我希望用Matlab中的Runge Kutta 4方法解决三个微分方程组的系统(Ode45
是不允许的)。
经过长时间的观察,我在网上找到的所有内容都是难以理解的例子,或者根本不包含例子的一般解释。我想要一个关于如何正确实现我的解决方案的具体示例,或者我可以构建的类似问题的解决方案。
我走得很远; 我当前的代码在大多数组件上吐出一个包含2个正确小数的矩阵,我很满意。
然而,当步长减小时,错误变得巨大。我知道我创建的for循环并不完全正确。我可能已经错误地定义了函数,但是我很确定如果对for循环做了一些小的改动就解决了问题,因为它似乎已经在当前状态下很好地解决了方程式系统。
clear all, close all, clc
%{
____________________TASK:______________________
Solve the system of differential equations below
in the interval 0<t<1, with stepsize h = 0.1.
x'= y x(0)=1
y'= -x-2e^t+1 y(0)=0 , where x=x(t), y=y(t), z=z(t)
z'= -x - e^t + 1 z(0)=1
THE EXACT SOLUTIONS for x y and z can be found in this pdf:
archives.math.utk.edu/ICTCM/VOL16/C029/paper.pdf
_______________________________________________
%}
h = 0.1;
t = 0:h:1
N = length(t);
%Defining the functions
x = zeros(N,1);%I am not entierly sure if x y z are supposed to be defined in this way.
y = zeros(N,1)
z = zeros(N,1)
f = @(t, x, y, z) -x-2*exp(t)+1;%Question: Do i need a function for x here as well??
g = @(t, x, y, z) -x - exp(t) + 1;
%Starting conditions
x(1) = 1;
y(1) = 0;
z(1) = 1;
for i = 1:(N-1)
K1 = h * ( y(i));%____I think z(i) is supposed to be here, but i dont know in what way.
L1 = h * f( t(i) , x(i) , y(i) , z(i));
M1 = h * g( t(i) , x(i) , y(i) , z(i));
K2 = h * (y(i) + 1/2*L1 + 1/2*M1);%____Again, z(i) should probably be here somewhere.
L2 = h * f(t(i) + 1/2*h, x(i)+1/2*K1 , y(i)+1/2*L1 , z(i)+1/2*M1);
M2 = h * g(t(i) + 1/2*h, x(i)+1/2*K1 , y(i)+1/2*L1 , z(i)+1/2*M1);
K3 = h * (y(i) + 1/2*L2 + 1/2*M2);%____z(i). Should it just be added, like "+z(i)" ?
L3 = h * f(t(i) + 1/2*h, x(i) + 1/2*K2 , y(i) + 1/2*L2 , z(i) + 1/2*M2);
M3 = h * g(t(i) + 1/2*h, x(i) + 1/2*K2 , y(i) + 1/2*L2 , z(i) + 1/2*M2);
K4 = h * (y(i) + L3 + M3);%_____z(i) ... ?
L4 = h * f( t(i)+h , x(i)+K3 , y(i)+L3, z(i)+M3);
M4 = h * g( t(i)+h , x(i)+K3 , y(i)+L3, z(i)+M3);
x(i+1) = x(i)+1/6*(K1+2*K2+2*K3+K4);
y(i+1) = y(i)+1/6*(L1+2*L2+2*L3+L4);
z(i+1) = z(i)+1/6*(M1+2*M2+2*M3+M4);
end
Answer_Matrix = [t' x y z]
答案 0 :(得分:4)
因此,您的主要问题是没有正确定义x
。你使用Runge Kutta 4(RK4)方法传播它的值,但从未真正定义它的衍生物是什么!
在这个答案的底部是一个函数,它可以采用任意给定数量的方程式及其初始条件。这已包含在内,以满足您对三个(或更多)的明确示例的需求方程。
作为参考,方程可以直接从标准的RK4方法中解除here。
这与您的相似,但使用稍微更清晰的命名约定和结构。
% Initialise step-size variables
h = 0.1;
t = (0:h:1)';
N = length(t);
% Initialise vectors
x = zeros(N,1); y = zeros(N,1); z = zeros(N,1);
% Starting conditions
x(1) = 1; y(1) = 0; z(1) = 1;
% Initialise derivative functions
dx = @(t, x, y, z) y; % dx = x' = dx/dt
dy = @(t, x, y, z) - x -2*exp(t) + 1; % dy = y' = dy/dt
dz = @(t, x, y, z) - x - exp(t) + 1; % dz = z' = dz/dt
% Initialise K vectors
kx = zeros(1,4); % to store K values for x
ky = zeros(1,4); % to store K values for y
kz = zeros(1,4); % to store K values for z
b = [1 2 2 1]; % RK4 coefficients
% Iterate, computing each K value in turn, then the i+1 step values
for i = 1:(N-1)
kx(1) = dx(t(i), x(i), y(i), z(i));
ky(1) = dy(t(i), x(i), y(i), z(i));
kz(1) = dz(t(i), x(i), y(i), z(i));
kx(2) = dx(t(i) + (h/2), x(i) + (h/2)*kx(1), y(i) + (h/2)*ky(1), z(i) + (h/2)*kz(1));
ky(2) = dy(t(i) + (h/2), x(i) + (h/2)*kx(1), y(i) + (h/2)*ky(1), z(i) + (h/2)*kz(1));
kz(2) = dz(t(i) + (h/2), x(i) + (h/2)*kx(1), y(i) + (h/2)*ky(1), z(i) + (h/2)*kz(1));
kx(3) = dx(t(i) + (h/2), x(i) + (h/2)*kx(2), y(i) + (h/2)*ky(2), z(i) + (h/2)*kz(2));
ky(3) = dy(t(i) + (h/2), x(i) + (h/2)*kx(2), y(i) + (h/2)*ky(2), z(i) + (h/2)*kz(2));
kz(3) = dz(t(i) + (h/2), x(i) + (h/2)*kx(2), y(i) + (h/2)*ky(2), z(i) + (h/2)*kz(2));
kx(4) = dx(t(i) + h, x(i) + h*kx(3), y(i) + h*ky(3), z(i) + h*kz(3));
ky(4) = dy(t(i) + h, x(i) + h*kx(3), y(i) + h*ky(3), z(i) + h*kz(3));
kz(4) = dz(t(i) + h, x(i) + h*kx(3), y(i) + h*ky(3), z(i) + h*kz(3));
x(i+1) = x(i) + (h/6)*sum(b.*kx);
y(i+1) = y(i) + (h/6)*sum(b.*ky);
z(i+1) = z(i) + (h/6)*sum(b.*kz);
end
% Group together in one solution matrix
txyz = [t,x,y,z];
您想要的代码可以“应用于任何方程式系统”。为了使你的脚本更有用,让我们利用向量输入,其中每个变量都在它自己的行上,然后将它变成一个函数。结果与Matlab自己的ode45
具有可比性(如何调用)。
% setup
odefun = @(t, y) [y(2); -y(1) - 2*exp(t) + 1; -y(1) - exp(t) + 1];
y0 = [1;0;1];
% ODE45 solution
[T, Y] = ode45(odefun, [0,1], y0);
% Custom RK4 solution
t = 0:0.1:1;
y = RK4(odefun, t, y0);
% Compare results
figure; hold on;
plot(T, Y); plot(t, y, '--', 'linewidth', 2)
您可以看到RK4
函数(下方)给出ode45
函数的相同结果。
函数RK4只是上述脚本的“精简”版本,它适用于您想要使用的许多公式。广泛使用时,您可能希望在函数中包含输入检查。为了清楚起见,我把它留了下来。
function y = RK4(odefun, tspan, y0)
% ODEFUN contains the ode functions of the system
% TSPAN is a 1D vector of equally spaced t values
% Y0 contains the intial conditions for the system variables
% Initialise step-size variables
t = tspan(:); % ensure column vector = (0:h:1)';
h = t(2)-t(1);% define h from t
N = length(t);
% Initialise y vector, with a column for each equation in odefun
y = zeros(N, numel(y0));
% Starting conditions
y(1, :) = y0(:)'; % Set intial conditions using row vector of y0
k = zeros(4, numel(y0)); % Initialise K vectors
b = repmat([1 2 2 1]', 1, numel(y0)); % RK4 coefficients
% Iterate, computing each K value in turn, then the i+1 step values
for i = 1:(N-1)
k(1, :) = odefun(t(i), y(i,:));
k(2, :) = odefun(t(i) + (h/2), y(i,:) + (h/2)*k(1,:));
k(3, :) = odefun(t(i) + (h/2), y(i,:) + (h/2)*k(2,:));
k(4, :) = odefun(t(i) + h, y(i,:) + h*k(3,:));
y(i+1, :) = y(i, :) + (h/6)*sum(b.*k);
end
end
答案 1 :(得分:0)
好的,事实证明这只是一个小错误,其中x变量没有被定义为y的函数(根据问题,x&#39;(t)= y。
所以:下面是一个关于如何在matlab中使用Runge Kutta 4求解微分方程系统的具体例子:
clear all, close all, clc
%{
____________________TASK:______________________
Solve the system of differential equations below
in the interval 0<t<1, with stepsize h = 0.1.
x'= y x(0)=1
y'= -x-2e^t+1 y(0)=0 , where x=x(t), y=y(t), z=z(t)
z'= -x - e^t + 1 z(0)=1
THE EXACT SOLUTIONS for x y and z can be found in this pdf:
archives.math.utk.edu/ICTCM/VOL16/C029/paper.pdf
_______________________________________________
%}
%Step-size
h = 0.1;
t = 0:h:1
N = length(t);
%Defining the vectors where the answer is stored.
x = zeros(N,1);
y = zeros(N,1)
z = zeros(N,1)
%Defining the functions
e = @(t, x, y, z) y;
f = @(t, x, y, z) -x-2*exp(t)+1;
g = @(t, x, y, z) -x - exp(t) + 1;
%Starting/initial conditions
x(1) = 1;
y(1) = 0;
z(1) = 1;
for i = 1:(N-1)
K1 = h * e( t(i) , x(i) , y(i) , z(i));
L1 = h * f( t(i) , x(i) , y(i) , z(i));
M1 = h * g( t(i) , x(i) , y(i) , z(i));
K2 = h * e(t(i) + 1/2*h, x(i)+1/2*K1 , y(i)+1/2*L1 , z(i)+1/2*M1);
L2 = h * f(t(i) + 1/2*h, x(i)+1/2*K1 , y(i)+1/2*L1 , z(i)+1/2*M1);
M2 = h * g(t(i) + 1/2*h, x(i)+1/2*K1 , y(i)+1/2*L1 , z(i)+1/2*M1);
K3 = h * e(t(i) + 1/2*h, x(i) + 1/2*K2 , y(i) + 1/2*L2 , z(i) + 1/2*M2);
L3 = h * f(t(i) + 1/2*h, x(i) + 1/2*K2 , y(i) + 1/2*L2 , z(i) + 1/2*M2);
M3 = h * g(t(i) + 1/2*h, x(i) + 1/2*K2 , y(i) + 1/2*L2 , z(i) + 1/2*M2);
K4 = h * e( t(i)+h , x(i)+K3 , y(i)+L3, z(i)+M3);
L4 = h * f( t(i)+h , x(i)+K3 , y(i)+L3, z(i)+M3);
M4 = h * g( t(i)+h , x(i)+K3 , y(i)+L3, z(i)+M3);
x(i+1) = x(i)+1/6*(K1+2*K2+2*K3+K4);
y(i+1) = y(i)+1/6*(L1+2*L2+2*L3+L4);
z(i+1) = z(i)+1/6*(M1+2*M2+2*M3+M4);
end
Answer_Matrix = [t' x y z]