chap3_9_1.m

利用MATLAB解决偏微分方程中的一些问题,如插值求解

function[e]=chap3_9_1(h,tao)
a=2;r=a*tao/h^2;
for k=1:1/tao+1
  for i=1:1/h+1
      U(k,i)=0;
  end
end
for i=1:1/h+1
    X(1,i)=(i-1)*h;U(1,i)=exp(X(1,i))*sin(0.5);
end

for k=1:1/tao+1
    T(1,k)=(k-1)*tao;U(k,1)=sin(0.5-T(1,k));U(k,1/h+1)=exp(1)*sin(0.5-T(1,k));
end
for k=2:1/tao+1
    for i=2:1/h
    U(k,i)=(1-2*r)*U(k-1,i)+r*(U(k-1,i-1)+U(k-1,i+1))+tao*(-exp(X(1,i))*(cos(0.5-T(1,k-1))+2*sin(0.5-T(1,k-1))));
end
end
for k=1:1/tao+1
  for i=1:1/h+1
      W(k,i)=exp(X(1,i))*sin(0.5-T(1,k));E(k,i)=abs(W(k,i)-U(k,i));
  end
end
a=max(E);
e=max(a);