1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Advanced Topic in Odes

  1. Mar 11, 2014 #1
    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.
     
    Last edited: Mar 11, 2014
  2. jcsd
  3. Mar 11, 2014 #2

    Mark44

    Staff: Mentor

    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.
     
  4. Mar 11, 2014 #3

    Mark44

    Staff: Mentor

    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.
     
  5. Mar 11, 2014 #4
    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.
     
  6. Mar 11, 2014 #5

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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.
    ... oh you found it - good :)
     
  7. Mar 11, 2014 #6
    that was nice poem. i like it thanks
     
  8. Mar 11, 2014 #7

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Imagine my disappointment when I found the thread was not about Odes :(
    ... 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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Advanced Topic in Odes
  1. Project Topics (Replies: 2)

  2. Advanced calc (Replies: 4)

  3. An ODE (Replies: 4)

  4. Advanced calculus (Replies: 5)

  5. Ode problem (Replies: 13)

Loading...