Stability Analysis for Implicit Euler Method with Negative Amplification Factor

[Marcoreus]

Fint the modified equation when the implicit Euler method is applied to y'= f(y). If f(y)=λy, where λ is negative. what is the effect on the amplication factor?
=>

y ' = λ * y
dy / dx = λ * y
dy / y = λ dx
ln y = λ* x + C
y = Ae^( λ* x ), the constant factor does not depend on λ.
i SOLVE THIS FOR THE ACTUAL SOLUTIONS
NOW,
The implicit Euler scheme is given by:
y_(n+1)= y_n +hf(y_n+1 , t_n+1 )
For f(y)=λ y, we have:
y_(n+1)= y_n +hf(y_n+1 , t_n+1 )= y_n + h λ y_n+1
Solving this for y_n+1 (in general, this is not possible), we arrive at:
y_n+1 = y_n / (1-h λ).....(eqn 1)

From (eqn 1), we can see that if |1−h λ|≥1, the solution is decaying (stable). Compare this to the actual solution of y(x)=Ae^(λ x).
If we have λ being negative, we would have:
y_n+1 = y_n / (1+h λ).....(eqn 2)

Compare this to the actual solution of y(x)=Ae^(−λ x). What conclusion can i draw?
Trying it for λ= ±1 , what happens to stability.

 

[Mark44]

Fint the modified equation when the implicit Euler method is applied to y'= f(y). If f(y)=λy, where λ is negative. what is the effect on the amplication factor?
=>

y ' = lambda * y
dy / dx = lambda * y
dy / y = lambda dx
ln y = lambda * x + C
y = [ C / lambda ]e^( lambda * x )

Your last equation is wrong. The equation before it is
ln y = λx + C
Exponentiating (making each side the exponent on e) gives
$$ e^{ln y} = e^{λx + C} = e^{λx} \cdot e^C$$
Can you finish this?

What you have done is solve the DE using separation. I don't know what the "implicit Euler method" is, so you might not have solved this in the intended way.

This is what I have done so far. Can anyone help me if Iam going right direction to answer the question?

 

[Mark44]

BTW, your title made me think you had a question about classical poetry. Odes are poems in praise of someone. ODEs are ordinary differential equations.

 

[Marcoreus]

Fint the modified equation when the implicit Euler method is applied to y'= f(y). If f(y)=λy, where λ is negative. what is the effect on the amplication factor?
=>

y ' = λ * y
dy / dx = λ * y
dy / y = λ dx
ln y = λ* x + C
y = Ae^( λ* x ), the constant factor does not depend on λ.
i SOLVE THIS FOR THE ACTUAL SOLUTIONS
NOW,
The implicit Euler scheme is given by:
y_(n+1)= y_n +hf(y_n+1 , t_n+1 )
For f(y)=λ y, we have:
y_(n+1)= y_n +hf(y_n+1 , t_n+1 )= y_n + h λ y_n+1
Solving this for y_n+1 (in general, this is not possible), we arrive at:
y_n+1 = y_n / (1-h λ).....(eqn 1)

From (eqn 1), we can see that if |1−h λ|≥1, the solution is decaying (stable). Compare this to the actual solution of y(x)=Ae^(λ x).
If we have λ being negative, we would have:
y_n+1 = y_n / (1+h λ).....(eqn 2)

Compare this to the actual solution of y(x)=Ae^(−λ x). What conclusion can i draw?
Trying it for λ= ±1 , what happens to stability.

 

[Simon Bridge]

Welcome to PF;

Ode to ODEs

O My sweet ODE
How is it you can be
So beautiful and yet so deadly
To me

The method's Implicit Euler,
But deep-fried brain is oilier
Numerical iterations don't clean me up!
I'm soilier.

Variables separate
As we cogitate
But it's not the method no
But it's not the method no

By brain hurts so sweetly when you are around
Even when you make by head to pound
But O the joy when the general solution
is found.

Sweet ODE.

"implicit eulers method" is a numerical method involving iteration.

y_(n+1)= y_n +hf(y_n+1 , t_n+1 )

... oh you found it - good :)

 

[Marcoreus]

that was nice poem. i like it thanks

 

[Simon Bridge]

Imagine my disappointment when I found the thread was not about Odes :(

we arrive at:
y_n+1 = y_n / (1-h λ).....(eqn 1)

From (eqn 1), we can see that if |1−h λ|≥1, the solution is decaying (stable). Compare this to the actual solution of y(x)=Ae^(λ x).
If we have λ being negative, we would have:
y_n+1 = y_n / (1+h λ).....(eqn 2)

Compare this to the actual solution of y(x)=Ae^(−λ x). What conclusion can i draw?
Trying it for λ= ±1 , what happens to stability.

... these are all good questions.
Before we pitch in ere, how about having a go at answering them?
Give it your best shot.

If you are stuck, try plotting a few points for some easy values of the parameters.