MATLAB code help. Backward Euler method - matlab

Question

Here is the MATLAB/FreeMat code I got to solve an ODE numerically using the backward Euler method. However, the results are inconsistent with my textbook results, and sometimes even ridiculously inconsistent. What is wrong with the code?
function [x,y] = backEuler(f,xinit,yinit,xfinal,h)
%f - this is your y prime
%xinit - initial X
%yinit - initial Y
%xfinal - final X
%h - step size
n = (xfinal-xinit)/h; %Calculate steps
%Inititialize arrays...
%The first elements take xinit and yinit corespondigly, the rest fill with 0s.
x = [xinit zeros(1,n)];
y = [yinit zeros(1,n)];
%Numeric routine
for i = 1:n
x(i+1) = x(i)+h;
ynew = y(i)+h*(f(x(i),y(i)));
y(i+1) = y(i)+h*f(x(i+1),ynew);
end
end

Answer 1

Your method is a method of a new kind. It is neither backward nor forward Euler. :-)
Forward Euler: y1 = y0 + h*f(x0,y0)
Backward Euler solve in y1: y1 - h*f(x1,y1) = y0
Your method: y1 = y0 +h*f(x0,x0+h*f(x0,y0))
Your method is not backward Euler.
You don't solve in y1, you just estimate y1 with the forward Euler method. I don't want to pursue the analysis of your method, but I believe it will behave poorly indeed, even compared with forward Euler, since you evaluate the function f at the wrong point.
Here is the closest method to your method that I can think of, explicit as well, which should give much better results. It's Heun's Method:
y1 = y0 + h/2*(f(x0,y0) + f(x1,x0+h*f(x0,y0)))

Answer 2

The only issue I can spot is that the line:
n=(xfinal-xinit)/h
Should be:
n = abs((xfinal-xinit)/h)
To avoid a negative step count. If you are moving in the negative-x direction, make sure to give the function a negative step size.
Your answers probably deviate because of how coarsely you are approximating your answer. To get a semi-accurate result, deltaX has to be very very small and your step size has to be very very very small.
PS. This isn't the "backward Euler method," it is just regular old Euler's method.
If this is homework please tag it so.

Answer 3

Have a look at numerical recipes, specifically chapter 16, integration of ordinary differential equations. Euler's method is known to have problems:
There are several reasons that Euler’s method is not recommended for practical use, among them, (i) the method is not very accurate when compared to other, fancier, methods run at the equivalent stepsize, and (ii) neither is it very stable
So unless you know your textbook is using Euler's method, I wouldn't expect the results to match. Even if it is, you probably have to use an identical step size to get an identical result.

Answer 4

Unless you really want to solve an ODE via Euler's method that you've written by yourself you should have a look at built-in ODE solvers.
On a sidenote: you don't need to create x(i) inside the loop like this: x(i+1) = x(i)+h;. Instead, you can simply write x = xinit:h:xfinal;. Also, you may want to write n = round(xfinal-xinit)/h); to avoid warnings.
Here are the solvers implemented by MATLAB.
ode45 is based on an explicit
Runge-Kutta (4,5) formula, the
Dormand-Prince pair. It is a one-step
solver – in computing y(tn), it needs
only the solution at the immediately
preceding time point, y(tn-1). In
general, ode45 is the best function to
apply as a first try for most
problems.
ode23 is an implementation of an
explicit Runge-Kutta (2,3) pair of
Bogacki and Shampine. It may be more
efficient than ode45 at crude
tolerances and in the presence of
moderate stiffness. Like ode45, ode23
is a one-step solver.
ode113 is a variable order
Adams-Bashforth-Moulton PECE solver.
It may be more efficient than ode45 at
stringent tolerances and when the ODE
file function is particularly
expensive to evaluate. ode113 is a
multistep solver — it normally needs
the solutions at several preceding
time points to compute the current
solution.
The above algorithms are intended to
solve nonstiff systems. If they appear
to be unduly slow, try using one of
the stiff solvers below.
ode15s is a variable order solver
based on the numerical differentiation
formulas (NDFs). Optionally, it uses
the backward differentiation formulas
(BDFs, also known as Gear's method)
that are usually less efficient. Like
ode113, ode15s is a multistep solver.
Try ode15s when ode45 fails, or is
very inefficient, and you suspect that
the problem is stiff, or when solving
a differential-algebraic problem.
ode23s is based on a modified
Rosenbrock formula of order 2. Because
it is a one-step solver, it may be
more efficient than ode15s at crude
tolerances. It can solve some kinds of
stiff problems for which ode15s is not
effective.
ode23t is an implementation of the
trapezoidal rule using a "free"
interpolant. Use this solver if the
problem is only moderately stiff and
you need a solution without numerical
damping. ode23t can solve DAEs.
ode23tb is an implementation of
TR-BDF2, an implicit Runge-Kutta
formula with a first stage that is a
trapezoidal rule step and a second
stage that is a backward
differentiation formula of order two.
By construction, the same iteration
matrix is used in evaluating both
stages. Like ode23s, this solver may
be more efficient than ode15s at crude
tolerances.

Answer 5

I think this code could work. Try this.
for i =1:n
t(i +1)=t(i )+dt;
y(i+1)=solve('y(i+1)=y(i)+dt*f(t(i+1),y(i+1)');
end

Answer 6

The code is fine. Just you have to add another loop within the for loop. To check the level of consistency.
if abs((y(i+1) - ynew)/ynew) > 0.0000000001
ynew = y(i+1);
y(i+1) = y(i)+h*f(x(i+1),ynew);
end
I checked for a dummy function and the results were promising.