False Position method for 2nd order Differential equation BVP

Question

Kevin Tovson 2015-11-19

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/256491-false-position-method-for-2nd-order-differential-equation-bvp

编辑： Torsten 2015-11-23

HW_8.pdf

I have a good code for the Runge-Kutta method to determine the function for a 2nd order differential equation, but I need to create a code that solves my boundary conditions. My boundaries are y(0)=25 and y(30)=20. I need to determine what y'(0)=? to get y(30)=20 using false position method. When y'(0)=0, y(30)=54.32. When y'(0)=-3, y(30)=-15.3. Basically what I need to do is use the y(30) values [-15.3,54.32] to find what y'(0) would give me y(30)=20 using false position. If anyone has any idea how to code this and can help I would appreciate it! I have put my RK code below. If you need to know more about the problem it is Problem 6 part d in the attached file.

function [KR4] = NewHighway(z1,z2)
C=.053;
f1 = z2 ; 
f2 = C*sqrt(1+z2.^2);
KR4 = [f1 ; f2] ;
end
clear all ;
h = 0.01 ; %step size
t = 0:h:30;
a = zeros(1,length(t)); 
b = zeros(1,length(t));
a(1) = 25 ; %initial values
b(1) = -3 ;
for i = 1:(length(t)-1) 
    k1 = NewHighway(a(i), b(i)); %steps for RK method
    k2 = NewHighway(a(i)+h/2*(k1(1)), b(i) + h/2*(k1(2)));
    k3 = NewHighway((a(i)+h/2*k2(1)), b(i) + h/2*(k2(2)));
    k4 = NewHighway((a(i)+h*k3(1)), b(i) + h*k3(2));
    a(i+1) = a(i) + (h/6)*(k1(1)+ 2*k2(1) + 2*k3(1) + k4(1)); %calculating new values using RK method
    b(i+1) = b(i) + (h/6)*(k1(2)+ 2*k2(2) + 2*k3(2) + k4(2));
end
plot(t,a)
a(3001) %the value of the function y at 30 (needs to be 20)

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Torsten 2015-11-20

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/256491-false-position-method-for-2nd-order-differential-equation-bvp#answer_200585

编辑：Torsten 2015-11-20

在 MATLAB Online 中打开

b30_target = 20;
b0_high    = 0;
b0_low     = -3;
b30_high   = 54.32;
b30_low    = -15.3;
solved     = 0;
while solved == 0 
    b0_test = b0_high + (b30_target-b30_high)/(b30_low-b30_high)*(b0_low-b0_high);
    % solve BVP with initial conditions a(1)=25 and b(1)=b0_test
    b30_test = b(end);
    if abs(b30_test-b30_target) < 0.1
      solved = 1;
    else
      if b30_test < b30_target
        b30_low = b30_test;
        b0_low  = b0_test;
      else
        b30_high = b30_test;
        b0_high = b0_test;
      end
    end 
end