = third temporary point t = 0.30 y = K 4 = new point t = 0.30 y 3 = Considerfluiddrainingoutof atank accordingtotheequation y = k y with k = 0.1 and y = 20 at t = 0. Use the 4th-order Runge Kutta method with step size h = 10 to estimate the values of y(t) (the height, in metres, of fluid in the tank) for t = 10, 20 and 30 measured in seconds. Running the (rk4) process gives... first step h = 10 t = 0 y 0 = K 1 = first temporary point t = 5 y m = K 2 = second temporary point t = 5 y m = K 3 = third temporary point t = 10 y = K 4 = new point t = 10 y 1 = h = 10 t = 10 y 1 = K 1 = first temporary point t = 15 y m = K 2 = second temporary point t = 15 y m = K 3 = third temporary point t = 20 y = K 4 = new point t = 20 y 2 = h = 10 t = 20 y 2 = K 1 = first temporary point t = 25 y m = K 2 = second temporary point t = 25 y m = K 3 = third temporary point t = 30 y = K 4 = new point t = 30 y 3 =

6 10. Consider the initial value problem: find y(x) such that y = 2xy 2 y(0) = 1. Carry out two steps of each of the following methods for solving the equation. (a) forward Euler with h = 1 (b) forward Euler with h = 0.5 (c) modified Euler with h = 1 (d) modified Euler with h = 0.5 (e) Runge Kutta with h = 1 (f) Runge Kutta with h = 0.5. Compare your results with the exact solution. (a) Running (fe) with h = 1... Unrealistic results. (b) Running (fe) with h = h x j y j f(y j,x j ) y(x j ) h x j y j f(y j,x j ) y(x j ) The final value is correct (by coincidence!). (c) Running (me) with h = 1... h x j y j f(y j,x j ) x m y m f(y m,x m ) y(x j ) All subsequent values will be zero. Not a realistic solution. (d) Running (me) with h = h x j y j f(y j,x j ) x m y m f(y m,x m ) y(x j ) These values are not particularly accurate but are pointing in the right direction. 6

7 (e) Running (rk4) with h = 1... h = 1.0 x = 0.00 y 0 = K 1 = first temporary point x = 0.50 y m = K 2 = second temporary point x = 0.50 y m = K 3 = third temporary point x = 1.00 y = K 4 = new point x = 1.00 y 1 = h = 1.0 x = 1.00 y 1 = K 1 = first temporary point x = 1.50 y m = K 2 = second temporary point x = 1.50 y m = K 3 = third temporary point x = 2.00 y = K 4 = new point x = 2.00 y = These values are not particularly accurate but are again pointing in the correct direction. (f) Running (rk4) with h = h = 0.5 x = 0.00 y 0 = K 1 = first temporary point x = 0.25 y m = K 2 = second temporary point x = 0.25 y m = K 3 = third temporary point x = 0.50 y = K 4 = new point x = 0.50 y 1 = h = 0.5 x = 0.50 y 1 = K 1 = first temporary point x = 0.75 y m = K 2 = second temporary point x = 0.75 y m = K 3 = third temporary point x = 1.00 y = K 4 = new point x = 1.00 y 2 = These values are much more accurate. 11. Express the following high-order equations as systems of first-order equations. (a) d2 y 2 +12y = 0; y(0) = 1, dy (0) = 0. (b) d2 y 2 +2dy 4y = x; y(0) = 0, dy (0) = 0. (c) d3 y 3 dy +y = dy x3, y(0) = 1, (0) = 0, d 2 y 2(0) = 0. (d) d2 p dt 2 = 1 dp (p+t) dt ; p(0) = 1, dp (0) = 2. dt (e) the coupled system d 2 y dt 2 2dz dt = y +z; y(t 0) = y 0, d 2 z dt 2 +2dy dt = y z; z(t 0) = z 0, dy dt (t 0) = v A, dz dt (t 0) = v B. 7

8 (a) Introducing z = dy dz so dy = z, y(0) = 1, dz = 12y, z(0) = 0. (b) Introducing z = dy dz so dy = z, y(0) = 0, dz = x+4y 2z, z(0) = 0. dz and w = (c) Introducing z = dy dy = z, y(0) = 1, dz = w, z(0) = 0, dw = x3 +z y, w(0) = 0. (d) Introducing q = dp dt dp dt = q, p(0) = 1, = d2 y gives the following coupled system. 2 = d2 y gives the following coupled system. 2 = d2 y gives the following system. 2 dq so dt = d2 p gives the following system. dt2 dq dt = q, q(0) = 2. p+t (e) Finally, introducing x = dy dz and w = gives the following system dt dt of four equations with associated initial conditions. dy dt = x, y(t 0) = y 0, dt = y +z +2w, x(t 0) = v A, dz dt = w, z(t 0) = z 0, dw dt = y z 2x, w(t 0) = v B. 8

9 12. Consider the coupled initial value problem dt = 3y, dy dt = 3x, x(0) = 1, y(0) = 0. Approximate the solution of this system using the Runge Kutta method with step size h = 0.1. Show that at the first step, the K values are (0,3), ( 0.45,3), ( 0.45,2.9325), ( ,2.865) giving new values x 1 and y 1 of and Show that the second step gives values x 2 and y 2 of and Compare these results with the exact values given by x = cos3t, y = sin3t. Running (rk4) with h = h = 0.1 t = 0.00 x 0 = y 0 = K f1 = K g1 = first temporary point t = 0.05 x m = y m = K f2 = K g2 = second temporary point t = 0.05 x m = y m = K f3 = K g3 = third temporary point t = 0.10 x = y = K f4 = K g4 = new point t = 0.10 x 1 = y 1 = h = 0.1 t = 0.10 x 1 = y 1 = K f1 = K g1 = first temporary point t = 0.15 x m = y m = K f2 = K g2 = second temporary point t = 0.15 x m = y m = K f3 = K g3 = third temporary point t = 0.20 x = y = K f4 = K g4 = new point t = 0.20 x 2 = y 2 = The exact solutions are the functions x(t) = cos3t, y(t) = sin3t. Thus when when t = 0.1, we have x = and y = and when t = 0.2 we have x = and y = We see that the numerical approximations are correct to 4 but not 5 decimal places. 13. Consider the coupled initial value problem dt = 1 3 y, dy dt = 6x3, x(0) = 1; y(0) = 3. Compute an approximation to the solution vector (x,y) at the final time t = 0.5 using the Runge Kutta method with step size h = 0.1. Then repeat the calculation for h = Compare the numerical results with the exact values given by x = 1/(t+1), y = 3/(t+1) 2. Editing rhs and running rk4 gives the following output: >> [time, zzrk4, nfeval] = rk4([1;-3],0.5,5); RK4 iteration in 2 dimensions... t x y all done 9

10 >> [time, zzrk4, nfeval] = rk4([1;-3],0.5,10); RK4 iteration in 2 dimensions... t x y all done Theexactvaluesarex(0.5) = andy(0.5) = The approximation errors for h = 0.1 are x(0.5) x 5 = and y(0.5) y 5 = For the smaller value h = 0.05, we have x(0.5) x 10 = and y(0.5) y 10 = The numerical results are much more accurate for the smaller value of h. 14. Consider the second-order initial value problem d 2 y dt 2 +y +ǫy3 = 0; y(0) = 1, dy dt (0) = 0 with the value ǫ = 0.1. (This is an example of a Duffing oscillator.) Compute three steps of Runge Kutta with a step size h = 0.1. The first step is to express the second-order problem as a coupled firstorder system. dy dt = v; y(0) = 1, dv dt = y 0.1y3 ; v(0) = 0. Editing rhs and running rk4 gives the following output: >> [time, zzrk4, nfeval] = rk4([1;0],0.3,3); RK4 iteration in 2 dimensions... t x y all done 10

11 15. A sphere floating in water (in the absence of friction and with its centre y above the water level) experiences a force due to gravity of 4 3 πr3 ρg and a force due to buoyancy of 1 3 ρ waterπ ( 2r 3 +y(y 2 3r 2 ) ) g. It s motion is determined by the ODE d 2 y dt 2 = g+gρ water 4ρ ] [2+ y3 r 3 3y r A 0.4 metre radius sphere with density 0.7 of that of water is initially at rest with its centre 25 cm below the level of the water. You may assume that g = 10 ms 2. (a) Express the equation of motion as a system of two first-order equations and state the appropriate initial conditions. (b) Take a step size of h = 0.1 and compute two steps of Runge Kutta to estimate the position of the centre of the sphere when t = 0.1 and when t = 0.2. The first step is to express the second-order problem as a coupled firstorder system: dy dt = v with y(0) = 0.25 dv dt = g +gρ water 4ρ [2+ y3 r 3 3y r ] with v(0) = 0 Next, we substitute the given values for the parameters to give dy dt dv 25 = 10+ dt 7 with v(0) = 0 = v with y(0) = 0.25 [ y 3 7.5y ] = y y Finally, editing rhs and running rk4 gives the following output: >> [time, zzrk4, nfeval] = rk4([-0.25;0],0.2,2); RK4 iteration in 2 dimensions... t x y all done The sphere is quickly floating back to the surface! Running rk4 for longer times suggests that the sphere tends to bob up and down in the water. This is illustrated by the following output. 11

12 >> [time, zzrk4, nfeval] = rk4([-0.25;0],10,100); RK4 iteration in 2 dimensions... t x y all done >> plot(time,zzrk4(1,:), -kx ), axis square 16. Compute a centered difference approximation to the BVP x 2 d2 y 2 4xdy +4y = x3, x (1,2); y(1) = 1, y(2) = 6, using a grid of size h = Compare your result with the exact solution values y(1.25) = 4.771, y(1.5) = , y(1.75) = We start by dividing the interval into 4 pieces so that x 0 = 1, x 1 = 1.25, x 2 = 1.5, x 3 = 1.75, x 4 = 2, and then we define the grid values y 0 = y(x 0 ) = 1; y 1 y(x 1 ), y 2 y(x 2 ), y 3 y(x 3 ), y 4 = y(x 4 ) = 6. 12

13 We also define the two finite difference approximations y (x j ) = 1 h 2(y j 1 2y j +y j+1 ), y (x j ) = 1 2h (y j+1 y j 1 ). To simplify notation define y 1 = A, y 2 = B and y 3 = C. The difference approximations at the point x 1 = 1.25 are y (x j ) B = 2B 2 and y (x j ) B 2A+1 = 32A + 16B Substituting these expressions into the ODE gives ( 32A+16B +16) (2B 2)+4A = that is, 46A+35B = The difference approximations at the point x 2 = 1.5 are y (x j ) C A 0.5 = 2C 2A and y (x j ) C 2B+A 0.25 = 16A 32B +16C. Substituting these 2 expressions into the ODE gives (16A 32B +16C)+4 1.5(2C 2A)+4B = that is, 24A 68B +48C = The difference approximations at the point x 3 = 1.75 are y (x j ) 6 B 0.5 = 12 2B and y (x j ) 6 2C+B = 96 32C + 16B. Substituting these expressions into the ODE gives (96 32C +16B) (12 2B)+4C = that is 35B 94C = Solving the 3 linear equations gives the results A = (= y 1 ), B = (= y 2 ) and C = (= y 3 ). Comparing with the exact values we see that the errors are of order of Compute a centered difference approximation to the BVP d 2 y 2 + π2 4 y = ǫy3, x (0,1); y(0) = 1, y(1) = 0, for thevalueǫ = 0.1, usingagrid ofsize h = 0.5. Compareyour computed result with the exact solution at the interior grid point. Dividing the interval into 2 pieces we introduce the unknown grid point value y 1 y(0.5). To simplify notation we let A = y 1. The difference approximations at the point x 1 = 0.5 are y (x j ) = 1 and y (x j ) 0 2A+1 = 4 8A. Substituting these expressions into the ODE gives 4 8A+(π 2 /4)A = 0.1A 3. This is a nonlinear equation. To solve it we set up a Newton iteration with f(a) = 4+(π 2 /4 8)A 0.1A 3 so that f (A) = π 2 / A 2. The iteration converges to the solution A = (= y 1 ). A f(a) f (A)

14 The exact solution solution to the BVP when ǫ = 0 is the function y(x) = cos( πx 2 ) so that y(0.5) = The difference approximation is surprisingly accurate. 18. An iron bar is kept at temperatures T 0 and T 1 at its two ends. The radiation of heat at point x is proportional its temperature AT(x) and this effect is balanced by thermal diffusion, that is, κ d2 T. Thus, when 2 the temperature T(x) has reached equilibrium, it satisfies the BVP κ d2 T 2 = AT, x (0,1); T(0) = T 0, T(1) = T 1. Write down a centered difference approximation to T(x) for values of x of 0.25, 0.5 and 0.75 and solve the resulting linear system in the case where A = 1, κ = 1, T 0 = 1 and T 1 = 2. Multiplying by minus one and substituting parameter values gives d 2 T T = 0 witht(0) = 1 andt(1) = 2. 2 Let s definegridvaluessothatt(0.25) P, T(0.5) QandT(0.75) R. Thedifferenceapproximationatthepointx 1 = 0.25isT (x j ) Q 2P = 16Q 32P +16. Substituting this expression into the ODE gives 16Q 33P +16 = 0. (a) Thedifferenceapproximationatthepointx 2 = 0.5isT (x j ) R 2Q+P = 16R 32Q+16P. Substituting this expression into the ODE gives 16P 33Q+16R = 0. (b) Thedifferenceapproximationatthepointx 3 = 0.75isT (x j ) 2 2R+Q = 32 32R+16Q. Substituting this expression into the ODE gives 16Q 33R+32 = 0. (c) Solving the 3 linear equations (a) (c) gives the results P = , Q = and R = Note that the exact solution to the BVP is T = 2e 1 e 2 1 ex + e2 2e e 2 1 e x. Thustheexact values aret(0.25) = , T(0.5) = , T(0.75) = so the numerical values are in good agreement. 14

15 19. Imagine two metal bars, one with ends maintained at temperature T = 600 and the other with a temperature maintained at T = 400. As well as thermal diffusion and general radiation, there is heat transferred from one bar to the other. The bars thus satisfy a coupled system of ODEs 0.01 d2 T T (T 1 T 2 ) = 0, 0.01 d2 T T (T 2 T 1 ) = 0, subject to T 1 (0) = T 1 (1) = 600 and T 2 (0) = T 2 (1) = 400. Take a grid with points x 0 = 0, x 1 = 0.25, x 2 = 0.5, x 3 = 0.75 and x 4 = 1 and compute a centered difference approximation to T 1 and T 2. First we introduce new functions y = 0.01T 1 and z = 0.01T 2 and substitute into the ODEs. This gives the rescaled equations: y 1.5y +0.5z = 0 z +0.5y 1.5z = 0 subject to y(0) = y(1) = 6 and z(0) = z(1) = 4. Let s define grid values so that A y(0.25), B y(0.5), C y(0.75), D z(0.25), E z(0.5) and F z(0.75). Substituting the second derivative difference approximation into the ODEs evaluated at the grid points gives a system of 6 algebraic equations: 6 2A+B A+0.5D = 0 67A+32B +D +192 = 0 C 2B +A B +0.5E = 0 32A 67B +32C +E = 0 B 2C C +0.5F = 0 32B 67C +F +192 = 0 4 2D +E D +0.5A = 0 A 67D +32E +128 = 0 D 2E +F E +0.5B = 0 B +32D 67E +32F = 0 E 2F F +0.5C = 0 C +32E 67F +128 = 0 Solving these equations gives A = 5.42, B = 5.23, C = 5.42, D = 3.73, E = 3.64, F = The corresponding temperature estimates are x = 0 x = 0.25 x = 0.5 x = 0.75 x = 1 Bar Bar