3rd order non-linear differential equation (shooting method)
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Hi, I have to solve the following equation --- f*f'' + 2*f''' = 0 with the boundary conditions --- for eta = 0: f = 0, f' = 0 and for eta = infinity (let's say it's 20): f' = 1 and I have a problem with the shooting method. If my initial solutions for f, f' and f'' are not near the real solution for the before mentioned then I don't get the right graph. Does anybody know what to do? Thanks in advance!
My code:
beskonacno = 20;
etaint = [0:.01:0.2 0.2:.02:0.8 0.8:.01:1]*20;
solinit = bvpinit(linspace(0,beskonacno,50000),@dbinit);
options = bvpset('stats','on');
sol = bvp4c(@difjedn,@rubuvj,solinit,options);
eta = sol.x;
f = sol.y;
figure
plot(eta,f(1,:),'b',eta,f(2,:),'m',eta,f(3,:),'k');
axis([0 beskonacno -1000 1000]);
title(' Rješenje ')
xlabel('\eta')
ylabel('f , df/d\eta , d^2f/d\eta^2')
legend('f','df/d\eta','d^2f/d\eta^2')
fi = deval(sol,etaint);
Tablica = table(etaint',fi(1,:)',fi(2,:)',fi(3,:)','VariableNames',{'eta' 'f' 'df' 'd2f'})
a = fi(3,1)
function dfdeta = difjedn(eta,f)
dfdeta(1) = f(2);
dfdeta(2) = f(3);
dfdeta(3) = -(f(1)*f(3))/2;
end
function res = rubuvj(f0,fbes)
res = [f0(1);
f0(2);
fbes(2)-1;]
end
function yinit = dbinit(x)
yinit = [x^4;
4*x^3;
12*x^2;
];
end
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Answers (1)
hashem
on 11 Jan 2024
let if you have many value for like f = -2s ,where s= -2.5,-2,-1.5,-1.-0.5,0 , what will be your code
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