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Second opinion on ODE interpretation sought.

  1. Jun 22, 2010 #1
    Hi, this is from a physics subforum of physicsforums:

    My calculus is not very good, but the above does not strike me as true and I would like a second opinion.

    If it is true, then my calculus is even worse than I thought and I need someone to explain to me how dr/dt=0 always implies that d^2r/dt^2 must be zero too.

    P.S. I am one of those "certain members", but I welcome re-education and conversion to the group of "uncertain members".
    Last edited: Jun 22, 2010
  2. jcsd
  3. Jun 22, 2010 #2


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    If dr/dt = 0 and you take a derivative of this equation, what do you get?

    (Note, however, that it is entirely possible that [itex]\frac{dr}{dt}(T) = 0[/itex] for a specific time T, but [itex]\frac{d^2r}{dt^2}(T) \neq 0[/itex], but this is for a single time. In a differential equation like quoted in the first post, setting dr/dt = 0 is setting it to zero for all time t, which implies d^2r/dt^2 is zero. This is what the author of your quoted post explains in this earlier post in that thread. Just because the initial condition is dr/dt(T=0) = 0 does NOT mean you can set dr/dt(t) = 0 in the differential equation)

    EDIT: Reading through more of that thread, you ask the question "Are you saying that you can not differentiate the left hand side of ax^2 + bx + c = 0 because the right hand side is zero?"; the answer is that you cannot differentiate both sides in this instance for the reason that it holds only for specific values of x, not all x. Above, when I suggest you differentiate dr/dt = 0, this you are allowed to do because I am saying r is a function of t that is equal to zero for all t:

    [tex]\frac{dr}{dt}(t) = 0,~\mbox{all}~t[/tex].

    Here, we are basically defining a function f(t) = dr/dt(t) and stating that it is 0 for all values of t.

    However, if dr/dt is not zero for all t but supposing there is a value t = T such that

    [tex]\frac{dr}{dt}(T) = 0[/tex]
    then in this case you cannot take a derivative of both sides, because here what we are saying is "assume there is a value t=T such that the function f(t) = dr/dt(t) is equal to zero when t=T"; you can then plug into f(t) the specific value T set f(T) = 0 to solve for T.

    To summarize:

    For an equation like [itex]ax^2 + bx + c = 0[/itex] you CANNOT take a derivative because in that equation you are solving for the value of x that makes it true; the equation holds only for this value. For a differential equation, e.g., [itex] a x''(t) + b x'(t) + cx(t) = 0[/itex], this you CAN take a derivative of because the equation holds for all t. Note, however, that you could not vary this equation with respect to x(t) because the RHS being zero only holds for the function X(t) that satisfies the differential equation.
    Last edited: Jun 22, 2010
  4. Jun 22, 2010 #3


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    The wording is ambiguous. "If you make dr/dt= 0 for any a< t< b" could be interpreted as "for some t" or as "for all t". If you interpret as "for all t between a and b" then certainly the second derivative is 0 since the first derivative is a constant. How you interpret it as "for some t" then the second derivative is not necessarily 0.
  5. Jun 22, 2010 #4
    Thanks Mute and HallsOf Ivy.

    Not being very mathematically gifted, it will take me a while to absorb all you have said. I think I have gathered enough, that I now realise that trying to take the derivative with respect to x of ax^2 +bx +c = 0 was not a good choice of example.

    To try and be more specific. This is from the other thread:

    Although Starthaus does not make it clear here, it is clear in his later comments that he is claiming that in this particular case when espen imposes dr/dt=0 that it follows that d^2r/dt^2 must also be zero. Espen and myself disagree, but Starthaus claims to have superior calculus skills, so we can not simply ignore his claims. In this specific example is Starthaus right in his claim? I thought the best place to ask would be in this subforum frequented by calculus specialists. I think I should make it clear that from the context of the other thread that espen is talking about dr/dt= at some instant t, without making any assumption about dr/dt=0 for all t. In other words, espen is making the claim that at the instant in the particles trajectory when dr/dt=0 (at the apogee) the acceleration of the particle is:

    [tex]\frac{\text{d}^2r}{\text{d}\tau^2} = -c^2\left(\frac{ r_s}{2r^2}-\frac{r_s^2}{2r^3}\right)\left(\frac{\text{d}t}{\t ext{d}\tau}\right)^2[/tex]

    (and of course espen's equation is only valid at the apogee) while starhaus is making the claim that at the apogee, the acceleration is:

    [tex]\frac{\text{d}^2r}{\text{d}\tau^2} = 0[/tex]

    My argument is that if the acceleration is zero at the instant the particle is at its apogee in the trajectory, then dr/dt will still be zero at the next instant and instant after that and so on, so that in effect the particle goes up but does not come down and simply hovers at its apogee. This seems a bit unphysical.

    I think I meant to say you can take the derivative of [itex] a x'(t) + b x(t) + c = 0[/itex]. Does that make more sense?
    Last edited: Jun 22, 2010
  6. Jun 22, 2010 #5


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    Starhaus is referring to the fact that if the function [itex]\frac{dr}{dt}[/itex] is identically zero for all time then the function [itex]\frac{d^2r}{dt^2}(t)[/itex] is identically zero for all time. This is true.

    It can also be true that if the value of the function [itex]\frac{dr}{dt}(t)[/itex] at some time t = T is zero while the value of the function [itex]\frac{d^2r}{dt^2}(t)[/itex] is not zero: [itex]\frac{dr}{dt}(T) = 0[/itex], [itex]\frac{d^2r}{dt^2}(T) \neq 0[/itex].

    The problem stated that at t = 0, the value of dr/dt is zero. So, if the differential equation were

    [tex]\frac{d^2r}{dt}(t) + A\frac{dr}{dt}(t) + B = 0[/tex]
    then this equation is true for all time t and this is the differential equation that must be solved. True, you could look at this at time t = 0, noting that dr/dt(t=0) = 0, which tells you

    [tex]\frac{d^2r}{dt}(t=0) + B = 0[/tex],
    i.e., the intial value of the acceleration is -B. However, this is not a differential equation. This equation only tells you the value of the acceleration at time t=0. It tells you nothing else. Interpreting this equation as a differential equation is nonsense because it's not a differential equation. This seems to be the problem in the other thread - it sounds like someone over there plugged the initial condition into the differential equation and then solved the resulting equation as if it were a differential equation, which it is not.

    Yes, you can take a derivative of that equation with respect to t because it hold for all t. However, note that you could not take a variational derivative that equation with respect to x(t) because that equation only holds for the function that is the solution of that differential equation. In the other thread you had written something like

    [tex]\mathcal L[x(t)] = \frac{1}{2} \dot{x}(t)^2 - \frac{1}{2}kx(t)^2 = 0[/tex]

    You then varied the lagrangian with respect to x and claimed you got the equation of motion for x(t), and starhaus told you it was incorrect. The reason it is incorrect is that saying L = 0 (or a constant) fixes what x(t) is. It demands that whatever x(t) is it has to satisfy [itex]0 = \frac{1}{2} \dot{x}(t)^2 - \frac{1}{2}kx(t)^2[/itex]. You can take a derivative of this expression with respect to t and it will still hold because it holds for all t, but you couldn't, for example, take a variational derivative with respect to x(t) because it doesn't hold for all x(t). This is analogous to taking a derivative (with respect to x) of the relation 0 = ax^2 + bx + c versus taking a derivative of the function f(x) = ax^2 + bx + c.

    When we vary a Lagrangian, it's because we're varying the integral over it, the action. The action is a functional of the function x(t) - that is, we give the action some function x(t), and the functional spits out a number. Different x(t) inputs will give different numbers output. So, we can vary the action with respect to x(t) because it is a functional of x(t), just as we can differentiate a function of x with respect to x. However, we can't vary the Euler Lagrange equations because they are just relations which define the specific function x(t) which extremizes the action, just as we can't take a derivative with respect to x of a quadratic equation because a quadratic equation is just a relation that defines the value x which solves that equation.

    I hope this makes sense...
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