一维扩散方程的有限差分法matlab 下载本文

一维扩散方程的有限差分法

——计算物理实验作业七

陈万 物理学2013级 13020011006

? 题目:

编程求解一维扩散方程的解

??u?2u??D2(0?x?a0,0?t?tmax)?x??t?u(x,t)|t?0?ex???u au?b?c(x?0)?111?n??u?au?b?c2(x?a0)22??n?取a1?1,b1?1,c1?0,a2?1,b2??1,c2?0,a0?1.0,tmax?10,D?0.1,h?0.1,??0.1。输出t=1,2,...,10时刻的x和u(x),并与解析解u=exp(x+0.1t)作比较。

? 主程序:

% 一维扩散方程的有限差分法 clear,clc;

%定义初始常量

a1 = 1; b1 = 1; c1 = 0; a2 = 1;b2 = -1; c2 = 0; a0 = 1.0; t_max = 10; D = 0.1; h = 0.1; tao = 0.1;

%调用扩散方程子函数求解

u = diffuse_equation(a0,t_max,h,tao,D,a1,b1,c1,a2,b2,c2);

? 子程序1:

function output = diffuse_equation(a0,t_max,h,tao,D,a1,b1,c1,a2,b2,c2)

% 一维扩散方程的有限差分法,采用隐式六点差分格式(Crank-Nicolson) % a0: x的最大值 % t:_max: t的最大值 % h: 空间步长 % tao: 时间步长 % D:扩散系数

% a1,b1,c1是(x=0)边界条件的系数;a2,b2,c2是(x=a0)边界条件的系数

x = 0:h:a0;

1

n = length(x); t = 0:tao:t_max; k = length(t);

P = tao * D/h^2; P1 = 1/P + 1; P2 = 1/P - 1;

u = zeros(k,n); %初始条件 u(1,:) = exp(x);

%求A矩阵的对角元素d d = zeros(1,n);

d(1,1) = b1*P1+h*a1; d(2:(n-1),1) = 2*P1; d(n,1) = b2*P1+h*a2;

%求A矩阵的对角元素下面一行元素e e = -ones(1,n-1); e(1,n-1) = -b2;

%求A矩阵的对角元素上面一行元素f f = -ones(1,n-1); f(1,1) = -b1;

R = zeros(k,n);%求R %追赶法求解 for i = 2:k

R(i,1) = (b1*P2-h*a1)*u(i-1,1)+b1*u(i-1,2)+2*h*c1; for j = 2:n-1

R(i,j) = u(i-1,j-1)+2*P2*u(i-1,j)+u(i-1,j+1); end

R(i,n) = b2*u(i-1,n-1)+( b2*P2-h*a2)*u(i-1,n)+2*h*c2; M = chase(e,d,f,R(i,:)); u(i,:) = M';

plot(x,u(i,:)); axis([0 a0 0 t_max]); pause(0.1) end

output = u;

% 绘图比较解析解和有限差分解 [X,T] = meshgrid(x,t); Z = exp(X+0.1*T);

surf(X,T,Z),xlabel('x'),ylabel('t'),zlabel('u'),title('解析解'); figure

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surf(X,T,u),xlabel('x'),ylabel('t'),zlabel('u'),title('有限差分解');

? 子程序2:

function M = chase(a,b,c,f)

% 追赶法求解三对角矩阵方程,Ax=f % a是对角线下边一行的元素 % b是对角线元素

% c是对角线上边一行的元素

% M是求得的结果,以列向量形式保存 n = length(b);

beta = ones(1,n-1); y = ones(1,n); M = ones(n,1);

for i = (n-1):(-1):1 a(i+1) = a(i); end

% 将a矩阵和n对应

beta(1) = c(1)/b(1); for i = 2:(n-1)

beta(i) = c(i)/( b(i)-a(i)*beta(i-1) ); end

y(1) = f(1)/b(1); for i = 2:n

y(i) = (f(i)-a(i)*y(i-1))/(b(i)-a(i)*beta(i-1)); end

M(n) = y(n);

for i = (n-1):(-1):1

M(i) = y(i)-beta(i)*M(i+1); end end

? 结果:

3