Numerical Analysis - 2nd Order Taylor Method to Solve IVP Problem With Code

Question

Maria Raheb 2020-4-10

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https://ww2.mathworks.cn/matlabcentral/answers/516804-numerical-analysis-2nd-order-taylor-method-to-solve-ivp-problem-with-code

评论： Guru Mohanty 2020-4-15

function [y1] = TaylorMethod(f,inter,y0,k)
t(1) = inter(1); %initial time
y1(1) = y0; y2(1) = 0; %setting the initial condition 
h = 0.1*(2.^-k); % setting the step size
n = 1/h; % number of steps in terms of the step size.
Ft = @(t,y) diff(f,t);
Fy = @(t,y) diff(f,y);
for i = 1:n
    t(i+1) = t(i) + h;    
    euler_y(i+1) = y(i) + h*f(t(i),y(i));
    y1(i+1) =  euler_y(i+1) + ((h^2)/2)*(Ft(t(i),y1(i)) + Fy(t(i),y1(i))*f(t(i),y1(i)));
end

Hello,

I am just playing around with some numerical methods and I seem to be having some issues with my code; I want my code to be as robust as possible, so I allocated the variables Fy and Ft to be the partial derivatives of a function 'f' inputed in the function TaylorMethod. However, I get an erro trying to do this.

The IVP I am trying to solve with my function is

y' = 5*(t^4)*y --> this is function 'f'

inter = [0 1]

y(0) = 1

Thank you in advance

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Maria Raheb 2020-4-10

f = @(t,y) 5*(t.^4).*y

As an example,

[y1] = TaylorMethod(f,[0 1],1,1)

Maria Raheb 2020-4-10

@darova

Thank you for the correction, but the problem is arising in my use of function handles for the partial derivatives of my function f.

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Answer 1

Guru Mohanty 2020-4-14

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/516804-numerical-analysis-2nd-order-taylor-method-to-solve-ivp-problem-with-code#answer_425983

在 MATLAB Online 中打开

I understand, you want to solve the IVP with 2ndorder Tayler Method. In your code the error occurs during computation of partial differentiation. You can also do the Partial differentiation using Symbolic toolbox and evaluate its value using subs function. Here is a sample code for it.

clc; clear all;
f =@(t,y)(5*(t^4)*y);%      --> this is function 'f'
inter = [0 1];
y0 = 1;
k=1;
Out = TaylorMethod(f,inter,y0,k);
plot(Out);
 
function [y1] = TaylorMethod(f,inter,y0,k)
    syms t y tp yp
    Ft1=subs(diff(f,t),{t,y},{tp,yp});
    Fy1=subs(diff(f,y),{t,y},{tp,yp});
    clear t y;
    t(1) = inter(1); %initial time
    y1(1) = y0;
    y2(1) = 0; %setting the initial condition 
    h = 0.1*(2.^-k); % setting the step size
    n = 1/h; % number of steps in terms of the step size.
    for i = 1:n-1
        t(i+1) = t(i) + h;    
        euler_y(i+1) = y1(i) + h*f(t(i),y1(i));
        y1(i+1) =  euler_y(i+1) + ((h^2)/2)*(subs(Ft1,{tp,yp}, {t(i),y1(i)}) + subs(Fy1,{tp,yp},{t(i),y1(i)})*f(t(i),y1(i)));
    end
end