Shooting Method On Harmonic Equation

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Alexander Kimbley
Alexander Kimbley el 19 de Feb. de 2019
Editada: Torsten el 28 de Feb. de 2019
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
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 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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