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What types of equations are more general than IVPs?

  1. Sep 8, 2010 #1
    I'm currently taking a Diff Eq course, where the instructor has included physics problems into the homework. I love solving Diff Eq problems but physics problems make me grit my teeth and here's an example of something I was mildly stumped by today that will help show my frustration.

    1. The problem statement, all variables and given/known data

    A ball whose mass is 1 kilogram rolls on a level surface, subject only to the force of friction. The initial velocity is 1 meter per second; after 1 second, the velocity is 0.8 meters per second. Assuming the magnitude of the frictional force is proportional to the velocity,

    (a) What is the friction constant?
    ....

    2. Relevant equations

    I already know that frictional force is expressible as F_f = kv, k being the coefficient of static friction, and v the velocity. I also already know Newton's first and second laws (who doesn't?) sum(F_i) = F and F = ma.

    3. The attempt at a solution

    Now, here is where my thinking started going crazy. I will example why in a moment, but first:

    I had noted that the initial force to send the ball into motion should be taken into account. Say F_0 = ma_0. Also, my diagram would show F_0 and F_f pointing in opposing directions, and that F_0 + F_f = F, by Newton's first law. Following that, I noted:

    At time t = 0, an instantaneous force F_0 = ma_0 is applied

    At time t > 0, a continuous force F_f = km is applied.

    Next, I began considering the piecewise function, to model the initial force:

    F_0(t) = m a_0 when t = 0
    F_0(t) = 0 when t > 0

    And I began carrying on in this very general, abstract away for the rest of the page. Giving an equation:

    F_0(t) + kv(t) = ma = mv'(t)

    But then I realized that this chapter is just supposed to be about separable equations, so something way simpler is required. However, I had a hard time convincing myself that something simpler could be used.

    This happened to be an odd problem, so rather than drive myself crazy with the way physics is worded, I just looked up the answer :(

    Here was my thinking process:

    The main problem I have with IVPs as a way to model certain things in physics is that it wants to model only instantaneous phenomenon, where dt is close enough to zero. But, as we all know, an impact does not occur in perfectly 0 seconds. Sometimes the brief moment of time is important to the actual result.

    What's worse, some "impacts" are actually forces that are applied for brief periods of time, and then released. Imagine a man pushing a box up a hill, stopping briefly, having the box slip a bit, and then resuming his push. How would that be modeled?

    So then, you, the veteran physicist, might say: Well, that's no longer one equation. It's more like a series of equations, marked off by discontinuous points when the forces are applied and released. It would be more like an algorithm then an equation.

    Well, I guess my bias is showing, as much of my love for mathematics is how much it resembles the logic of computer science. But having a general algorithm for solving force body diagrams would satisfy much of my insecurities on the subject.

    I'm not even asking to be given all of the numbers. In fact, I'm not even asking to be given the equations. I would at least like to know the relationship between the equations so I can begin filling in the details.

    More than anything, I want a rigorous proof that a set of equations and variables truly represent the solution to a problem.

    If I need to widdle down my model to something reasonable, at least let me know what those assumptions are, so I can write down in my problem "I took the assumption that ....".
     
  2. jcsd
  3. Sep 8, 2010 #2

    collinsmark

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    Hello Farleyknight,

    Don't get too discouraged. I think you might be over-thinking the problem a little. It turns out that this problem actually is pretty simple. You're right that some IVPs require instantaneous changes (particularly in acceleration) and often involve the Heaviside step function as a part of setting the differential equation (and this is particularly true when one uses the Laplace transform). My point is that I don't think that this is one of those problems. (For the rest of this message, I'm assuming that your coursework does not require you to solve this problem using the Laplace transform.)

    You can solve this problem pretty simply (at least the way it was phrased in the problem statement) by assuming the ball was rolling, and experiencing friction proportional to its velocity, since the "beginning of time" itself (t = -∞), and will continue until the "end of time" (t = ∞). (Or more precisely, the ball always has been, and always will be, rolling and experiencing friction.)

    One of the things inherent to differential equations are the solutions have arbitrary constants. That's where your "initial conditions" fit in. And don't let the name fool you, initial conditions don't have to occur "initially," at the beginning of time. They can be applied at any time. That goes for t=0 sec, t= -1 sec, t = 1 sec, or even t = 10100 seconds. You can use your initial conditions to make the "arbitrary constants" less arbitrary (i.e. make them actual constants).

    One method to solve this differential equation (in the problem statement) is to assume that the frictional force always has been and always will be proportional to the velocity. With that in mind, set up your differential equation and solve it*.

    You'll end up with a formula for velocity. And that formula will contain one arbitrary constant. Use one of your initial conditions to make that arbitrary constant an actual constant. Use your second initial condition to solve for k.

    *When setting up your differential equation, pay attention to the sign. Because it is a frictional force, the acceleration will be in the opposite direction as the velocity. In other words, make sure to set up your equation such that [itex] v [/itex] is in the negative direction compared to [itex] \dot v [/itex]

    [Edit: By the way, I know you're looking for a more procedural/algorithmic approach to setting up differential equations. I wish I could be of more help to you on that level. I'm not sure if I'm really qualified to answer, but let me give it a shot. If there must be discontinuous forces or other one-time-changing things in your equations, make use the Heaviside step function, ramp function, impulse function, etc., as appropriate, to model the system. Just don't make your equations model a system that is more complex than what it needs to be, given the problem statement. Try to keep it as simple as possible. When it comes to solving the differential equations I can say this: some methods for solving differential equations work better on some problems than other methods, and other methods on other problems. Some problems can't be solved using some methods, but can be using other methods. As you continue the course, I'm anticipating you will like the Laplace transform. Using the Laplace transform, you can solve pretty much any linear, time invariant system in a way that is very procedural and sort of resembles an algorithm. Of course, you might was well throw the Laplace transform out the window when dealing with nonlinear differential equations.]
     
    Last edited: Sep 8, 2010
  4. Sep 9, 2010 #3
    Yes, I realized just a few hours ago that an IVP is roughly equivalent to the piecewise equation:

    y(t) = C when t = a
    y(t) = F(t, y) otherwise

    where y(a) = C is the initial value. In the case of the ball being struck, the equation

    F_0(t) + kv(t) = ma = mv'(t)

    Is valid, provided that the term kv(t) = 0 when t = 0, yielding our IVP. In this case, kv(t) is zero because the amount of time caused by the strike is so small that the force of friction during that "instant" is neglectible.

    There is probably no "proof" that this is true, but I suspect that my prof would simply argue that it's an approximation, so good enough for the course.

    I'm still bothered by the fact that you physicist use Occam's Razor as an axiom for producing models :( Maybe that's why you guys http://xkcd.com/397/" [Broken].

    But I can't blame you. You rarely have to re-build the entire system of fundamental laws for which you rely on. For some of us computer scientists and logicians, we can never be sure exactly which axiomatic basis we are using. Sure, they are all equivalent, but they produce slightly different results in some cases. Thus producing crazy abstract models of very simple phenomenon (like category theorists do) is actually justified.
     
    Last edited by a moderator: May 4, 2017
  5. Sep 9, 2010 #4

    collinsmark

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    But I still would not set the equation up that way. That's because

    [tex] kv(0) \ne 0 [/tex]

    That's because if it was equal, it would imply that v = 0 (because k is a constant), The actual number comes from one of your initial conditions,

    [tex] kv(0) = k(1 \ [\mathrm{m/s}]) [/tex]

    But you haven't figured out what k is yet, so you don't know the actual number just yet. But you can get to that later if you wanted to. (Although it's not necessary to solve for the force in this part of the problem anyway.)

    ================================
    Let's take a step back.

    Allow me to tell you how I would approach the problem. You are instructed to assume "the magnitude of the frictional force is proportional to the velocity." Don't worry about your initial conditions until after you solve the differential equation. The problem statment said nothing about the forces suddenly being applied or released at any specific point in time, so there is no reason to model it that way. So model the differential equation simply:

    [tex] m \dot v = -kv [/tex]

    Pretty simple huh.

    Rearranging gives us

    [tex] \dot v = -\frac{k}{m}v [/tex]

    Now use whatever method you want to (or are supposed to ) to solve this differential equation. The solution is of the general form,

    [tex] v(t) = Cf(k,t) [/tex]

    Where C is some arbitrary constant. But you know that 1 [m/s] = Cf(k,0). Solve for C. Now you know what C is (numerically, even).

    Next, you also know that

    [tex] 0.8 \ [\mathrm{m/s}] = Cf(k,1 \ [\mathrm{s}]). [/tex]

    Since you already know what C is, you can solve for k and get a numeric answer. You've finished part (a) of the problem.

    Well, I don't think I'm qualified to say too much about that that. Except to say that if you want to model more intricacies of the physical system, it is always possible to make the models more complicated and precise. But if these intricacies are negligible -- or more importantly if your textbook is not asking for you to model them -- there's really not much point in 'inventing' them.
     
    Last edited by a moderator: May 4, 2017
  6. Sep 9, 2010 #5
    Don't worry, I've already solved the problem. It was pretty straight forward. But the idea of using piecewise functions to model discontinuous forces was not my idea. My prof has thrown a few in, and here's another problem I'm working on to show what I mean:

    =========================================================================================================

    In this exercise, we must modify our model for the motion of a falling body to take into account a variable mass. Let m(t) denote the mass of a falling body. Then the velocity of the body is a solution of the following ODE, derived from Newton’s second law of motion:

    m(t)v' = -m(t)g - k|v|v

    A rocket has a mass of 6 kilograms, including 2 kilograms of fuel. The fuel burns uniformly for 20 seconds, leaving no residue, and produces a constant thrust of 120 newtons. After 20 seconds, the fuel is spent, and the only forces are gravity and air resistance. The force of
    air resistance is proportional to the square of the velocity, with drag coefficient k = 0.02 kilograms per meter. Write initial value problems to determine the velocity of the rocket as a function of time for

    (a) 0 < t < 20
    (b) 20 < t < T_1, where T_1 is the time that the rocket reaches its maximum altitude, and for
    (c) T_1 < t < T_2, where T_2 is the time that the rocket hits the ground.

    =========================================================================================================

    Now, unless I'm mistaken, we would have to use a piecewise function to model the thrust. Is that correct?
     
    Last edited: Sep 9, 2010
  7. Sep 10, 2010 #6

    collinsmark

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    Yes, that would be one way to do it. And it would probably be the way I would do it too. The initial conditions of some pieces rely on the answers from previous pieces.

    Now you could set up the whole thing as one big differential equation relying on the Heaviside step function, u (it's often called H, but I'll use 'u' here. Sometimes the unit step function notation u skips the 1/2 part when t = 0 and either stays at 0 or goes straight to 1, depending on the author. The Heaviside step function H implicitly is 1/2 when t = 0. But sometimes the Heaviside step function is denoted as u by some authors).

    [tex] u(t) = \left\{\begin{array}{cc}
    0,&\mbox{ if } x< 0\\
    \frac{1}{2}, & \mbox{ if } x=0\\
    1, & \mbox{ if } x>0
    \end{array}\right [/tex]

    Suppose a ball is floating through frictionless space, when suddenly, at t = 0, it encounters a constant force F0. Then, after 20 sec it encounters yet another force (frictional this time) that is proportional to its velocity. You could model that as

    [tex] m \dot v = u(t) F_0 -u(t - 20 [\mathrm{sec}])kv [/tex]

    There are methods of differential equations capable of dealing with the Heaviside step function (or other variants of the unit step function) and other types of discontinuities. But you will find that usually the way that you need to work with them is that the solutions to the differential equations (whether they be interim or final) are broken up under different conditions. And mathematically this is really not much different than breaking up the differential equation into separate differential equations at the beginning. The difference is that these methods break things up in the middle, or at the end (which may involve step functions as part of the solutions), before everything is combined into a single solution. But generally speaking yes, you usually need to break something up somewhere, whenever there are such discontinuities.

    [Edit: And to be clear, some of these methods do their "breaking up" simply by containing multiple terms in the single equation. It doesn't always involve multiple equations. You'll see this in the Laplace transform methods for solving differential equations, which handle step functions fairly elegantly (well once you get used to it that is). But as I mentioned earlier, the Laplace transform is only useful for linear differential equations. So whenever you have something like a [itex] m \dot v = -k|v|v [/tex] term, which is nonlinear (due to the v2 part; common when dealing with air resistance), the Laplace transform doesn't apply. But my point is that there are some methods of solving differential equations that can handle IVPs without breaking everything up into pieces first.]
     
    Last edited: Sep 10, 2010
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