Question

如何求解具有z(inf)等边界条件的2 ^nd阶微分方程？

2(x+0.1)·z'' + 2.355·z' - 0.71·z = 0
z(0) = 1
z(inf) = 0
z'(0) = -4.805

我无法理解在z(inf)函数中使用边界值ode45()的位置。我在以下条件[z(0) z'(0) z(inf)]中使用，但这并不能提供准确的输出。

function [T, Y]=test()

    % some random x function 
    x = @(t) t;

    t=[0 :.01 :7];


    % integrate numerically
    [T, Y] = ode45(@linearized, t, [1 -4.805 0]);

    % plot the result
    plot(T, Y(:,1))

    % linearized ode
    function dy = linearized(t,y)

       dy = zeros(3,1);
       dy(1) = y(2);
       dy(2) = y(3);
       dy(3) = (-2.355*y(2)+0.71*y(1))/((2*x(t))+0.2);

    end

end

请帮我解决这个微分方程。

Answer 1

你手上似乎有一个相当高级的问题，但对MATLAB和/或ODE理论知之甚少。如果你愿意，我很乐意解释更多，但这应该是聊天（我会邀请你）或通过个人电子邮件（我的姓氏来自Google DOT com最受欢迎的邮件服务）< / p>

现在您已经澄清了一些事情并解释了整个问题，事情变得更加清晰，我能够找到合理的解决方案。我认为以下内容至少是您需要做的总体方向：

function  [tSpan, Y2, Y3] = test

    %%# Parameters

    %# Time parameters
    tMax  = 1e3;
    tSpan = 0 : 0.01 : 7;

    %# Initial values
    y02 = [1 -4.805];    %# second-order ODE
    y03 = [0 0 4.8403];  %# third-order ODE

    %# Optimization options
    opts = optimset(...
        'display', 'off',...
        'TolFun' , 1e-5,...
        'TolX'   , 1e-5);


    %%# Main procedure

    %# Find X so that z2(t,X) -> 0 for t -> inf
    sol2 = fminsearch(@obj2, 0.9879680932400429, opts);

    %# Plug this solution into the original
    %# NOTE: we need dense output, which is done via deval()
    Z = ode45(@(t,y) linearized2(t,y,sol2), [0 tMax], y02);

    %# plot the result
    Y2 = deval(Z,tSpan,1);
    plot(tSpan, Y2, 'b');


    %# Find X so that z3(t,X) -> 1 for t -> inf
    sol3 = fminsearch(@obj3, 1.215435887288112, opts);

    %# Plug this solution into the original
    [~, Y3] = ode45(@(t,y) linearized3(t,y,sol3), tSpan, y03);

    %# plot the result
    hold on, plot(tSpan, Y3(:,1), 'r');

    %# Finish plots
    legend('Second order ODE', 'Third order ODE')
    xlabel('T [s]')
    ylabel('Function value [-]');

    %%# Helper functions

    %# Function to optimize X for the second-order ODE
    function val = obj2(X)
        [~, y] = ode45(@(t,y) linearized2(t,y,X), [0 tMax], y02);
        val = abs(y(end,1));
    end

    %# linearized second-order ODE with parameter X
    function dy = linearized2(t,y,X)
        dy = [
            y(2)
            (-2.355*y(2) + 0.71*y(1))/2/(X*t + 0.1)
            ];
    end


    %# Function to optimize X for the third-order ODE
    function val = obj3(X3)
        [~, y] = ode45(@(t,y) linearized3(t,y,X3), [0 tMax], y03);
        val = abs(y(end,2) - 1);
    end

    %# linearized third-order ODE with parameters X and Z
    function dy = linearized3(t,y,X)
        zt = deval(Z, t, 1);
        dy = [
            y(2)
            y(3)
            (-1 -0.1*zt + y(2) -2.5*y(3))/2/(X*t + 0.1)
            ];
    end

end

Answer 2

正如我在上面的评论中所说，我认为你混淆了几件事。我怀疑这是要求的：

function [T,Y] = test

    tMax = 1e3;

    function val = obj(X)
        [~, y] = ode45(@(t,y) linearized(t,y,X), [0 tMax], [1 -4.805]);
        val = abs(y(end,1));
    end

    % linearized ode with parameter X
    function dy = linearized(t,y,X)
        dy = [
            y(2)
            (-2.355*y(2) + 0.71*y(1))/2/(X*t + 0.1)
            ];
    end

    % Find X so that z(t,X) -> 0 for t -> inf
    sol = fminsearch(@obj, 0.9879);

    % Plug this ssolution into the original 
    [T, Y] = ode45(@(t,y) linearized(t,y,sol), [0 tMax], [1 -4.805]);

    % plot the result
    plot(T, Y(:,1));

end

但我不能让tMax超过1000秒再收敛。您可能会遇到ode45能力w.r.t的限制。准确度（切换到ode113），或者您的初始值不够准确等等。

求解边界条件为z（inf）= 0的二阶微分方程

2 个答案: