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//Example 18.5
//Bender Schmidt Method
//Page no. 625
clc;clear;close;
h=0.1;k=0.005;
r=k/h^2;
printf('\n j |\ti -->\t')
for i=0:10
printf(' %i\t',i)
end
printf('\n | |\tx -->\t')
for i=0:10
printf('%.3f\t',(i)/10)
end
printf('\n------------------------------------------------------------------------------------------------------')
for j=1:9
printf('\n %i |\t\t',j-1)
for i=1:11
if i==1 | i==11 then
u(j,i)=0;
elseif j==1 then
u(j,i)=sin(%pi*(i-1)/10)
else
u(j,i)=u(j-1,i)/2+(u(j-1,i-1)+u(j-1,i+1))/4
end
printf('%.3f\t',u(j,i))
end
end
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