下面的MATLAB代码是使用有限差分在β平面上模拟赤道开尔文波。使用前向时间居中空间区分第一和第二时间步长，主循环具有居中的时间中心空间方案。浅水方程的解在x中是周期性的，并且在域的顶部和底部具有实心边界。
dt = 2000 % time step (s)
dx = 40000 % x grid size (m)
dy = 40000 % y grid size (m)
ce = 10 % phase speed of the wave (m per second)
g = 9.81; % acceleration due to gravity (m s^-2)
h = ce/g; % equivalent depth (m) - not used
X = 40000000 % distance in the x direction (m)
Y = 6000000 % distance in the y direction (m)
x = [0:dx:X];
y = transpose([-Y/2:dy:Y/2]);
y1 = y.^2;
beta = 2.3e-11; % value of the Coriolois parameter divided by the radius of the Earth (m^-1 s^-1)
N = X/(ce*dt) % the number of timesteps to complete one latitudinal circle
I = 1+X/dx; % number of grid-steps in the x-direction for the u and v components
J = 1+Y/dy; % number of grid-steps in the y-direction for the u and v components
A=ones(1,I);
a = beta*y;
B = a*A;
na = [1:N]; % set up the grid for the time array
na2 = [2:N-1]; % set up the grid for the interiror of the time array
ia = [1:I]; % set up the grid for the x array of the u and v components
ia2 = [2:I-1]; % set up the grid for the interior for the x array of the u and v components
ja = [1:J]; % set up the grid for the y array of the u and v components
ja2 = [2:J-1]; % set up the grid for the interior points of the y array of the u and v components
C=B;
for n=1:N
C(:,:,N) = B;
end
b = (g*dt)/dx; % simplification of some terms for the future equations
c = ((ce^2)*dt)/(g*dx); %simplification of some terms for the future equations
d = (g*dt)/dy; %simplification of some terms for the future equations for v component
e = ((ce^2)*dt)/(g*dy);
u0 = 1; % amplitude of the perturbation zonal velocity at the Equator
Z0 = (u0*ce)/g;
lambda = 40000000; % typical wavelength of equatorial Kelvin waves
k = (2*pi)/lambda; % wavenumber
u(ja,ia,na) = NaN; % set up the array u of correct dimensions
v(ja,ia,na) = NaN;
Z(ja,ia,na) = NaN;
% Initial Conditions
u_i(ja,ia) = u0*exp(-(beta*y1/(2*ce)))*cos(k*x);
v_i(ja,ia) = 0;
Z_i(ja,ia) = Z0*exp(-(beta*y1/(2*ce)))*cos(k*x);
%FTCS for timestep 1 of u
u(1,1,1) = u_i(1,1)-0.5*b*(Z_i(1,2)-Z_i(1,I))+dt*B(1,1)*v_i(1,1);
u(ja2,1,1) = u_i(ja2,1)-0.5*b*(Z_i(ja2,2)-Z_i(ja2,I))+dt*B(ja2,1).*v_i(ja2,1);
u(J,1,1) = u_i(J,1)-0.5*b*(Z_i(J,2)-Z_i(J,I))+dt*B(J,1)*v_i(J,1);
u(1,I,1) = u_i(1,I)-0.5*b*(Z_i(1,1)-Z_i(1,I-1))+dt*B(1,I)*v_i(1,I);
u(ja2,I,1) = u_i(ja2,I)-0.5*b*(Z_i(ja2,1)-Z_i(ja2,I- 1))+dt*B(ja2,I).*v_i(ja2,I);
u(J,I,1) = u_i(J,I)-0.5*b*(Z_i(J,1)-Z_i(J,I-1))+dt*B(J,I)*v_i(J,I);
u(1,ia2,1) = u_i(1,ia2)-0.5*b*(Z_i(1,ia2+1)-Z_i(1,ia2-1))+dt*B(1,ia2).*v_i(1,ia2);
u(ja2,ia2,1) = u_i(ja2,ia2)-0.5*b*(Z_i(ja2,ia2+1)-Z_i(ja2,ia2-1))+dt*B(ja2,ia2).*v_i(ja2,ia2);
u(J,ia2,1) = u_i(J,ia2)-0.5*b*(Z_i(J,ia2+1)-Z_i(J,ia2-1))+dt*B(J,ia2).*v_i(J,ia2);
%FTCS for timestep 1 of v (use backward difference for the top boundary and forward difference for the bottom boundary)
v(J,ia,1) = 0; % Boundary Conditions for the top of the domain
v(1,ia,1) = 0; % Boundary Conditions for the bottom of the domain
v(1,1,1) = v_i(1,1)-d*(Z_i(2,1)-Z_i(1,1))-dt*B(1,1)*u_i(1,1);
v(ja2,1,1) = v_i(ja2,1)-0.5*d*(Z_i(ja2+1,1)-Z_i(ja2-1,1))-dt*B(ja2,1).*u_i(ja2,1);
v(J,1,1) = v_i(J,1)-d*(Z_i(J,1)-Z_i(J-1,1))-dt*B(J,1)*u_i(J,1);
v(1,I,1) = v_i(1,I)-d*(Z_i(2,I)-Z_i(1,I))-dt*B(1,I)*u_i(1,I);
v(ja2,I,1) = v_i(ja2,I)-0.5*d*(Z_i(ja2+1,I)-Z_i(ja2-1,I))-dt*B(ja2,I).*u_i(ja2,I);
v(J,I,1) = v_i(J,I)-d*(Z_i(J,I)-Z_i(J-1,I))-dt*B(J,I)*u_i(J,I);
v(1,ia2,1) = v_i(1,ia2)-d*(Z_i(2,ia2)-Z_i(1,ia2))-dt*B(1,ia2).*u_i(1,ia2);
v(ja2,ia2,1) = v_i(ja2,ia2)-0.5*d*(Z_i(ja2+1,ia2)-Z_i(ja2-1,ia2))-dt*B(ja2,ia2).*u_i(ja2,ia2);
v(J,ia2,1) = v_i(J,ia2)-d*(Z_i(J,ia2)-Z_i(J-1,ia2))-dt*B(J,ia2).*u_i(J,ia2);
%FTCS for timestep 1 of Z (boundaries are discretised over a single spatial
%step
Z(1,1,1) = Z_i(1,1)-0.5*c*(u_i(1,2)-u_i(1,I))-e*(v_i(2,1)-v_i(1,1));
Z(ja2,1,1) = Z_i(ja2,1)-0.5*c*(u_i(ja2,2)-u_i(ja2,I))-0.5*e*(v_i(ja2+1,1)-v_i(ja2-1,1));
Z(J,1,1) = Z_i(J,1)-0.5*c*(u_i(J,2)-u_i(J,I))-e*(v_i(J,1)-v_i(J-1,1));
Z(1,I,1) = Z_i(1,I)-0.5*c*(u_i(1,1)-u_i(1,I-1))-e*(v_i(2,I)-v_i(1,I));
Z(ja2,I,1) = Z_i(ja2,I)-0.5*c*(u_i(ja2,1)-u_i(ja2,I-1))-0.5*e*(v_i(ja2+1,1)-v_i(ja2-1,1));
Z(J,I,1) = Z_i(J,I)-0.5*c*(u_i(J,1)-u_i(J,I-1))-e*(v_i(J,I)-v_i(J-1,I));
Z(1,ia2,1) = Z_i(1,ia2)-0.5*c*(u_i(1,ia2+1)-u_i(1,ia2-1))-e*(v_i(2,ia2)-v_i(1,ia2));
Z(ja2,ia2,1) = Z_i(ja2,ia2)-0.5*c*(u_i(ja2,ia2+1)-u_i(ja2,ia2-1))-0.5*e*(v_i(ja2+1,ia2)-v_i(ja2-1,ia2));
Z(J,ia2,1) = Z_i(J,ia2)-0.5*c*(u_i(J,ia2+1)-u_i(J,ia2-1))-e*(v_i(J,ia2)-v_i(J-1,ia2));
% Define the second timestep for the u component using a FTCS scheme
u(1,1,2) = u(1,1,1)-0.5*b*(Z(1,2,1)-Z(1,I,1))+dt*C(1,1,1)*v(1,1,1);
u(ja2,1,2) = u(ja2,1,1)-0.5*b*(Z(ja2,2,1)-Z(ja2,I,1))+dt*C(ja2,1,1).*v(ja2,1,1);
u(J,1,2) = u(J,1,1)-0.5*b*(Z(J,2,1)-Z(J,I,1))+dt*C(J,1,1)*v(J,1,1);
u(1,I,2) = u(1,I,1)-0.5*b*(Z(1,1,1)-Z(1,I-1,1))+dt*C(1,I,1)*v(1,I,1);
u(ja2,I,2) = u(ja2,I,1)-0.5*b*(Z(ja2,1,1)-Z(ja2,I- 1,1))+dt*C(ja2,I,1).*v(ja2,I,1);
u(J,I,2) = u(J,I,1)-0.5*b*(Z(J,1,1)-Z(J,I-1,1))+dt*C(J,I,1)*v(J,I,1);
u(1,ia2,2) = u(1,ia2,1)-0.5*b*(Z(1,ia2+1,1)-Z(1,ia2-1,1))+dt*C(1,ia2,1).*v(1,ia2,1);
u(ja2,ia2,2) = u(ja2,ia2,1)-0.5*b*(Z(ja2,ia2+1,1)-Z(ja2,ia2-1,1))+dt*C(ja2,ia2,1).*v(ja2,ia2,1);
u(J,ia2,2) = u(J,ia2,1)-0.5*b*(Z(J,ia2+1,1)-Z(J,ia2-1,1))+dt*C(J,ia2,1).*v(J,ia2,1);
% Define the second timestep for the v component using a FTCS scheme
v(J,ia,2) = 0; % Boundary Conditions for the top of the domain
v(1,ia,2) = 0; % Boundary Cponditions for the bottom of the domain
v(1,1,2) = v(1,1,1)-d*(Z(2,1,1)-Z(1,1,1))-dt*C(1,1,1)*u(1,1,1);
v(ja2,1,2) = v(ja2,1,1)-0.5*d*(Z(ja2+1,1,1)-Z(ja2-1,1,1))-dt*C(ja2,1,1).*u(ja2,1,1);
v(J,1,2) = v(J,1,1)-d*(Z(J,1,1)-Z(J-1,1,1))-dt*C(J,1,1)*u(J,1,1);
v(1,I,2) = v(1,I,1)-d*(Z(2,I,1)-Z(1,I,1))-dt*C(1,I,1)*u(1,I,1);
v(ja2,I,2) = v(ja2,I,1)-0.5*d*(Z(ja2+1,I,1)-Z(ja2-1,I,1))-dt*C(ja2,I,1).*u(ja2,I,1);
v(J,I,2) = v(J,I,1)-d*(Z(J,I,1)-Z(J-1,I,1))-dt*C(J,I,1)*u(J,I,1);
v(1,ia2,2) = v(1,ia2,1)-d*(Z(2,ia2,1)-Z(1,ia2,1))-dt*C(1,ia2,1).*u(1,ia2,1);
v(ja2,ia2,2) = v(ja2,ia2,1)-0.5*d*(Z(ja2+1,ia2,1)-Z(ja2-1,ia2,1))-dt*C(ja2,ia2,1).*u(ja2,ia2,1);
v(J,ia2,2) = v(J,ia2,1)-d*(Z(J,ia2,1)-Z(J-1,ia2,1))-dt*C(J,ia2,1).*u(J,ia2,1);
% Define the second timestep for the Z component using a FTCS scheme
Z(1,1,2) = Z(1,1,1)-0.5*c*(u(1,2,1)-u(1,I,1))-e*(v(2,1,1)-v(1,1,1));
Z(ja2,1,2) = Z(ja2,1,1)-0.5*c*(u(ja2,2,1)-u(ja2,I,1))-0.5*e*(v(ja2+1,1,1)-v(ja2-1,1,1));
Z(J,1,2) = Z(J,1,1)-0.5*c*(u(J,2,1)-u(J,I,1))-e*(v(J,1,1)-v(J-1,1,1));
Z(1,I,2) = Z(1,I,1)-0.5*c*(u(1,1,1)-u(1,I-1,1))-e*(v(2,I,1)-v(1,I,1));
Z(ja2,I,2) = Z(ja2,I,1)-0.5*c*(u(ja2,1,2)-u(ja2,I-1,1))-0.5*e*(v(ja2+1,I,1)-v(ja2-1,I,1));
Z(J,I,2) = Z(J,I,1)-0.5*c*(u(J,1,1)-u(J,I-1,1))-e*(v(J,I,1)-v(J-1,I,1));
Z(1,ia2,2) = Z(1,ia2,1)-0.5*c*(u(1,ia2+1,1)-u(1,ia2-1,1))-e*(v(2,ia2,1)-v(1,ia2,1));
Z(ja2,ia2,2) = Z(ja2,ia2,1)-0.5*c*(u(ja2,ia2+1,1)-u(ja2,ia2-1,1))-0.5*e*(v(ja2+1,ia2,1)-v(ja2-1,ia2,1));
Z(J,ia2,2) = Z(J,ia2,1)-0.5*c*(u(J,ia2+1,1)-u(J,ia2-1,1))-e*(v(J,ia2,1)-v(J-1,ia2,1));
for na = 2:N
% U component
u(1,1,na+1) = u(1,1,na-1)-b*(Z(1,2,na)-Z(1,I,na))+dt*C(1,1,na)*v(1,1,na)*2;
u(ja2,1,na+1) = u(ja2,1,na-1)-b*(Z(ja2,2,na)-Z(ja2,I,na))+dt*C(ja2,1,na).*v(ja2,1,na)*2;
u(J,1,na+1) = u(J,1,na-1)-b*(Z(J,2,na)-Z(J,I,na))+dt*C(J,1,na)*v(J,1,na)*2;
u(1,I,na+1) = u(1,I,na-1)-b*(Z(1,1,na)-Z(1,I-1,na))+dt*C(1,I,na)*v(1,I,na)*2;
u(ja2,I,na+1) = u(ja2,I,na-1)-b*(Z(ja2,1,na)-Z(ja2,I-1,na))+dt*C(ja2,I,na).*v(ja2,I,na)*2;
u(J,I,na+1) = u(J,1,na-1)-b*(Z(J,1,na)-Z(J,I-1,na))+dt*C(J,I,na)*v(J,I,na)*2;
u(1,ia2,na+1) = u(1,ia2,na-1)-b*(Z(1,ia2+1,na)-Z(1,ia2-1,na))+dt*C(1,ia2,na).*v(1,ia2,na)*2;
u(ja2,ia2,na+1) = u(ja2,ia2,na-1)-b*(Z(ja2,ia2+1,na)-Z(ja2,ia2-1,na))+dt*C(ja2,ia2,na).*v(ja2,ia2,na)*2;
u(J,ia2,na+1) = u(J,ia2,na-1)-b*(Z(J,ia2+1,na)-Z(J,ia2-1,na))+dt*C(J,ia2,na).*v(J,ia2,na)*2;
% V component
v(J,ia,na) = 0; % Boundary Conditions for the top of the domain
v(1,ia,na) = 0; % Boundary Cponditions for the bottom of the domain
v(1,1,na+1) = v(1,1,na-1)-2*d*(Z(2,1,na)-Z(1,1,na))-dt*C(1,1,na)*u(1,1,na)*2;
v(ja2,1,na+1) = v(ja2,1,na-1)-(d*(Z(ja2+1,1,na)-Z(ja2-1,1,na)))-dt*C(ja2,1,na).*u(ja2,1,na)*2; %*
v(J,1,na+1) = v(J,1,na-1)-2*d*(Z(J,1,na)-Z(J-1,1,na))-dt*C(J,1,na)*u(J,1,na)*2;
v(1,I,na+1) = v(1,I,na-1)-2*d*(Z(2,I,na)-Z(1,I,na))-dt*C(1,I,na)*u(1,I,na)*2;
v(ja2,I,na+1) = v(ja2,I,na-1)-(d*(Z(ja2+1,I,na)-Z(ja2-1,I,na)))-dt*C(ja2,I,na).*u(ja2,I,na)*2; %*
v(J,I,na+1) = v(J,I,na-1)-2*d*(Z(J,I,na)-Z(J-1,I,na))-dt*C(J,I,na)*u(J,I,na)*2;
v(1,ia2,na+1) = v(1,ia2,na-1)-2*d*(Z(2,ia2,na)-Z(1,ia2,na))-dt*C(1,ia2,na).*u(1,ia2,na)*2;
v(ja2,ia2,na+1) = v(ja2,ia2,na-1)-d*(Z(ja2-1,ia2,na)-Z(ja2-1,ia2,na))-dt*C(ja2,ia2,na).*u(ja2,ia2,na)*2;
v(J,ia2,na+1) = v(J,ia2,na-1)-2*d*(Z(J,ia2,na)-Z(J-1,ia2,na))-dt*C(J,ia2,na).*u(J,ia2,na)*2;
disp(v(60,1,na))
% Z component
Z(1,1,na+1) = Z(1,1,na-1)-c*(u(1,2,na)-u(1,I,na))-2*e*(v(2,1,na)-v(1,1,na));
Z(ja2,1,na+1) = Z(ja2,1,na-1)-c*(u(ja2,2,na)-u(ja2,I,na))-e*(v(ja2+1,1,na)-v(ja2-1,1,na));
Z(J,1,na+1) = Z(J,1,na-1)-c*(u(J,2,na)-u(J,I,na))-2*e*(v(J,1,na)-v(J-1,1,na));
Z(1,I,na+1) = Z(1,I,na-1)-c*(u(1,1,na-1)-u(1,I-1,na))-2*e*(v(2,I,na)-v(1,I,na));
Z(ja2,I,na+1) = Z(ja2,I,na-1)-c*(u(ja2,1,na)-u(ja2,I-1,na))-e*(v(ja2+1,I,na)-v(ja2+1,I,na));
Z(J,I,na+1) = Z(J,I,na-1)-c*(u(J,1,na)-u(J,I-1,na))-2*e*(v(J,I,na)-v(J-1,I,na));
Z(1,ia2,na+1) = Z(1,ia2,na-1)-c*(u(1,ia2+1,na)-u(1,ia2-1,na))-2*e*(v(2,ia2,na)-v(1,ia2,na));
Z(ja2,ia2,na+1) = Z(ja2,ia2,na-1)-c*(u(ja2,ia2+1,na)-u(ja2,ia2-1,na))-e*(v(ja2+1,ia2,na)-v(ja2-1,ia2,na));
Z(J,ia2,na+1) = Z(J,ia2,na-1)-c*(u(J,ia2+1,na)-u(J,ia2-1,na))-2*e*(v(J,ia2,na)-v(J-1,ia2,na));
end
我遇到的问题如下：一旦模型进入for循环部分，v的值开始在东边界和西边界增长，当它们应该停留在O（10 ^ -3）到10 ^ -4）。正是这些值在第11个时间步时开始影响u和Z，并继续进入模型。
如果我将2个相关方程中的v（：，：，na-1）（在代码中用*符号表示）乘以0.1和0.5之间的任何常数，则模型产生我预期的结果。
我希望我可以称之为我的问题的解决方案，但我想知道社区中的任何人是否可以帮助我找到问题的实际来源？