I have shown that for the given ODE system, that when we apply the forward Euler method to something like y′=Ay y(t_0)=y_0 t in (t_0,T] where A is diagonalizable; A=PDP^(−1), that the solution given by forward Euler is u_n=P(I+hD)^nP^(−1)u_0 where we consider the Forward Euler method to be y_(k+1)=y_k+hy′(x_k,y_k).

clovnerie0q 2022-09-27 Answered
I have shown that for the given ODE system, that when we apply the forward Euler method to something like
y = A y y ( t 0 ) = y 0 t ( t 0 , T ]
where A is diagonalizable; A = P D P 1 , that the solution given by forward Euler is
u n = P ( I + h D ) n P 1 u 0
where we consider the Forward Euler method to be y k + 1 = y k + h y ( x k , y k ) . N
Now, I want to show that if the real parts of λ k (the eigenvalues of A and thus the entries of D) are negative, then h < 2 R e λ k | λ k | 2 for all k implies that u n 0.
Now, in order to show this, based on what we have obtained for the formula for u n , we need only show that for each | 1 + h λ k | < 1. I am having a really tough time doing this, and have mostly just tried to play around with AM-GM stuff.
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Answers (1)

beshrewd6g
Answered 2022-09-28 Author has 12 answers
You have to find the set | 1 + z | < 1 in the complex plane. Squaring this inequality results in
1 + 2 R e ( z ) + | z | 2 < 1 R e ( z ) < 1 2 | z | 2
Now insert z = h λ k and that h is a real number...
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has local truncation error O ( h 2 ) and global error O ( h ) .
For higher order ODEs one can rewrite the ODE as a first order system and then apply Euler as before.
Should the local truncation error still be O ( h 2 )?
I tried this for a second order ODE
y ( t ) = f ( t , y ( t ) , y ( t ) ) , t [ a , b ] , y ( a ) = α , y ( a ) = β
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This of course contradicts my expectation that the error would still be O ( h 2 ).
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Let y ( t ) satisfy the differential equation
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. Then the backward Euler method for n > 1 and h > 0
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Solve the following ODE by using the method of undetermined coefficients in which Euler's formula needs to be utilized:
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My approach:
The formula can be written as y ( t ) = y h ( t ) + y p ( t ) where y h ( t ) is the "homogeneous version" of the ODE and y p ( t ) is the particular solution that we'll obtain via the basic rule of the method of undetermined coefficients.

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Putting r ( t ) = sin ( t ) = 0 in the original equation, the ODE we need to solve is:
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y p ( t ):
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substituting these equations into the original equation and then simplifying gives us:
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And in conclusion, we can write that the solution to the given ODE is:
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How would we be able to derive this conclusion via Euler's formula? Thanks in advance.