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//Example 19.13
//Hyperbolic Differential Equations in 2D or 3D
//Page no. 675
clc;clear;close;
deff('y=f(x,y)','y=x*(2-x)*y*(2-y)')
c2=3;k=0.4;h=0.4;c2=3;s2=0.5
for l=0:11
if l==0 then
printf('\n t = %i\n\n i\t x\t|\tj -->\t',l)
for i=0:5
printf(' %i\t',i)
end
printf('\n |\t |\t|\ty -->\t')
for i=0:5
printf('%.3f\t',i*k)
end
x=0;
printf('\n---------------------------------------------------------------------------------------')
for i=1:6
y=0;
printf('\n %i\t%.3f\t|\t\t',i-1,x)
for j=1:6
if i==1 | i==6 then
u(i,j)=0;
elseif j==1 | j==6 then
u(i,j)=0
else
u(i,j)=f(x,y)
end
printf('%.3f\t',u(i,j))
y=y+k;
end
x=x+h
end
u2=u;
else
printf('\n\n\n t = %i\n\n i\t x\t|\tj -->\t',l)
for i=0:5
printf(' %i\t',i)
end
printf('\n |\t |\t|\ty -->\t')
for i=0:5
printf('%.3f\t',i*k)
end
x=0;
printf('\n---------------------------------------------------------------------------------------')
for i=1:6
y=0;
printf('\n %i\t%.3f\t|\t\t',i-1,x)
for j=1:6
if i==1 | i==6 then
u(i,j)=0;
elseif j==1 | j==6 then
u(i,j)=0
elseif l==1
u(i,j)=s2*(u1(i+1,j)+u1(i-1,j)+u1(i,j+1)+u1(i,j-1)-4*u1(i,j))+2*u1(i,j)
else
u(i,j)=s2*(u1(i+1,j)+u1(i-1,j)+u1(i,j+1)+u1(i,j-1)-4*u1(i,j))+2*u1(i,j)-u2(i,j)
end
printf('%.4f\t',u(i,j))
y=y+k;
end
x=x+h
end
end
if l>1 then
u2=u1
end
u1=u;
end
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