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Exact differential equations

  1. Dec 21, 2006 #1
    Hello!

    From what I have read, a differential equation is exact if d^2f/dxdy = d^2f/dydx, is this correct? My professor has given us examples of dif. equations that are not exact. However I remember from last semester that Schwartz theorem states that the order in which you differentiate does not matter, the result is always the same. So d^2f/dxdy should always be equal to d^2f/dydx, so all diferential equations should be exact. This is obviously not right, but I don't understand why, can someone please explain?

    Thank you!
     
  2. jcsd
  3. Dec 21, 2006 #2
    the test for whether or not a differential equation is exact is to differetiate the term with the dx in it, with respect to y, and the differential equation with the dy term with respect to x

    in other words if you have an equation of the form

    Mdx+Ndy=0

    then you differentiate M with respect to y and N with respect to x.

    this is actually an extention of the idea of curl, if the curl of a vector field is 0 than its integrable. the derivative test is merely a simplified curl calculation
     
  4. Dec 21, 2006 #3
    But the term in dx (M) is already the first derivative of the function with respect to x. So differentiating M with respect to y is the second derivative d^2f/dxdy . Is this not correct?
     
  5. Dec 22, 2006 #4

    HallsofIvy

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    Your statement of the definition of "exact differential equation" is incorrect since you are ASSUMING the existance of f.

    A differential equation of the form g(x,y)dx+ h(x,y)dy= 0 (and any first order differential equation can be put in that form) is "exact" if and only if
    [tex]\frac{\partial g}{\partial y}= \frac{\partial h}{\partial x}[/tex]

    If there exist a function f(x,y) such that [itex]\frac{\partial f}{\partial x}= g(x,y)[/itex] and [itex]\frac{\partial f}{\partial y}= h[/itex], then those two partial derivatives of g and h are the "mixed second derivatives" of f and so must be equal (as long as they are continuous).

    Conversely, if those partial derivatives of g and h are equal, then there must exist a function f(x,y) having g and h as its partial derivatives. In that case, g(x,y)dx+ h(x,y)dy is the "total differential" of f and so the equation g(x,y)dx+ h(x,y)dy= 0 is the same as df= 0 and f(x,y)= constant.

    For example, if (2x- y)dx+ (y2- x)dy= 0, we have g= 2x- y and h(x,y)= y2- x. Then gy= -1= hx and so we know the equation is "exact"- the left hand side is an exact differential of some function f(x,y). We must then have fx[/sup]= 2x- y and intgrating, while treating y as a constant (since partial derivative treats y as a constant), f(x,y)= x2- xy+ p(y) (we are treating y as a constant so the "constant of integration" may depend on y). Differentiating that with respect to y give fy= -x+ p'(y)= y2- x. The "-x" terms cancel (they HAD to do that since p depends only on y and it is the fact that gy= hx that guarenties that) and we have p'(y)= y2 so p= (1/3)y3+ C (Since p was a function of y only, that "C" really is a constant). Putting that into f(x,y)= x2- xy+ p(y) we have f(x,y)= x2- xy+ (1/3)y3+ C. The equation df= 0 says that f is a constant and so, incorporating the "C" into the constant on the right side, f(x,y)= x2- xy+ (1/3)y3= Constant is the general solution to the differential equation.

    On the other hand, if (2x- y)dx+ (y2)dy= 0, g(x,y)= 2x-y, h(x,y)= y2+ x. gy= -1 but hx= +1 so this equation is NOT exact. There does NOT exist a function "f(x,y)" having those partial derivatives so the equality of mixed second derivitives does not apply.
     
  6. Dec 22, 2006 #5
    Ok, I think I almost understand.
    Your explanation brings up another question though. We learned how to calculate differential equations that are not exact, but if there does not exist a function f(x,y) with those partial derivatives, then how can you calculate it?
     
  7. Dec 22, 2006 #6

    mathwonk

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    boy this browser is touchy today

    i think you meant how can you solve a non exact diff eq.

    sometimes you can find an "integrating factor" i.e. an auxiliary function which makes the non exact equation become exact after multilying by it.


    and a little remark of interest to global geomjetrs. the condition hurkyl gave is necessary for finding a function f with the given partials, and LOCALLY sufficient, but not globally so. you must also check the "periods" are zero.

    i.e. the necessary and sufficient condition is that the integral of your differential around every loop is zero. there are of course infinitely many loops in any region, so this is impractical to verify. however the theorem says that if the necessary condition above is satisfied, then in fact inmtegration is a deformation invariant, and even a cobordism invariant.

    i.e. if ∂N/∂x = ∂M/∂y or whatever, then two loops that togetehr for the boundary of a piece of the region, have the same integral.. so you only need to verify the periods are zero when taken around the boundary curves of your region, for the differential to be exact.

    for example the region outside the unit circle has the unit circle as boundaryt curve. not the differential (ydx - xdy)/(x^2+y^2), does satisfy the necessary condition, but is not eaxact because it ahs integral 2pi around the unit circle.

    it is said to be closed but not truly exact. so in elementary diff eq course theya ctullay use the wrong terminology and do a slightly poor job of rpeparing us for later courses in diff top.

    merry xmas.
     
  8. Dec 22, 2006 #7

    mathwonk

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    4th try

    i think you meant how can you solve a non exact diff eq.

    sometimes you can find an "integrating factor" i.e. an auxiliary function which makes the non exact equation become exact after multiplying by it.


    and a little remark of interest to global geometers. the condition hurkyl gave is necessary for finding a function f with the given partials, and LOCALLY sufficient, but not globally so. you must also check the "periods" are zero.

    i.e. the necessary and sufficient condition is that the integral of your differential around every loop is zero. there are of course infinitely many loops in any region, so this is impractical to verify. however the theorem says that if the necessary condition above is satisfied, then in fact integration is a deformation invariant, and even a cobordism invariant.

    i.e. if ∂N/∂x = ∂N/∂y or whatever, then two loops that together form the boundary of a piece of the region, have the same integral.. so you only need to verify the periods are zero when taken around the boundary curves of your region, for the differential to be exact.

    for example the region outside the unit circle has the unit circle as boundary curve. note the differential (ydx - xdy)/(x^2+y^2), does satisfy the necessary condition, but is not exact because it has integral 2pi around the unit circle.

    it is said to be closed but not truly exact. so in elementary diff eq course they actually use the wrong terminology and do a slightly deceptive job of preparing us for later courses in diff top.

    merry xmas.
     
  9. Dec 22, 2006 #8
    I know how to solve them, but I don't understand why they are solvable in the first place. The differencial is composed of the partial derivatives of a function, right? From HallsofIvy's post:
    "There does NOT exist a function "f(x,y)" having those partial derivatives so the equality of mixed second derivitives does not apply."
    This is what I don't understand. When we are trying to solve a non exact dif eq (or any dif eq), we are trying to find the function which has that differencial, is this right? If no such function exists (because the mixed second derivatives are not equal), then how can we find it? ( regardless of the method )
     
  10. Dec 22, 2006 #9
    Also, something else I don't understand closely related to this ( I know these are very basic abstract conceptual questions that are difficult to explain, so I'm sorry)

    We usually solve differentials like these: (something in x and y)dx + (something in x and y)dy = Q(x) and we get an answer like (something in x and y) = c
    My question is, both equations (the dif eq, and the solution) seem to have only one independent variable , so then how come there exist partial derivatives? Should there not exist only one derivative? If so, then how come the differential exists, since it is made up of the partial derivatives of the original function?
     
    Last edited: Dec 22, 2006
  11. Dec 23, 2006 #10

    HallsofIvy

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    You seem to be confusing two different functions. If I have a differential equation of the form dy/dx= g(x,y) or g(x,y)dx+ h(x,y)dy= 0 (of course, either of those can be put in the other form) then I am looking for a function y(x) that satisfies that equation.

    If a differential equation of the form g(x,y)dx+ h(x,y)dy= 0 is exact then there exist a function f(x,y) such that df= g(x,y)dx+ h(x,y)dy. That f is NOT the y(x) that satisfies the equation. IF the equation is exact then you can find the f(x,y) and so y is implicitely defined by the equation f(x,y)= 0. We can, sometimes, solve that for y but most often we leave the solution in the form f(x,y)= 0. But still, it is y(x) that is the solution to the differential equation, not f(x,y).

    If the f(x,y) does not exist, then there are other ways to solve for y(x). In particular, one can show that there always exists an "integrating factor", a function u(x,y) such that equation we get by multiplying g(x,y)dx+ h(x,y)dy= 0 by it, u(x,y)g(x,y)dx+ u(x,y)h(x,y)dy= 0, is exact. Of course, the "f(x,y)" we get from this new equation has nothing to do with the original equation except that the function y(x) define implicitely by it satisfies the orginal equation.
     
  12. Dec 29, 2006 #11

    Thank you for your effort but there are still some things that do not make sense in my mind about the whole concept.

    First: With a differential equation of the sort: form dy/dx= g(x,y) or g(x,y)dx+ h(x,y)dy= 0
    dy/dx is the derivative of y, right? And the fact that it is not a partial derivative implies that y is a function of x, a one variable function. How can that then be equal to a function of two variables x and y, g(x,y)?
    Also, if there is a term in dy, does this not imply that the original function has been differentiated with respect to y? How can a function y(x) then satisfy the equation, if y has been differentiated with respect to itself? How is this possible?

    Regarding your second paragraph:
    When you say "IF the equation is exact then you can find the f(x,y) and so y is implicitely defined by the equation f(x,y)= 0." Does this not imply that f is a function of one variable, and so how can there be two partial derivatives that give rise to the differential equation with terms in dx and dy?
     
    Last edited: Dec 29, 2006
  13. Dec 29, 2006 #12

    HallsofIvy

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    Because y is a function of x. Replace the y in g(x,y) by y as a function of x. For example, dy/dx= y+ x has y= 2ex-x+1 as a solution. dy/dx= y+ x= ex+1 is a function of x.

    ??? What do you mean by "a term in dy"? dy/dx= 1 (which has y= x+ C as general solution) has "a term in dy" but that's because y is being differentiated not because anything is differentiated with respect to y. Now the form f(x,y)dx+ g(x,y)dy= 0 is often referred to as the "symmetric form" because you can think of y as a function of x or think of x as a function of y.

    Okay, yes, for a given function y(x) you could replace y in f(x,y) with that function (exactly like I said above). However, when you are solving the equation, you don't yet know y as a function of x. So it's better to think of this f as a function of two independent variables, x and y. After you have found f(x,y), setting it equal to a constant (since df= 0) gives an equation you can solve for y as a function of x (or x as a function of y).
     
    Last edited: Dec 30, 2006
  14. Dec 30, 2006 #13


    Okay, that cleared some doubts..
    Now, if y is a function of x, why not have everything as a function of x, just y(x)? Why do we need g(x,y), why not just g(y) since y is already a function of x?
    Also, if it's all a function of one variable, why can we not just integrate it straight away to get the solution? Why can we only do that with exact differential equations?
    When I say a term in dy, I mean the following: When you have a two variable equation, the differential is the partial derivatives of f each multiplied by dx, dy, etc, right? That then leads to the differential equation where we have to find the original equation from which we calculated the differential, is this not what we do with differential equations? This is one of my main doubts. If that's the case, then the original function is a two variable function, so then how can there be a derivative of y, dy/dx? Should it not be a partial derivative, since in the original equation z=f(x,y) we can solve for y, and it has two partial derivatives, one with respect to x, and one with respect to z. It's the dx's and the number of variables and the actual task we are performing that confuses me. I have learnt how to solve the exercices, but I should be able to deduce the method logically and I can not because I don't understand what is being done:frown:
     
  15. Dec 30, 2006 #14
    Also, for example, sometimes my professor has an equation (...)dx + (...)dy and he substitutes y for something. He then calculates the differential of y like: dy=(...)dx + (...)du and substitutes for the dy in the original equation.
    Are these two dy's not two different things? I thought the dy in the first equation was an infinitesimal distance but in the second it is the differential which can be as large as we want? How can we replace one with the other, how are they the same?
     
  16. Dec 30, 2006 #15

    mathwonk

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    gosh i just lost my whole post again. when will ilearn.?
     
  17. Dec 30, 2006 #16

    mathwonk

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    exact differentials

    i will, start over in a copmpletely different vein. the point is to use geometry.


    let a curve be given in the plane. then differentiation means finding the family of tangent lines to this curve.

    the opposite problem, a differential equaton, means given a family of lines, one through each point in the plane, to find a curve such that at each point of the curve, the line through that point is tangent to the curve.

    this problem is always solvable if the family of lienbs moves nicely, certainly smoothloy, and i think con tinuously.

    one dimension up however it is not as easy. consider a smootha urface in three space, and ask to calculate the family of tabngent planes to this surface. thi is the problem of differenbtiation.

    the inverse problem of a differential equation, means given a family of planes in three space, one through each point, to try to find a surface such that ate ach point of the surface the plane through that point is tangent to the surface.

    this si not always possible. indeed it is possible if and only if the family of planes defiens a exact differential.

    i.e. suppose our family of planes through points (x,y,z) were define by just the two direction vectors v and w, where v(x,y) and w(x,y) depend on only x andy. i.e. everywhere over the point (x,y) we put the plane with direction vectors v(x,y) and w(x,y).

    then we lok for a surface lying over th x,y plane, i.e. the graph of a function z = f(x,y), such that for everty x,y the tangehnt plaen to th surface z = f(x,y) is given at (x,y,f(x,y)), by the direction vectors v(x,y) and w(x,y).

    this means the partials of f must be the slopes defiend by v and w.

    well i am losing concentration. but one can study this question by looking a various lines in the plane, and trying to solve the equation of only producing curves over these liens.

    this emans looking at y or x asa fucntion of the other. but it may not be possible to put these curves togetehr into a surface.

    the point is, one stduies probklems in several dimensionsk, i.e. z = f(x,y) by setting something equal to a constant, like f(x,y) = 0, and then solving for y to get y = g(x), and then one can pretend y is a fucntion of x, instead of both x and y being fucntion of z.

    i apologize, i have to go, and cannot polish this up now.
     
  18. Dec 30, 2006 #17

    HallsofIvy

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    Unless you are willing to do a lot of really hard work with abstract logic, do NOT think of dy as an "infinitesmal distance". dy is a differential- that's all! We can replace one dy by another because they ARE the same thing.
     
  19. Jan 1, 2007 #18
    The point of solving a differential equation is ALWAYS to find the original function that satisfies the equation, is this correct? I'm still having difficulties understanding when that original equation is a one variable, or two variable function.
    For example:

    In the case (...)dx + (...)dy = Q(x) or 0, then is the original equation a one or two variable function? If it is a one variable function, then how come there are terms in dx and dy? If the original equation was for example x^2+2y=0, then the term in dx would be 2x and the term in dy would be 2? What is the point of this? Why is the original equation not in the form y=y(x) (y=(-x^2)/2) and we just integrate the differential equation? Since we can do the derivative of any function of one variable, we can likewise integrate that function to get the original, why not do this? So if we had 2x dx + 2dy = 0, why not do y' = -x (which is what we get when we change the form of the 2x dx + 2dy =0) and simply integrate that (which immediately gives the original equation)? Why can't we do this for any first order differential equation?
     
    Last edited: Jan 1, 2007
  20. Jan 1, 2007 #19

    mathwonk

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    suppose we have an equation like y = sqrt(1-x^2). does that look like an equation in one variable? so that dy/dx makes sense?

    what if we write it as y^2 = 1-x^2, and then as x^2 + y^2 = 1. then we can set f(x,y) = z, where f(x,y) = x^2 + y^2, and set z = 1.

    then it looks like a function of two variables, but with the value set equal to 1.

    the point is that every equation can be looked at in many ways.

    a plane curve with equation y = (plus or minus)sqrt(1-x^2), there the graph of two functions, each of one variable,

    can also be looked at as the image of a parametrized curve, where x(t) = cos(t), and y(t) = sin(t).

    or it is the locus where z = 1, of the function f(x,y) = x^2 + y^2. in this light it falls into a family of "parallel" curves with equations f= c, for various c.

    each way of representing it has advantages. It also gives us various ways to look at the derivative. we can ask for dy/dx as z is kepot = 1. i.e. as we move along the curve x^2+y^2=1 we can ask how the x and y coordinates are relkated.

    or we can ask for the "normal" vector perpendicular to that curve, which has coordinates <curlydz/curlydx, curlydz/curlydy>.

    then there are relation between those different derivatives, like dividing the coordinates of the normal vector to get (plus or minus) the derivative dy/dx.
     
  21. Jan 1, 2007 #20

    mathwonk

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    you can also look at the graph of the function z = x^2+y^2 in three space, where the original curve appears as a slice by the plane z=1.

    it is a rich game, and they are playing with it.


    there is a curve, and when they look at dy/dx they are moving along the curve. when they look at dz/dx znd dz/dy, they are actually moving the whole curve parallel to itself.
     
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