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// example 7.5
// solve the boundary value problem u''=u+x;
// u(x=0)=u(0)=0; u(x=1)=u(4)=0; h=1/4;
// we know; u''=(u(j-1)-2*u(j)+u(j+1))/h^2;
// 1) second order method;
x=0:1/4:1;
u0=0;
u4=0;
u=[u0 u1 u2 u3 u4];
// hence;
disp('(u(j-1)-2*u(j)+u(j+1))/h^2=u(j)+x(j)') // for j=1,2,3;
disp('for j=1 -16*u0+33*u1-16*u2=-.25')
disp('for j=2 -16*u1+33*u2-16*u3=-.50')
disp('for j=3 -16*u2+33*u3-16*u4=-.75')
// hence solving for u1,u2,u3) ,
u1=-.034885;
u2=-.056326;
u3=-.050037;
disp(x);
disp(u);
// 2) numerov method;
x=0:1/4:1;
u0=0;
u4=0;
u=[u0 u1 u2 u3 u4];
// since according to numerov method we get the following system of equations;
disp('(191*u(j-1)-394*u(j)+191*u(j+1)=x(j-1)+10*x(j)+x(j+1)') // for j=1,2,3;
disp('for j=1 191*u0-394*u1+191*u2=3')
disp('for j=2 191*u1-394*u2+191*u3=6')
disp('for j=3 191*u2-394*u3+191*u4=9')
// hence solving for u1,u2,u3 ,
u1=-.034885
u2=-.056326
u3=-.050037
disp(x);
disp(u);
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