Plot polynomial using Eueler's method

syms x;

y=x.^3 + x.^2- 12*x; %polynomial function

dy= diff(y,x)%derivative of y

N=100;

h=0.1;

x=-4:0.1:3;

y(1)=-10;

y=zeros(size(x));

EM=[]; %empty array to store values

for n=1:N

y(n+1)=y(n)+h*dy(n);

x(n+1)=n*h;

EM(x,y)=[x,y];

end

plot(EMsub,"r-")

Hello, this is the code that i have so far. I have tried writing a code to plot f(x)=x^3+x^2-12x using Euler's method in matlab. I can seem to figure out why the code is not running. At line 12 i keep getting the following error:

Error using symengine

Unable to convert expression containing symbolic variables into double array. Apply 'subs' function first to substitute values for variables.

I am not exactly sure how to fix this error or what is wrong. Could someone explain why its not working and what i am doing wrong here? Any help would be appreciated.

Answer 1

[Davide Masiello]

so, initially, you calculate the analytical derivative of the function using the symbolic environment

syms x
y   = x^3+x^2-12*x;                    % polynomial function
yp = diff(y,x)                           % symbolic derivative of y 

yp = 

You'll need this symbolic derivative to be turned into a matlab function, which you can to like this

dydx = matlabFunction(yp)

dydx = function_handle with value:
    @(x)x.*2.0+x.^2.*3.0-1.2e+1

Now, you can solve the ODE represented by dydx using Euler's method.

h       = 0.1;
xn      = -5:h:5;
yn      = zeros(size(xn));
yn(1)   = -40;

for idx = 2:length(xn)
    yn(idx) = yn(idx-1)+h*dydx(xn(idx-1));
end

Now, you can plot both numerical and analytical solutions for comparison

fplot(y)
hold on
plot(xn,yn,'k--')
legend('Analytical solution','Euler''s method')

[E]

Thank you so much. I have a question though. Why wasn't my previous code running? What was i missing? I also noticed the changes in the for loop. From what i know Euler's method changes using step sizes and increasing (n+1) in a range of n=1:N. Why start at 2, and remove 1 instead of adding it?

[Davide Masiello]

There are a number of problems with your code, I can try and summarise them below

syms x;
y=x.^3 + x.^2- 12*x;
dy= diff(y,x)

f= @(x,y)3*x.^2+2*x-12; % no need to write the derivative function yourself

N=100; % no need to specify the number of points if you specify interval h and x0 and xf
h=0.1;
x=-4:0.1:3;
y(1)=-10; % you should specify the initial condition after initializing the solution array y
y=zeros(size(x)); % you already used y as a symbolic variable, you need to use a different name for the numerical one
EM=[]; % no need for this, the values will fill up the array you initialised in the previous line.
for n=1:N  %since you have n+1 below, the indexes cannot span 1:N, but either 1:N-1 or 2:N
    y(n+1)=y(n)+h*dy(n);
    x(n+1)=n*h;     % x is defined already, no need to recompute it
    EM(x,y)=[x,y];  % you can't use x and y as indexes
end
plot(EMsub,"r-")

[John D'Errico]

@E

Your code was not terrible. But you need to worry more about what variables you use, and about moving from the symbolic side of MATLAB, to the numerical side. For example,

syms x;
y=x.^3 + x.^2- 12*x; %polynomial function
dy= diff(y,x)%derivative of y 

dy = 

At this point, y and dy are symbolic expressions in MATLAB.

whos y dy

  Name      Size            Bytes  Class    Attributes

  dy        1x1                 8  sym                
  y         1x1                 8  sym                

But then later on, you tried to use y to store a vector of numbers, as the solution to the ODE. That might cause a problem. ;-) Be careful about re-using the same variable names. Yes, you did reassign y. But what did you do?

y(1)=-10;

So, first, you assigned the initial condition to y. But that does not actually do what you think. It overwrites the symbolic function y that you created.

y

y = 

And THEN you essentially did this:

x=-4:0.1:3;
y=zeros(size(x));

So you have now lost the information about the initial condition you tried to enforce, because you have now just overwritten the variable y.

As far as the code by Davide goes, you ask why N was started at 2. Since you already know the value of y(1), you don't need to use the Euler step to compute y(1) at all. So that only applies to the values y(2) through y(N).

But as I said, your code was not too far from actually working. I would say you mainly need to take a little more care, thinking about what your variables store, and exactly what they mean at that point in time. As well, I'd be willing to use new variable names, to avoid confusion in your code.

[E]

Thank you so much everyone for your helpful feedback. Seeing my code not working and not knowing why was truly frustrating. Matlab is a challenge. I read your comments and i think i understand where my code was running into issues. Thank you again for your help on this.

[Steven Lord]

Since it sounds like you're new to MATLAB, I suggest you start with the free MATLAB Onramp tutorial to quickly learn the essentials of MATLAB.