我正在尝试在边界条件x(0)= 1,x(3)= 3的情况下,对0“至3”的u“(x)= exp(x)进行求解,并将其绘制出来。我能够绘制近似解与精确解,但该图看起来有点偏离:
% Interval
a=0;
b=3;
n=10;
% Boundary vals
alpha=1;
beta=3;
%grid size
h=(b-a)/(n+1);
%Matrix generation
m = -2;
u = 1;
l = 1;
% Obtained from (y(i-1) -2y(i) + y(i+1)) = h^2 exp(x(i)
M = (1/h^2).*(diag(m*ones(1,n)) + diag(u*ones(1,n-1),1) + diag(l*ones(1,n-1),-1));
B=[];
xjj=[];
for j=1:n
xjj=[xjj,j*h];
if j==1
B=[B,f(j*h)-(alpha/h^2)];
continue
end
if j==n
B=[B,f(j*h)-(beta/h^2)];
continue
else
B=[B,f(j*h)];
end
end
X=M\B';
x=linspace(0,3,101);
plot(xjj',X,'r*')
hold on
plot(x,exp(x),'b-')
我感谢所有建议和解释。这是我遵循的方案:http://web.mit.edu/10.001/Web/Course_Notes/Differential_Equations_Notes/node9.html
答案 0 :(得分:1)
您可以将大循环简化为
x=linspace(a,b,n+2);
B = f(x(2:end-1));
B(1)-=alpha/h^2;
B(n)-=beta/h^2;
确切的解是u(x)=C*x+D+exp(x)
,边界条件给出D=0
和3*C+exp(3)=3 <=> C=1-exp(3)/3
。
将此精确解与数值解相对应,可以很好地适应如此大的步长:
f=@(x)exp(x)
a=0; b=3;
n=10;
% Boundary vals
alpha=1; beta=3;
%grid
x=linspace(a,b,n+2);
h=x(2)-x(1);
% M*u=B obtained from (u(i-1) -2u(i) + u(i+1)) = h^2 exp(x(i))
M = (1/h^2).*(diag(-2*ones(1,n)) + diag(1*ones(1,n-1),1) + diag(1*ones(1,n-1),-1));
B = f(x(2:end-1));
B(1)-=alpha/h^2; B(n)-=beta/h^2;
U=M\B';
U = [ alpha; U; beta ];
clf;
plot(x',U,'r*')
hold on
x=linspace(0,3,101);
C = 1-exp(3)/3
plot(x,exp(x)+C*x,'b-')