Question

Please make a little adjust for the following script then make it work for

Runge Kutta method

Consider the initial value problem

du/dt=t^5   u(0)=0

0<=t<=1

Q1. how we change the below script to make it to solve the above IVP and the answer should be near 4

here is the script

f =@(t ,y)(t^5); % define f by f(t,y)=y^5
for k =1:10
[ tlist , ylist ]= RKfour174 (f ,0 ,1 ,0, 2^k);
error1 (k)= max ( abs ( ylist - ( tlist )));
hlist (k) =1/2^k;
end
x= log ( hlist );
y= log ( error1 );
polyfit (x ,y ,1)

function [tlist,ylist] =RungMethod(f,t0,tfinal ,y0,N)     what is the input for N, n is a stepsize

Answer 1

clear all
close all
%function for which Solution have to do
fun=@(t,y) t.^5;
%exact solution
y_ext=@(t) t.^6/6;
%initial guess
tinit=0; yinit=0;
tend=1;

for k=1:10
    [t_rk,y_rk]=RK4(fun,tinit,yinit,tend,2^k);
    y_exact=double(y_ext(t_rk));
    error(k)=norm(y_exact-y_rk);
    hlist(k)=1./2.^k;
    fprintf('\tFor n=%d value of u(%2.2f) is %f\n',2^k,t_rk(end),y_rk(end))
end
x=log(hlist);
y=log(error);

plot(x,y)
ylabel('log(error)')
xlabel('log(hlist)')
title('error vs. step size plot')
p=polyfit(x,y,1);
fprintf('\npolyfit values are p=%f and %f\n',p(1),p(2))

%%Matlab function for Runge Kutta Method
function [t_rk,y_rk]=RK4(f,tinit,yinit,tend,n)
    %Euler method
    % h amount of intervals
    t=tinit;         % initial t
    y=yinit;         % initial y
    t_eval=tend;     % at what point we have to evaluate
    h=(t_eval-t)/n; % Number of steps
    t_rk(1)=t;
    y_rk(1)=y;
    for i=1:n
    %RK4 Steps
       k1=h*double(f(t,y));
       k2=h*double(f((t+h/2),(y+k1/2)));
       k3=h*double(f((t+h/2),(y+k2/2)));
       k4=h*double(f((t+h),(y+k3)));
       dy=(1/6)*(k1+2*k2+2*k3+k4);
       t=t+h;
       y=y+dy;
       t_rk(i+1)=t;
       y_rk(i+1)=y;
    end
end
%%%%%%%%%%%%%%%%%%%%%% End of Code %%%%%%%%%%%%%%%%%%%%%%%