Results 1 to 6 of 6

Math Help - Differencial Equation Theory Question

  1. #1
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9

    Differencial Equation Theory Question

    I was thinking about posting this as a problem of the week question, but it seems non-elementary.

    I believe it is time for me to learn the theory of differencial equations.

    Q.What does it mean to solve a differencial equation?

    This question is much more complicated than it seems, you cannot simply say a function whose derivative(s) satisfies the conditions, no!
    ---
    Consider,
    y'=1/sqrt(1-x^2)
    We can show that,
    y=asin(x) is a solution on the open interval (-1,1)
    Then we can show that y=asin(x)+C is too a solution.
    Then we can use MVT to prove no other solutions exist.

    But that is not true!
    What about,
    f=asin(x) defined on (-1/2,1/2)
    Certainly,
    y not = to f
    So the question remains what does it mean to solve a differencial equation?

    Does it mean a function defined on some open interval that satisfies this differencial equation for all point in the interval? If so, then that can lead to problem that I will soon show.

    Or do we need to define an open interval to begin with.
    For example,
    Solve,
    y'=1/sqrt(1-x^2) on interval (-1,1)
    Then all solutions are,
    y=asin(x)+C

    I think the latter is much simpler and more elegant.
    ---
    To illustrate the elegance of my second defintion is stated in this obvious theorem.

    Theorem:
    The differencial equation,
    y'=f(x)
    where f(x) is a countinous function on some open interval
    Has infinitely many solutions.
    If F(x) is a solution then all solutions are {F(x)+C}.

    Proof:
    It is sufficient to show that at least one function F(x) exists then using MVT we can show (omitted) that all solutions must be the set {F(x)+C}.
    To do that we use the second fundamental theorem,
    Define the function,
    g(x)=INT_{x_0}^x f(x) dx for any point x_0 in the open interval.
    Since f(x) is continous on the open interval the derivative of g(x) on any point in the open interval is f(x).
    Q.E.D.

    Look how simpler it is to state an open interval to solve a solution on it.
    ---
    Linear differencial equations do not bother me so much because they have a good formal explanation in my book.
    However, seperable equations are really giving me theory problems.

    Consider,
    y'=yx
    Again what does it mean to "solve" this equation?

    Do we say a function definied on some open interval that satisfies this equation? (Because that is what the book seems to assume). But leads to major problems, observe.

    The way people procede is to divide by y obtaining,
    y'/y=x
    Thus two functions equal implies integrals equal thus,
    INT y'/y = INT x
    Using the substitution rule,
    1/2y^2=1/2x^2+C
    Thus,
    y^2=x^2+C
    Thus,
    y=+\-sqrt(x^2+C)

    Here is the obvious problem. We can only divide by y if we know the function is non-vanishing. If we assume that then the solution must take form,
    y=+\-sqrt(x^2+C)
    We note that C cannot be 0 that leads to |x| which is not everywhere differenciable. And if C<0 then it is differenciable on some finite open interval not the full line, so what doth we do? The only reasonable answer is C>0. If we check this we see it is a solution, good.

    But that is only for when y is non-vanishing.

    If y is vanishing at some point. We have two possibilities:
    1)y is always vanishing.
    2)y is non vanishing at some point, thus non vanishing in some neighborhood (property of continous functions).

    If we accept statement 1) then,
    y=0 is a solution.
    Also,
    y=0 on any finite or half-finite interval.

    If we accept statement 2) then,
    [y=+/-sqrt(C+x^2) on neighborhood otherwise y=0]
    But leads to problem since there is a transition from non-vanishing to vanishing there is a point where non-differnciabilitiy occurs thus that cannot be a solution.
    ---
    I hope you can see how difficult a seemingly simple differencial equation can be if we do not state an open interval.

    Thus, the question reamins, what does it mean to "solve" a differencial equation.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    An ordinary diff. eq. (ODE) is an equation of the form: f(x,y',y'',...y^(n)) = 0.
    Of course, y has to be n times differentiable on at least an open interval.

    The derivative of a function f is defined at a point, with the limit definition. If it exists, we call f differentiable at that point. If this holds for all points on an interval, we say f is differentiable on that interval. Since ODE's are 'built' with these derivatives, 'solving' a ODE is done, by definition, locally. However, we usually try to make the interval(s) I where the solution holds, as big as possible.

    Example: y' cos²x = 1 is given for all x in R, yet the solution y = tan(x) is only valid in all open intervals (pi/2+k*pi,pi/2+(k+1)pi), k in Z of course.

    In general, there's an infinite number of solutions to an ODE (of course, no solutions are possible as well!). Usually, we'll be interested in a particular solution, satisfying certain (initial and/or boundary) conditions. In this case, there are important theorems which give guarantees that locally, there exists a unique solution for a point in the plane; usually referred to as the "existence" and "uniqueness theorem". Note that these theorems gives a sufficient condition for a (unique) solution to exist, but they don't tell you how to find it! Similar theorems exist in other fields/applications as well.

    Although it's possible that no solution can be given in "closed form", as a finite composition of elementary functions etc, it may still exist. Solutions are often given in implicit form as well, i.e. as f(x,y) = 0 rather than y = f(x). The implicit function theorem deals with these kind of functions, and how they can be locally unique, possibly solved to one of the variables under certain conditions.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by TD! View Post
    An ordinary diff. eq. (ODE) is an equation of the form: f(x,y',y'',...y^(n)) = 0.
    Of course, y has to be n times differentiable on at least an open interval.

    The derivative of a function f is defined at a point, with the limit definition. If it exists, we call f differentiable at that point. If this holds for all points on an interval, we say f is differentiable on that interval. Since ODE's are 'built' with these derivatives, 'solving' a ODE is done, by definition, locally. However, we usually try to make the interval(s) I where the solution holds, as big as possible.

    Example: y' cos²x = 1 is given for all x in R, yet the solution y = tan(x) is only valid in all open intervals (pi/2+k*pi,pi/2+(k+1)pi), k in Z of course.

    In general, there's an infinite number of solutions to an ODE (of course, no solutions are possible as well!). Usually, we'll be interested in a particular solution, satisfying certain (initial and/or boundary) conditions. In this case, there are important theorems which give guarantees that locally, there exists a unique solution for a point in the plane; usually referred to as the "existence" and "uniqueness theorem". Note that these theorems gives a sufficient condition for a (unique) solution to exist, but they don't tell you how to find it! Similar theorems exist in other fields/applications as well.

    Although it's possible that no solution can be given in "closed form", as a finite composition of elementary functions etc, it may still exist. Solutions are often given in implicit form as well, i.e. as f(x,y) = 0 rather than y = f(x). The implicit function theorem deals with these kind of functions, and how they can be locally unique, possibly solved to one of the variables under certain conditions.
    But you do agree with me if a person says,
    y'=xy
    That is not good enough.
    He should specify the open interval?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    I would say it depends: if you're looking for a solution on on specific interval, yes. But generally, you're supposed to find a the general solution (and possibly the singular solution(s)) - whici is usually not unique. However, (implicitly) you would try to find it on a (union of) interval(s) which is as large as possible.

    As for your specific example, I didn't look into it at first but now you mention it again, I think your solution isn't correct. I'll show how it is usually done, agreeing though that it's mathematically not the nicest way to write it, since we're splitting the 'dy' and 'dx' which is more subtle than it seems.

    y' = xy
    dy/dx = xy
    dy/y = xdx
    ln|y| = x&#178;/2 + c'
    y = c.exp(x&#178;/2)

    So in this specific case, the solution holds on whole IR. You went wrong here:
    Quote Originally Posted by ThePerfectHacker
    INT y'/y = INT x
    Using the substitution rule,
    1/2y^2=1/2x^2+C
    In any case, you can always try solving the ODE and then afterwards, check on what interval your solution holds.

    Edit: if you dislike the y' = dy/dx and splitting the dy and dx, substitution is fine too:

    y' = xy
    y'/y = x
    (ln|y|)' = x
    ln|y| = INT x dx = x&#178;/2 + C
    y = c.exp(x&#178;/2)
    Last edited by TD!; October 14th 2006 at 03:28 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    I'm going to quickly butt in here with a question.

    Quote Originally Posted by TD! View Post
    In any case, you can always try solving the ODE and then afterwards, check on what interval your solution holds.
    I believe that for a linear DEq this always works. For non-linear DEqs I have heard that you can solve for a solution that looks general, but find out that other solutions exist for specific initial conditions. As I don't know much about solving non-linear DEqs I am wondering if there is another way to approach the domain question? Thanks!

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    I have no difficulty with linear diffencial equations because the definitions are rigours for me.

    I discussed it with my professor (I realized my solution mistake, TD!, but that was not my theme of my argument) and he said I am right. He should mention things like that in class but since it is 29 engineers against 1 mathematician he chose not too.

    For exmaple,
    y=e^(x^2/2) definied on (-1,1) only is not included in,
    Ce^(x^2/2) on whole line.

    And it even my be possible to definie piecewise function using that set and create a bigger mess.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. differencial equation
    Posted in the Calculus Forum
    Replies: 4
    Last Post: August 27th 2008, 12:43 PM
  2. differencial equation
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 8th 2008, 11:03 AM
  3. differencial equation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 5th 2008, 10:31 AM
  4. differencial equation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 27th 2007, 10:15 AM
  5. differencial equation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 20th 2007, 04:14 PM

Search Tags


/mathhelpforum @mathhelpforum