Question

2. Use an RK4 shooting method with a step size of h - 0.01 to find the unique negative solution to the boundary value problem non ul") -u)- 05 x 1 u(0)0, u(1) - 1 1 + x Hi Then give the approximate value of u(0.5

Answer 1

Matlab code shooting method using RK4 for 2nd order differential equation clear all close all Program for shooting method f-e(x,yl,y2) g-e(x,y1,y2) y2; y12-(h/(1+x)); %function %function (i) (11) tall guesses for y2(1) using shooting method a ini [0:.005:10] y1end-1; for ii-1:length(a ini) x(1)-0 ; yl (1)-0 ; y2(1)-a-ini (ii); %initial h-0.01; x_in-x (1) x_max-1; nm(x-max-x-in) /h ; %number of steps conditions 8step length %initial x Final x %Runge Kutta 4 iterations for i-1:n ko=h * f ( x ( ǐ ) , yl (i),y2 (i)); k1=h * f ( x ( ǐ ) + ( 1 /2 ) *h,y1 (i)t(1/2)*korV2(i)+(1/2)*10); 11=h*g ( x ( i ) + (1 /2 ) *ћ, yl (i)+(1/2)"k0y2(i)t(1/2)+10); 13-h*g (x(i)+h,yl (i)+k2,y2 (i)+12); x i+1)-x inti*h; yl (i+1)-double(yl (i) + (1/6)" (k0+2*k 1 +2 * k2 +k3) ); y2 (i+1)-double (y2(i)+(1/6) (10+2 11+2 12+13) end yy1 (ii)-y1 (end)i end pinterp1 yy1, a ini,yl end) fprintf("\tThe value of U2 (O) using shooting method is %f.\n',p); solution using value of y2 (1) x(1,-0 ; y 1 (1) h-0.01 x in-x(1); 0 ; y2(1) %initial conditions p; step length %initial x %Final x n«(x, max-x-in) /h; %number of steps %Runge Kutta 4 iterations

for i= 1 : n k2-h * f (X ( ¡ ) + ( 1 /2 ) *h,yl(i)t(1/2)"klg2(i)t(1/2)+11); x(i+1)=x in+1*h; yl (i+1)-double (yl(i)+(1/6) (k0+2*kl+2*k2+k3) y2(i+1)-double ( y2 (i)t(1/6)" (10+2+1 1+2+12+13) ); end plotting the solution figure (2) plot (x,yl) title 'Plotting of u (x) using shooting method') xlabel(' x' ) ylabel(u(x) ps-find ( x-0 . 5 ) ; fprintf("At The approximate value for U%f) is %f. n',x(ps), yl (ps)) 88888888888888888888 End of Code 옹88888888888888888888 The value of U2 (0) using shooting method is 1.029696. The approximate value for U(0.500000) is 0.503450

Plotting of u(x) using shooting method 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 .1 0 03 04 05 0608 0.9 Published with MATLAB R2018a

SDE n. du() 2. てつさん

%%Matlab code shooting method using RK4 for 2nd order differential equation
clear all
close all

%Program for shooting method

f=@(x,y1,y2) y2;                        %function (i)
g=@(x,y1,y2) y1^2-(x./(1+x));           %function (ii)

%all guesses for y2(1) using shooting method
a_ini=[0:.005:10]; y1_end=1;

for ii=1:length(a_ini)
    x(1)=0;y1(1)=0;y2(1)=a_ini(ii); %initial conditions
    h=0.01;                 %step length
    x_in=x(1);              %Initial x
    x_max=1;          %Final x
    n=(x_max-x_in)/h; %number of steps

    %Runge Kutta 4 iterations
    for i=1:n

        k0=h*f(x(i),y1(i),y2(i));
        l0=h*g(x(i),y1(i),y2(i));
        k1=h*f(x(i)+(1/2)*h,y1(i)+(1/2)*k0,y2(i)+(1/2)*l0);
        l1=h*g(x(i)+(1/2)*h,y1(i)+(1/2)*k0,y2(i)+(1/2)*l0);
        k2=h*f(x(i)+(1/2)*h,y1(i)+(1/2)*k1,y2(i)+(1/2)*l1);
        l2=h*g(x(i)+(1/2)*h,y1(i)+(1/2)*k1,y2(i)+(1/2)*l1);
        k3=h*f(x(i)+h,y1(i)+k2,y2(i)+l2);
        l3=h*g(x(i)+h,y1(i)+k2,y2(i)+l2);
        x(i+1)=x_in+i*h;
        y1(i+1)=double(y1(i)+(1/6)*(k0+2*k1+2*k2+k3));
        y2(i+1)=double(y2(i)+(1/6)*(l0+2*l1+2*l2+l3));
      
    end
    yy1(ii)=y1(end);
end
p = interp1(yy1,a_ini,y1_end);

fprintf('\tThe value of U2(0) using shooting method is %f.\n',p);

%solution using value of y2(1)

    x(1)=0;y1(1)=0;y2(1)=p; %initial conditions
    h=0.01;            %step length
    x_in=x(1);         %Initial x
    x_max=1;           %Final x
    n=(x_max-x_in)/h; %number of steps

    %Runge Kutta 4 iterations
    for i=1:n

        k0=h*f(x(i),y1(i),y2(i));
        l0=h*g(x(i),y1(i),y2(i));
        k1=h*f(x(i)+(1/2)*h,y1(i)+(1/2)*k0,y2(i)+(1/2)*l0);
        l1=h*g(x(i)+(1/2)*h,y1(i)+(1/2)*k0,y2(i)+(1/2)*l0);
        k2=h*f(x(i)+(1/2)*h,y1(i)+(1/2)*k1,y2(i)+(1/2)*l1);
        l2=h*g(x(i)+(1/2)*h,y1(i)+(1/2)*k1,y2(i)+(1/2)*l1);
        k3=h*f(x(i)+h,y1(i)+k2,y2(i)+l2);
        l3=h*g(x(i)+h,y1(i)+k2,y2(i)+l2);
        x(i+1)=x_in+i*h;
        y1(i+1)=double(y1(i)+(1/6)*(k0+2*k1+2*k2+k3));
        y2(i+1)=double(y2(i)+(1/6)*(l0+2*l1+2*l2+l3));
    end
    %plotting the solution
    figure(2)

    plot(x,y1)
    title('Plotting of u(x) using shooting method')
    xlabel('x')
    ylabel('u(x)')
    ps=find(x==0.5);
    fprintf('\t The approximate value for U(%f) is %f.\n',x(ps),y1(ps))
  

    %%%%%%%%%%%%%%%%%%%% End of Code %%%%%%%%%%%%%%%%%%%%%