Homework 14 solutions

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Valentine Stafford
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1 Section 9.1: Ex 1,5,10,15,16 Section 9.: Ex 1,8,9; AP 1-5,9 Section 9.3: AP 1-5 Section 9.1 Homework 14 solutions 1. (a) Show that y( is a solution to the differential equation by substitution. (b) Find a Lipshitz constant L for the rectangle R = {(t,y):0 t 3, 0 y 5}. y = t y, y( = Ce -t + t t +. (a) y ( = -Ce -t + t ; t y = t (Ce -t + t t + ) = -Ce -t + t. (b) f(t,y 1 ) f(t,y ) = t y 1 (t y ) = y 1 y, so a Lipshitz constant is L = Repeat problem 1 for y = ty, y( = 1/(C-t ). (a) y ( = t/(c-t ), ty = t/(c-t ). (b) f(t,y 1 ) f(t,y ) = ty 1 ty = t(y 1 y ) = t(y 1 +y )(y 1 -y ). In the rectangle t 3, and y 1 +y 10, so f(t,y 1 ) f(t,y ) (3)(10) y 1 -y = 60 y 1 -y, so a Lipshitz constant is L = Consider y = 3/ y 1/3, with y(0) = 0. (a) Verify that y( = 0 for t 0 is a solution. Direct substitution yields 0 = 0. (b) Verify that y( = t 3/ for t 0 is a solution. y ( = 3/ t 1/ ; 3/ y 1/3 = 3/ (t 3/ ) 1/3 = 3/ t 1/. (c) Does this violate Theorem 9.? Why? Note that f y = ½ y -/3, which grows without bound as y approaches 0. If we try to find a Lipshitz constant L, we need y 1/3 1 y 1/3 L y 1 y for some constant L and all y 1 and y in a rectangle containing t=0, y=0. So let y = 0, and note that 1/ 3 1/ 3 y1 y 1, / 3 y1 y y1 which exceeds all bounds as y 1 approaches zero. Therefore there does not exist a Lipshitz constant for this problem, and the hypotheses of Theorem 9. are violated. 15. Find the solution to Integrating both sides, y' where erf is the error function. t / e, y(0) = 0. t s / y( e ds erf (, 0

2 16. Consider the 1 st order equation y ( + p(y( = q( Show that the general solution is given by p( dt p( dt y( e e q( dt C. This is the formula for integrating the equation by finding an integrating factor. It can be verified by differentiating y( and substituting the result into the differential equation. Section (a) Find the first two iterations of the Euler method with h = 0., and the first four iterations using h = 0.1. (b) Compare with the exact solution at y(0.4). (c) Does the F.G.E. behave as expected when the step-size is halved? y ( = t y with y(0) = 1, y( = -e -t + t t +. The Euler iteration is y k+1 = y k + h(t k y k ). With h = 0., y 1 = (0 1) = 0.8, y = (0. 0.8) = (.76) = so the F.G.E. =.0417 With h = 0.1, y 1 = (0 1) = = 0.9 y = ( ) = (.89) = =.811 y 3 = , y 4 = so the F.G.E. =.00, and the F.G.E. is approximately halved when the step-size is halved, as expected. 8. Show that Euler s method fails to approximate the solution y( = t 3/ to y = 3/ y 1/3, y(0) = 0 The Euler method gives y k+1 = y k 1/3 + h 3/ y k 1/3 = (1+3/ h) y k 1/3. Therefore, all iterates will be zero if y k = Can Euler s method be used to solve the I.V.P y = 1 + y over [0, 3] with y(0) = 0? The exact solution is y( = tan(. NO. The Euler method gives y k+1 = y k + h(1+y k ), which will always be positive after the first iteration. It cannot capture the behavior of the exact solution past the asymptote at t = ð/.

3 Algorithms and Programs 1-5. (a) Solve the given equations using the Euler method with step sizes of h=0.1 and h=0.05. (b) Compare the exact solution with the approximation. (c) Does the F.G.E. behave as expected when h is halved? (d) Plot the two approximations and the exact solution on the same coordinate system. Problems 1-5 in 9.3 are to repeat the above with the Heun method. 1. y = t y, y(0) = 1. y = 3y + 3t, y(0) = 1 3. y = -ty, y(0) = 1 4. y = e -t y, y(0) = 1/10, 5. y = ty, y(0) = 1. The Euler method is an O(h) method, so when the step size is halved we expect the F.G.E. to also be approximately halved. The Heun method is an O(h ) method, so when the step size is halved we expect the F.G.E. to decrease approximately ¼. This is what we observe for the given problems. The figure below shows the Euler and Heun approximations, along with the exact solution and F.G.E. s. The code to produce this figure is below, which uses the codes eulerint.m and heunint.m, similar to the codes in the text. for j=1:5 if j==1; f=@(t,y)t.^-y; a=0; b=; ya=1; g=@(-exp(-+t.^ - *t+; fl = 't^ - y'; elseif j==; f=@(t,y)3*y+3*t; a=0;b=;ya=1; g=@(4/3*exp(3*-t-1/3; fl = '3y + 3t'; elseif j==3; f=@(t,y)-t*y; a=0; b=; ya=1; g=@(exp(-t.^/); fl = '-ty'; elseif j==4; f=@(t,y)exp(-*-*y; a=0; b=; ya=1/10; g=@(1/10*exp(-*+t.*exp(- *; fl = 'e^{-t} - y'; elseif j==5; f=@(t,y)*t.*y.^; a=0; b=.9; ya=1; g=@(1./(1-t.^); fl = 'ty^'; end % step sizes h = 0.1 and h = 0.05 h = 0.1; M = (b-a)/h; E1 = eulint(f,a,b,ya,m); H1 = heunint(f,a,b,ya,m);%h1=tayode(f,a,b,ya,m,4); h = 0.05; M = (b-a)/h; E = eulint(f,a,b,ya,m); H = heunint(f,a,b,ya,m); tt = linspace(a,b,100); % <<< plot Euler approximation >>> subplot(5,,*j-1) plot(e1(:,1),e1(:,),e(:,1),e(:,),tt,g(t) if j==1; legend('h=0.1','h=0.05','exact','location','northwest'); end h = get(gca,'ylim'); err1 = abs(e1(max(size(e1)),)-g(b)); err = abs(e(max(size(e)),)-g(b)); text((b-a)*3/5,h(1)+.75*(h()-h(1)),sprintf('fge(.1)=%.3f\nfge(.05)=%.5f',err1,err)) ylabel(sprintf('y'' = %s',fl),'fontsize',14) % <<< plot Heun approximation >>> subplot(5,,*j) plot(h1(:,1),h1(:,),h(:,1),h(:,),tt,g(t) h = get(gca,'ylim'); err1 = abs(h1(max(size(h1)),)-g(b)); err = abs(h(max(size(h)),)-g(b)); text((b-a)*3/5,h(1)+.75*(h()-h(1)),sprintf(' FGE(.1)=%f\nFGE(.05)=%f',err1,err)) end subplot(5,,1); title('euler approximation','fontsize',16) subplot(5,,); title('heun approximation','fontsize',16)

5 9. A model for epidemics is written as y = ky(l-y), y(0) = y 0 (a) Use L = 5,000, k = , and h = 0. with the initial condition y(0) = 50, and the Euler approximation code to solve the equation over [0, 60]. (b) Plot the graph of the approximation from part (a). (c) Estimate the average number of infected individuals by taking the mean of the ordinates from the approximation in part (a). (d) Estimate the average by fitting a curve to the data and using the fact that the average 1 b of a function over [a, b] is f ( x) dx. b a a A script is seen below. The function can be interpolated with a spline, and the integral of the spline can be found using the Matlab commands spline, ppval, and quad. We find that the average is given in either case by Mean of ordinates: Average of spline interpolant: L=5000; k=.00003; f=@(t,y)k*y*(l-y); a=0; b=60; ya=50; M=300; E = eulint(f,a,b,ya,m); plot(e(:,1),e(:,)) sf = spline(e(:,1),e(:,)); splineintegral = quad(@(x)ppval(sf,x),a,b); disp(sprintf('mean of ordinates: %.f',mean(e(:,)))) disp(sprintf('average of spline interpolant: %.f',splineintegral/(b-a)))

Euler s Methods (a family of Runge- Ku9a methods)

Euler s Methods (a family of Runge- Ku9a methods) ODE IVP An Ordinary Differential Equation (ODE) is an equation that contains a function having one independent variable: The equation is coupled with an