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I Integration - chain rule / functional

  1. Oct 19, 2016 #1
    I have ## \int_{t = 0}^{t = 1} \frac{1}{x} \frac{dx}{dt} dt = \int_{t = 0}^{t = 1} (1-y) dt ## [1]
    The LHS evaluates to ## ln \frac{(x(t_0+T))}{x(t_0)} ##, where ##t_{1}=t_{0}+T##

    My issue is that, asked to write out the intermediatary step, I could not. I am unsure how you do this when the limits aren't expressed in terms of ##x## here. So I can see that ##1/x dx = ln x ##, but I'm unsure of what has happened to ## '\frac {dt}{dt}' ## and how the integration limits are done properly.

    I think the two below points tie in, and link to where my understanding is lacking with this:

    1) Am I correct in thinking that, given ##dx/dt = x(1-y) ##, in order to get to expression [1] you can not simply, formally, 'seperate variables' - see point below, but rather you should divide by ##x## and then integrate both sides w.r.t ##t##.

    2) From high school up to now, I have used seperation of variables to solve such things as : ##dx/dt=b(x)## => ##dx/b(x)=dt##

    However my lecturer today told me that is inproper and rather one should use a integration factor - i think understanding this point ties with/ may help my initial question?

    Many thanks in advance.
     
    Last edited by a moderator: Oct 19, 2016
  2. jcsd
  3. Oct 19, 2016 #2

    Mark44

    Staff: Mentor

    Without knowing the relationship between x and t, I don't know how one would evaluate the integral on the left, since it isn't known what x(1) and x(0) are.
    Well, the above should be ##\int 1/x dx = \ln|x| + C##
     
  4. Oct 19, 2016 #3
    Sorry, I missed the constant C, yes.

    my question wasn't about evaluating x(1) and x(0)? it was about how you get from line 1 to line 2 and 'transform' from the limits in ##t## to ##x## limits, I'm only ever used to seeing ##x## limits in such an expression... (evaluates was probably the wrong word...)
     
  5. Oct 19, 2016 #4

    Mark44

    Staff: Mentor

    You can't transform the t limits to x limits unless you know how x and t relate; that is, unless you know a formula for x(t).
     
  6. Oct 19, 2016 #5
    I mean how its gone to ##x(t_{1})## and ##x(t_{0}##), not explicitly what they are.

    The only other information that I have not included in the OP is that ##T## is the period of the solution so the LHS evaluates to ##0## ...
    ##x## and ##t## relate by the equation in 1) in the OP and we are also given ##dy/dt=K(-x+xy)##
     
  7. Oct 19, 2016 #6

    Mark44

    Staff: Mentor

    When you change the variable of integration, by for example, using a substitution, you usually need to change the limits of integration. The exception is when you "undo" the substitution after finding an antiderivative.

    Here's a very simple example,
    $$\int_{x = 0}^1 (x + 2)^3 dx$$
    Let u = x + 2
    Then du = dx
    And x = 0 => u = 2, x = 1 => u =3
    Carrying out the substitution and changing the limits of integration:
    $$\int_{x = 0}^1 (x + 2)^3 dx = \int_{u = 2}^3 u^3 du \\
    = \frac{u^4}{4}|_2^3 = \frac{81}{4} - \frac{16}{4} = \frac{65} 4$$
    I don't see how equation 1 shows a specific relationship between x and t.
     
  8. Oct 19, 2016 #7
    yeh that example is fine. but i'm still not understanding what has happened in my initial question...
     
  9. Oct 19, 2016 #8
    Apologies in the OP the limits over the intergral should read as ## ^{t_{1}} _{t_{0}} \int ##, where ##t_{1}=t_{0}+T##
     
  10. Oct 19, 2016 #9

    Mark44

    Staff: Mentor

    It's customary to put the integration limits to the right of the integral sign, like so:
    ##\int_{t_0}^{t_1} f(t) dt##
     
  11. Oct 19, 2016 #10
    okay. i was concerning the change in limits, as not being explicit..
    still no idea how to approach my initial question.
     
  12. Oct 19, 2016 #11

    Mark44

    Staff: Mentor

    In my example there is a clear relationship between u and x; namely, u(x) = x + 2.
    In your integral, there is no such relationship, at least one that I can see.
     
  13. Oct 19, 2016 #12
    I agree.
    my OP is what we done in lectures however.
     
  14. Oct 19, 2016 #13

    Stephen Tashi

    User Avatar
    Science Advisor

    Can you quote the statements that came before the equation in your OP ?
     
  15. Oct 21, 2016 #14
    Literally all information has been posted above. But I shall summarise again:

    System of equations:

    ##dx/dt=x-xy## [1]
    ##dy/dt= K(-y+xy)## [2], ##K## a constant.

    The lecturer makes the statement that many of the solutions of ##x(t)## and ##y(t)## are periodic.
    Let the period be ##T##.

    My lecturer then jumps to the line:

    ## \int _{t_{0}}^{t_{0}+T} 1/x (dx/dt) dt = \ln (\frac{x(t_{0}+T)}{(x(t_{0})})=0= \int _{t_{0}} ^{t_{0}+T} (1-Ky) dt ##
     
    Last edited by a moderator: Oct 21, 2016
  16. Oct 21, 2016 #15

    Mark44

    Staff: Mentor

    The integral on the left could be rewritten as ##\int_{x(t_0)}^{x(t_0 + T)} \frac {dx} x##.
    Since x(t) is assumed to be periodic with period T, then this integral is 0, since ##x(t_0) = x(t_0 + T)##.
    However, saying that both x(t) and y(t) are periodic doesn't necessarily imply that both functions have the same period.
     
    Last edited: Oct 21, 2016
  17. Oct 21, 2016 #16
    Alright cheers. any hints please?
     
    Last edited by a moderator: Oct 21, 2016
  18. Oct 21, 2016 #17

    Mark44

    Staff: Mentor

    Hints about what? I explained why the integral on the left (i.e., ##\int_{x(t_0)}^{x(t_0 + T)} \frac {dx} x##) is zero. As I already said ##x(t_0) = x(t_0 + T)##, assuming that x(t) is periodic with period T, Then ##\frac {x(t_0 + T)}{x(t_0)} = 1##, so ln of this fraction is zero.
    If both x(t) and y(t) have the same period, then ##\int_{x(t_0)}^{x(t_0 + T)} ydt## would also be zero, and I leave it as an exercise for you to explain why ##\int_{x(t_0)}^{x(t_0 + T)} (1 - Ky)dt## is also zero.
     
  19. Oct 22, 2016 #18
    My initial question.
    I know why the integral evaluates to 0.
    and this wasn't my question
    it says 'asked to show the intermediatary step, I could not'
     
  20. Oct 22, 2016 #19

    Mark44

    Staff: Mentor

    What do you mean by the intermediate step? ("intermediatary" is not a word...)
     
  21. Oct 22, 2016 #20
    why are you bothering to reply if your not addressing the question? you're sharing your general knowledge on other parts of the question...
     
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