I'm really new to Matlab, so this may be ridicously easy what I'm about to ask, but bare with me please.
I'm trying to integrate the Harmonic equation y'' +(a^2)*y=0 with a=2.4, with BC y(0)=y'(pi)=0. I'm doing this to try and "shoot" for the actual value of y at y=pi which we obviously can find analytically but I need to get my head around the code so I can apply this to a more complicated problem.
Thanks.

 Respuesta aceptada

Torsten
Torsten el 20 de Feb. de 2019

1 voto

function main
ydot0_start = 1.0;
a = 2.4;
iflag = 0;
sol = fzero(@(x)fun_shooting(x,a,iflag),ydot0_start);
iflag = 1;
y_at_pi = fun_shooting(sol,a,iflag)
end
function res = fun_shooting(x,a,iflag)
fun_ode = @(t,y)[y(2);-a^2*y(1)];
tspan = [0,pi];
y0 = [0;x];
[t,y] = ode45(fun_ode,tspan,y0);
if iflag == 0
res = y(end,2);
else
res = y(end,1);
end
end

8 comentarios

Alexander Kimbley
Alexander Kimbley el 20 de Feb. de 2019
Thank you so much! However I've tried using the code & i get the error message "function definition not supported in this context. Create functions in code file." I have MATLAB_R2018b, do I need to download another package or is it somehting else?
Thanks.
Torsten
Torsten el 21 de Feb. de 2019
If you save the above file as "main.m", load it in MATLAB and run it, there should be no problems.
Alexander Kimbley
Alexander Kimbley el 25 de Feb. de 2019
Brilliant, thanks.
Alexander Kimbley
Alexander Kimbley el 26 de Feb. de 2019
Hi, just to double check what are the conditions youve used on Y? Y(0)=0 and what other?
Thanks.
Torsten
Torsten el 27 de Feb. de 2019
I used y(0)=0 and y'(0)=x in "fun_shooting" and adjust x such that y'(pi)=0.
Alexander Kimbley
Alexander Kimbley el 27 de Feb. de 2019
ahh okay, brilliant thanks, you've been a great help!!
Alexander Kimbley
Alexander Kimbley el 27 de Feb. de 2019
So how would I go about changing the conditions to y(0)=0, y'(0)=1, where we alter 'a' to get y(pi)=0 to 10^-8 accuaracy say, assuming we dont actually know the true value of a.
Thanks.
Torsten
Torsten el 28 de Feb. de 2019
Editada: Torsten el 28 de Feb. de 2019
function main
a0 = 4.5;
a = fzero(@fun_shooting,a0)
end
function res = fun_shooting(x)
fun_ode = @(t,y)[y(2);-x^2*y(1)];
tspan = [0,pi];
y0 = [0;1];
[t,y] = ode45(fun_ode,tspan,y0);
res = y(end,1);
end

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Preguntada:

el 19 de Feb. de 2019

Editada:

el 28 de Feb. de 2019

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