Question

MATH LAB CODE PLS

ter Problems 1. Use finite differences to approximate solutions to the linear BVPs for n=9, 19, and 39. (a) | y" = y + ſet { y(0) = 0) 1 y(1) = ſe (b) | y" = (2 + 442) y { y(0) = 1 y(1) = e Plot the approximate solutions together with the exact solutions (a) y(t) = te'/3 and (b) y(t) = et', and display the errors as a function of t in a separate semilog plot

Answer 1

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Following is the answer:

(a) Consider the following linear Boundary Value Problem, y(0) = 0 »(1) =- The above differential equation at 1, in the discretized form is given by, Wo.-2w,+w,, - h’w – 25 €* =0 W-(2w; +hº)w, + w., _ 24?<" =0 3 The first and last equations along with the boundary conditions are given by, ya-(2+hº)w; - 24*2" + w, = 0 Www-(2+hº)w.-25 <"+y = 0

Here, ya and by are the boundary conditions.

Differentiate y, -(2+h?) w; -25% e" + W, = 0) with respect to w; . Thus, -(2+h+)=0 y(0) = 0, y(1) --- Use the following MATLAB program to approximate solutions for the linear BVP y" = y + . This program is saved as 'nlbvpfdi.m'. % This program is used to find the approximate solutions for % the boundary value problem(BVP) by using the Finite Difference Method % The interval(inter), boundary values(bv), number of steps(n) are inputs % The solution(w) is an output % Command used: w=nlbvpfdi [O 1].[O exp(1)/3],9) function W=nlbvpfd1(inter, bv,n) a=inter(1); b=inter(2),ya=bv(1); yb=bv(2); h=(b-a)/(n+1); % This gives the step size h W=zeros(n,1); % This initializes solution array w for i=1:20 W=W-jac(Winter, bv,n)\f(Winter, bv,n); end plot([a a+(1:n)*h b],[ya W' yb]); % This gives plot w with boundary data xlabel('t'),ylabel('y') title('Approximate solutions of BVP by the Finite Difference Method')

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function y=f(Winter, bv,n) y=zeros(n,1); h=(inter(2)-inter(1)/(n+1); y(1)=bv(1)-(2+h^2)*W(1)-2*h^2*exp(h)/3 +W(2); y(n)=W(n-1)-(2+h^2)*W(n)-(2*h^2/3)*exp(h+bv(2): for i=2:n-1 y(i)=W(i-1)-(2+h^2)*W(1)-(2*(h^2)/3)*exp(h)+W(i+1); end %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function a=jac(Winter, bv,n) a=zeros(n,n); h=(inter(2)-inter(1)/(n+1); for i=1:n a(i,i)=-2-h^2; end for i=1:n-1 a(i,i+1)=1; a (i+1,1)=1; end Consider n=9 Use the MATLAB command 'w=nlbvpfd1( [0 1], [O exp(1)/3],9)' to approximate solutions for the linear BVP y = y +-e',y - for n=9 See the screenshot given below. 3

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The resulting solution curve yt is :

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The exact solution for the linear BVP Y" = y + e',y(0) = 0, y(1)=-e is given by, v(t) = zce Use the following MATLAB program to plot the approximate solutions together with the exact solutions y(t) = -te'. This program is saved as 'nlbupfdia.m! % This program is used to find the approximate solutions for % the boundary value problem(BVP) by using the Finite Difference Method % The interval(inter), boundary values(bv), number of steps(n) are inputs % The solution(w) is an output % Command used: w=nlbvpfdi( [01],[O exp(1)/3],9) function W=nlbvpfdia(inter, bv,n) a=inter(1); b=inter(2);ya=bv(1); yb=bv(2); h=(b-a)/(n+1); % This gives the step size h W=zeros(n,1); % This initializes solution array w for i=1:20 % loop of Newton step W=W-jac(Winter, bv,n)\f(Winter, bv,n); end for i=1:11 z(i)=0; t(i)=0; end

for i=0:10 t(i+1)=t(i+1)+i/10; z(i+1)=z(i+1)+t(i+1)*exp(t(i+1))/3; end plot([a a+(1:n)*h b].[ya W' yb]);hold on plot(t,z) xlabel('t'),ylabel(y) title('Approximate solutions of BVP by the Finite Difference Method') function y=f(Winter,bv,n) y=zeros(n,1); h=(inter(2)-inter(1))/(n+1); y(1)=bv(1)-(2+h^2)*W(1)-2*h^2*exp(h)/3 +W(2); y(n)=W(n-1)-(2+h^2)*W(n)-(2*h^2/3)*exp(h)+by(2); for i=2:n-1 yO=W(1-1)-(2+h^2)*W(1)-(2*(h^2)/3)*exp(h)+W(i+1); end function a=jac(W,inter,bv,n) a=zeros(n,n); h=(inter(2)-inter(1))/(n+1); for i=1:n a(i,i)=-2-h^2; end for i=1:n-1

a(i,i+1)=1; a (i+1,1)=1; end Use the MATLAB command 'w=nlbvpfdia( [0 1],[0 exp(1)/3],9)' to plot the approximate solutions together with the exact solutions y The screenshot of the output of the above program is given below.

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Use the following MATLAB program to calculate errors and make a semilog plot for the errors. This program is saved as 'nlbvpfdial.m! % This program is used to find the approximate solutions for % the boundary value problem(BVP) by using the Finite Difference Method % The interval(inter), boundary values(bv), number of steps(n) are inputs % This program gives semilog plot for errors. % Command used: w=nlbvpfd1( [0 1], [0 exp(1)/3],9) function W=nlbvpfdial(inter, bv,n) a=inter(1); b=inter(2),ya=bv(1); yb=bv(2); h=(b-a)/(n+1); % This gives the step size h Wrzeros(n,1); % This initializes solution array w for i=1:20 % loop of Newton step W=W-jac(Winter, bv,n)\f(W,inter, bv,n); end for i=1:10 z(i)=0; t(i)=0; err(i)=0; end W(10)=bv(2); for i=1:10 t(i)=t(i)+i/10;

z(i)=z(i)+t(i)*exp(t1))/3; err(i)=err(i)+abs(w(1)-z(i)); end disp([' Errors=']) disp(err) semilogy(t.err) xlabel('t'),ylabel('Error') title('A semilog plot for the errors') %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function y=f(winter, bv,n) y=zeros(n,1); h=(inter(2)-inter(1))/(n+1); y(1)=bv(1)-(2+h^2)*W(1)-2*h^2*exp(h)/3 +W(2); y(n)=W(n-1)-(2+h^2)*W(n)-(2*h^2/3)*exp(h)+bv(2); for i=2:n-1 y(i)=W(i-1)-(2+h^2) *W(1)-(2*(h^2)/3)*exp(h)+W(i+1); end %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function a=jac(Winter,bv,n) a=zeros(n,n); h=(inter(2)-inter(1))/(n+1); for i=1:n a(i,i)=-2-h^2;

end for i=1:n-1 a(i,i+1)=1; a (i+1,1)=1; end Use the MATLAB command 'w=nlbvpfdia( [01],[O exp(1)/3],9)' to calculate errors and make a semilog plot for the errors. The screenshot of the output of the above program is given below.

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(b) Consider the following linear Boundary Value Problem, y" = (2+41?)y y(0)=1 y(1)=e The above differential equation at 1, in the discretized form is given by, W.-2w, + W-1 w, = 0 W-, -2w; + W-,-h (2+4h?)w; = 0 W*-(2+ 2h? + 4h* )w, +W;+1 = 0 The first and last equations along with the boundary conditions are given by, y. -(2+ 2h +4h*) w; +w, = 0 W-1-(2+ 2h +4h*)w, + y) = 0

Here, ya and by are the boundary conditions.

Use the following MATLAB program to approximate solutions for the linear BVP y" = (2+41° ) y, y(0)=1, y(1)=e. This program is saved as 'nlbvpfdib.m. % This program is used to find the approximate solutions for % the boundary value problem(BVP) by using the Finite Difference Method % The interval(inter), boundary values(bv), number of steps(n) are inputs % The solution(w) is an output % Command used: w=nlbvpfd1( [01],[1 exp(1)],9) function W=nlbvpfdi(inter, bv,n) a=inter(1); b=inter(2);ya=bv(1); yb=bv(2); h=(b-a)/(n+1); % This gives the step size h Wrzeros(n,1); % This initializes solution array w for i=1:20 % loop of Newton step W=W-jac(Winter, bv,n)\f(Winter, bv,n); end plot([a a+(1:n)*h b].[ya W' yb]); % This gives plot w with boundary data xlabel('x'),ylabel('y') title('Approximate solutions of BVP by the Finite Difference Method') %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function y=f(Winter,bv,n) y=zeros(n,1); h=(inter(2)-inter(1))/(n+1);

y(1)=bv(1)-(2+2+h^2+4*h^4)*W(1)+W(2); y(n)=W(n-1)-(2+2*h^2+4**40*W(n)+bv(2); for i=2:n-1 y(i)=W(i-1)-(2+2*h^2+4*h^4)*W(1)+W(i+1); end %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function a=jac(Winter,bv,n) a=zeros(n,n); h=(inter(2)-inter(1))/(n+1); for i=1:n a(i,i)=-2-2*h^2-4*h^4; end for i=1:-1 a(i,i+1)=1; a (i+1,i)=1; end

Consider n=9. Use the MATLAB command 'w=nlbvpfdib( [01],[1 exp(1)],9)' to approximate solutions for the linear BVP y" = (2+41?) y, y(0) = 1, y(1) = e for n=9 .See the screenshot given below.

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The resulting solution curve y(t) is given below:

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The exact solution for the linear BVP y" = (2+41?)y, y(0) = 1, y(1)= e is given by, y(t)=en Use the following MATLAB program to plot the approximate solutions together with the exact solutions y(t)=e. This program is saved as 'nlbvpfdibi.m! % This program is used to find the approximate solutions for % the boundary value problem(BVP) by using the Finite Difference Method % The interval(inter), boundary values(bv), number of steps(n) are inputs % The solution(w) is an output % This program plots the approximate and the exact solutions. % Command used: w=nlbvpfd1 [O 1],[O exp(1)/3],9) function W=nlbvpfd1b1(inter,bv,n) a=inter(1); b=inter(2),ya=bv(1); yb=bv(2); h=(b-a)/(n+1); % This gives the step size h Wrzeros(n,1); % This initializes solution array w for i=1:20 % loop of Newton step W=W-jac(Winter, bv,n)\f(W,inter,bv,n); end for i=1:11 z(i)=0; t(i)=0; end

for i=0:10 t(i+1)=t(i+1)+i/10; z(i+1)=z(i+1)+exp(t(i+1)^2); end disp(['The exact solutions are']) disp(z) plot([a a+(1:n)*h b],[ya W' yb]);hold on plot(t,z) xlabel('t'),ylabel('') title('Plot of the approximate and the exact solutions) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function y=f(Winter, bv,n) y=zeros(n,1); h=(inter(2)-inter(1))/(n+1); y(1)=bv(1)-(2+2*h^2+4*h^4)*W(1)+W(2); y(n)= (n-1)-(2+2*h^2+4*5*40*W(n)+bv(2); for i=2:n-1 y(i)=W(i-1)-(2+2*h^2+4*h^4)*W(1)+W(i+1); end %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function a=jac(W,inter,bv,n) a=zeros(n,n); h=(inter(2)-inter(1))/(n+1);

for i=1:n ali,i)=-2-2*h^2-4*h^4; end for i=1:n-1 a(i,i+1)=1; a (i+1,1)=1; end Use the MATLAB command 'w=nlbvpfdib( [01],[1 exp(1)],9)' to plot the approximate solutions together with the exact solutions y(t) = e. See the screenshot given below.

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t(i)=t(i)+i/10; z(i)=z(i)+t(i)*exp(ti))/3; err(i)=err(i)+abs(w(1)-z(i)); end disp('Errors=']) disp(err) semilogy(t,err) xlabel('t'),ylabel('Error') title('A semilog plot for the errors') %%%%%%%%%%%%%%%%%%%%%%%%%%%%% function y=f(Winter,bv,n) y=zeros(n,1); h=(inter(2)-inter(1))/(n+1); y(1)=bv(1)-(2+2*h^2+4*h^4)*W(1+W(2); y(n)=W(n-1)-(2+2*1^2+4*5*40*W(n)+bv(2); for i=2:n-1 y(i)=W(i-1)-(2+2*h^2+4*h^4)*W(1)+W(i+1); end %%%%%%%%%%%%%%%%%%%%%%%%%% function a=jac(W,inter, bv,n) a=zeros(n,n); h=(inter(2)-inter(1))/(n+1); for i=1:n

a(i,i)=-2-2*h^2-4*h^4; end for i=1:n-1 a(i,i+1)=1; a (i+1,1)=1; end

Use the MATLAB command 'w=nlbvpfd1b2( [01],[1 exp(1)],9)' to calculate errors and make a semilog plot for the errors. See the screenshot given below.

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Consider n=19 Use the MATLAB command 'w=nlbvpfdib( [01],[1 exp(1)],19) to approximate solutions for the linear BVP b) = 1, y(1)= e for n=19 See the screenshot given below.

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Consider n = 39 Use the MATLAB command 'w=nlbvpfdi( [01],[1 exp(1)],39)' to approximate solutions for the linear BVP y" = (2+41?)y, y(0)=1, y(1)= e for n = 39 .See the screenshot given below.

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