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heat equation for infinite bar(rod)

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Nicoleta
Nicoleta am 8 Jan. 2015
Kommentiert: Torsten am 9 Jan. 2015
%Ecuatia caldurii pt bara infinita function ec_cald_bara_infin a=0;b=50;L=10;T=15;Ns=20;M=50;N=50;t=linspace(0,T,50); x=linspace(0,10,50); c=linspace(a,10,b); s=linspace(0,10,50); %calcul Y for n=1:N for m=1:M
Y=f(s).*exp(-( x(n)-s).^2/(4*a*a*t(m))); y=f(s); l=length(y) z=exp(-( x(n)-s).^2/(4*a*a*t(m))); p=length(z) D=Y.*z; r=length(c)
q=trapz(D,c);
%calcul u(t,n)
A=1/(2*a*sqrt(pi*t(m)));
w=length(A)
u(m,n)=A.*q;
end
end
figure(1) surf(t,x,u'); figure(2) for m=1:M plot(x,u(m,:))
axis([-L,L,min(min(u)), max(max(u))]);
film(m)=getframe; end movie(film,3,40); function f=f(s) f=s.*(L-s); end end
I tried figuring out this problem and all I managed to do is this code. Is it correct?
  1 Kommentar
Torsten
Torsten am 9 Jan. 2015
I see the code, but from the title of your request, I can't deduce the problem.
Best wishes
Torsten.

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