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POWhy does the numeric error in my implementation of Runge-Kutta not decrease like N^a?
 

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  1. POWhy does the numeric error in my implementation of Runge-Kutta not decrease like N^a?
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    <p>I'm trying to determine how many steps it takes for the <a href="http://mathworld.wolfram.com/Runge-KuttaMethod.html" rel="nofollow">Runge-Kutta Method</a> ("RK4") to be within 0.01% of the exact solution of an ordinary differential equation. I'm comparing this to the <a href="http://mathworld.wolfram.com/EulerForwardMethod.html" rel="nofollow">Euler method</a>. Both should result in a straight line on a loglog plot. My Euler solution seems to be correct but I am getting a curved solution for RK. They are based on the same code so I am completely confused about the problem. </p> <p>EDIT: Sorry for removing the wikipedia links. It wouldn't let me keep more than one link since I'm a new user. However, both methods are detailed on Wikipedia similarly to my implementation.</p> <p>Should someone wish to tackle my problem, the code is below and graphs are in <a href="http://dl.dropbox.com/u/5962491/td4.4.doc" rel="nofollow">this Word file on dropbox.com</a>. Yes this is a homework problem; I'm posting this because I'd like to understand what is wrong in my thought process.</p> <pre><code>f = @(x,y) x+y; %this is the eqn (the part after the @(t,y) </code></pre> <p>This is my RK4 code:</p> <pre><code>k1=@(x,y) h*f(x,y); k2=@(x,y) h*f(x+1/2*h,y+1/2*k1(x,y)); k3=@(x,y) h*f(x+1/2*h,y+1/2*k2(x,y)); k4=@(x,y) h*f(x+h,y+k3(x,y)); clear y x exact i x(1)=0; y(1)=2; xn=1; x0=0; exact=3*exp(xn)-xn-1; %exact solution at x=1 %# Evaluate RK4 with 1 step for x=0...1 N=1; %# number of steps h=(xn-x0)/N; %# step size i=1; y(i+1)=y(i)+1/6*k1(x(i),y(i))+1/3*k2(x(i),y(i))+1/3*k3(x(i),y(i))+1/6*k4(x(i),y(i)); RK4(N)=y(i+1); %# store result of RK4 in vector RK4(# of steps) E_RK4(N)=-(RK4(N)-exact)/exact*100;%keep track of %error for each N Nsteps_RK4(N)=N; %# repeat for increasing N until error is less than 0.01% while -(RK4(N)-exact)/exact &gt; 0.0001 i=1; N=N+1; h=(xn-x0)/N; for i=1:N y(i+1)=y(i)+1/6*k1(x(i),y(i))+1/3*k2(x(i),y(i))+1/3*k3(x(i),y(i))+1/6*k4(x(i),y(i)); x(i+1)=x(i)+h; end RK4(N)=y(i+1); Nsteps_RK4(N)=N; E_RK4(N)=-(RK4(N)-exact)/exact*100; %# keep track of %error for each N end Nsteps_RK4(end); </code></pre> <p>This is my Euler code: </p> <pre><code>%# Evaluate Euler with 1 step for x=0...1 clear y x i x(1)=0; y(1)=2; N=1; %# number of steps h=(xn-x0)/N; %# step size i=1; y(i+1)= y(i)+h*f(x(i),y(i)); Euler(N)=y(i+1); %# store result of Euler in vector Euler(# of steps) E_Euler(N)=-(Euler(N)-exact)/exact*100;%# keep track of %error for each N Nsteps_Euler(N)=N; %# repeat for increasing N until error is less than 0.01% while -(Euler(N)-exact)/exact &gt; 0.0001 i=1; N=N+1; h=(xn-x0)/N; for i=1:N y(i+1)= y(i)+h*f(x(i),y(i)); x(i+1)=x(i)+h; end Euler(N)=y(i+1); Nsteps_Euler(N)=N; E_Euler(N)=-(Euler(N)-exact)/exact*100; %# keep track of %error for each N end </code></pre>
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      1. POWhy does the numeric error in my implementation of Runge-Kutta not decrease like N^a?
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    1. COHow is the function `f` defined?
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      1. POWhy does the numeric error in my implementation of Runge-Kutta not decrease like N^a?
      1. USstardt
    2. COHi, as stardt said you should tell us which `f`you are supposed to use. Also, can you make sure the Wikipedia article links I added correctly reflect the version of RK4 you are using?
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      1. POWhy does the numeric error in my implementation of Runge-Kutta not decrease like N^a?
      1. USJonas Heidelberg
    3. COThere are many problems with the code you posted. `h` is used in the anonymous functions before it is defined. You should use `abs()` in the `while` conditions because I don't see how you can guarantee that the estimate will always be less than the exact solution. You access the wrong element of the `y` vector after computing the values in the `for` loop. The line after `end` should be `RK4(N) = y(N+1)`. The same mistakes are in the euler code.
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      1. POWhy does the numeric error in my implementation of Runge-Kutta not decrease like N^a?
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