I have the following system: y′=y+e^x y(0)=0 The problem asks for applying Euler's method and then finding an expression for the global error.

Kendra Hudson

Kendra Hudson

Answered question

2022-09-03

I have the following system:
y = y + e x
y ( 0 ) = 0
The problem asks for applying Euler's method and then finding an expression for the global error. Finally, supposing that
lim h > 0 1 ( 1 + h e h ) x h x h ( e h ( 1 + h ) ) = 1
Check that the global error tends to 0 as h tends to 0 as well.
Let's first apply Euler's method.
η 0 = y 0 = 0
x i = x 0 + i h = i h
η i + 1 = η i + h f ( x i , η i )
Since x i = i h, we have η i + 1 = η i + h ( η i + e i h . Factoring by η i :
η i + 1 = η i ( 1 + h ) + h e i h
And now I don't how should I proceed. Should I solve the iteration? I've done problems where I can express η i + 1 in terms of η 0 by iterations, but here I can't do so because the h e i h prevents that (or at least I haven't managed to do it).
I know that global error is defined as e ( x , h ) = η ( x , h ) y ( x ) but I don't know how exactly should I use that formula. My guess it that I have to solve the differential equation and use the solution y ( x ), but I don't know how to solve that system or what exactly is η ( x , h ).
Any help regarding to solving this problem will be highly appreciated.
Thanks!

Answer & Explanation

Christina Matthews

Christina Matthews

Beginner2022-09-04Added 16 answers

The integrating factor for the exact solution is e x , so that
( e x y ( x ) ) = 1 y ( x ) = y ( 0 ) e x + x e x .
This resonance behavior will make the analysis of the numerical solution a little more complicated.

The Euler approximations have the general form y i = A ( 1 + h ) i + B e i h . Inserting into the recursion one finds
B e h = ( 1 + h ) B + h B = h e h 1 h
and from the initial condition
y 0 = A + B y i = ( y 0 B ) ( 1 + h ) i + B e i h = y 0 ( 1 + h ) i + h ( ( 1 + h ) i e i h ) e h 1 h
Now you can insert the expansions for
( 1 + h ) i = exp ( i ln ( 1 + h ) ) = exp ( i h 1 2 i h 2 + 1 3 i h 3 + . . . )
to find the lower error order terms.of the difference to the exact solution.

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