Shooting Method On Harmonic Equation

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Alexander Kimbley
Alexander Kimbley 2019-2-19
编辑： Torsten 2019-2-28
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.

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Torsten
Torsten 2019-2-20
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
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Alexander Kimbley
Alexander Kimbley 2019-2-27
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 2019-2-28
编辑：Torsten 2019-2-28
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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