Shooting Method On Harmonic Equation

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.

 采纳的回答

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

8 个评论

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.
If you save the above file as "main.m", load it in MATLAB and run it, there should be no problems.
Hi, just to double check what are the conditions youve used on Y? Y(0)=0 and what other?
Thanks.
I used y(0)=0 and y'(0)=x in "fun_shooting" and adjust x such that y'(pi)=0.
ahh okay, brilliant thanks, you've been a great help!!
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.
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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