Results 1 to 6 of 6
Like Tree1Thanks
  • 1 Post By HallsofIvy

Math Help - points in proof of homogeneous DE

  1. #1
    Member
    Joined
    Aug 2011
    Posts
    84

    points in proof of homogeneous DE

    I have a question about the proof of this elementary DE, the calculations are easy
    \frac{dy}{dt}+a(t)y=0\rightarrow \left | y(t)\right|=exp(-\int a(t)dt+ca)\rightarrow |cexp(-\int a(t)dt)|=c
    -not sure if you can add the integral into the absolute value
    y(t)exp(\int a(t)dt) is a continuous function of time
    -I'm not sure how this is assumed to be continuous in time?
    If the absolute value of a function g(t) is constant then g must be constant,
    Proof:
    If g isn't constant there exists t1,t2 for which g(t1)=c, g(t2)=-c. By the intermediate value theorem g must achieve all values between -c and +c which is impossible if |g(t)|=c
    -but if g(t)=x or d then why does it necessarily equal c and -c
    Last edited by mathnerd15; May 28th 2014 at 11:08 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Aug 2011
    Posts
    84

    Re: points in proof of homogeneous DE

    I misposted the equation it should read
    \frac{dy}{dt}+a(t)y=0\rightarrow \left | y(t)\right|=exp(-\int a(t)dt+c)\rightarrow |y(t)exp(-\int a(t)dt)|=c
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,370
    Thanks
    1314

    Re: points in proof of homogeneous DE

    -I'm not sure how this is assumed to be continuous in time?
    You are given that dy/dt exists, y is differentiable so continuous. Any integral is continuous, the exponential of a continuous function is continuous, and the product of two continuous functions is continuous. That's why y(t)exp(\int a(t)dt) is continuous.

    If the absolute value of a function g(t) is constant then g must be constant
    No, that's not true. The absolute value of "g(t)= 1 if t is rational, g(t)= -1 if t is irrational" is the constant 1.

    If g isn't constant there exists t1,t2 for which g(t1)=c, g(t2)=-c.
    Again, no. The function g(x)= 1 if x is rational, 2 if x is irrational has NO pair t1 and t2 such that g(t1)= c, g(t2)= -c.

    By the intermediate value theorem g must achieve all values between -c and +c which is impossible if |g(t)|=c
    -but if g(t)=x or d then why does it necessarily equal c and -c
    Are you assuming that g is continuous? You didn't say that.

    It is true that if the absolute value of a continuous function is constant, then the function is constant.
    (But "If g isn't constant there exists t1,t2 for which g(t1)=c, g(t2)=-c." is not necessarily true even for continuous functions.)
    Last edited by HallsofIvy; May 29th 2014 at 07:19 AM.
    Thanks from mathnerd15
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Aug 2011
    Posts
    84

    Re: points in proof of homogeneous DE

    then this is an error in the Braun Springer text in the early chapters. I looked at this a LONG time but instantly thought how can he come up with g(t1)=c, g(t2)=-c from assuming g isn't constant. is this a really bad book? it covers a lot of nice applications and has some interesting problems

    it bothered me so much I drank a beer and vomited. how can there be an error in this text????
    Last edited by mathnerd15; May 30th 2014 at 11:53 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Aug 2011
    Posts
    84

    Re: points in proof of homogeneous DE

    I mean you have a transcendental like y=sin(x)+1 and no value of y(t)=-c
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Aug 2011
    Posts
    84

    Re: points in proof of homogeneous DE

    what is Braun saying here in the early chapters?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Colored points proof help
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: October 20th 2012, 10:39 AM
  2. A proof concerning homogeneous DE's and integrating factors
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: March 23rd 2011, 05:34 AM
  3. Interior points proof
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 13th 2011, 06:59 PM
  4. Replies: 1
    Last Post: December 15th 2010, 05:43 AM
  5. points proof
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 21st 2006, 11:34 PM

Search Tags


/mathhelpforum @mathhelpforum