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On First Order Non-linear PDEs

  1. Sep 10, 2011 #1
    Let us consider the following partial differential equation:

    [tex]{(}\frac{\partial z}{\partial x}{)}^2{+}{(}\frac{\partial z}{\partial y}{)}^{2}{=}{1}[/tex] ---------- (1)
    The general solution[you will find in the texts: http://eqworld.ipmnet.ru/en/solutions/fpde/fpde3201.pdf is given by:
    [tex]{z}{=}{ax}{+}\sqrt{1-a^2}{y}{+}{B}[/tex]; a and B are constants.--------(2)
    Now let me search for a solution in the form:
    [tex]{z}{=}{ax}{f}{(}{z}{)}{+}\sqrt{1-a^2}{y}{+}{B}[/tex]


    [tex]\frac{\partial z}{\partial x}{=}{ax}{f'}{(}{z}{)}\frac{\partial z}{\partial x}{+}{a}{f}{(}{z}{)}[/tex]

    [tex]\frac{\partial z}{\partial y}{=}\sqrt{1-a^2}{+}{ax}{f'}{(}{z}{)}\frac{\partial z}{\partial y}[/tex]

    Substituting the above values into (1) we get
    [tex]\frac{{(}{a}{f}{(}{z}{)}{)}^{2}}{{(}{1}{-}{ax}{f'}{(}{z}{)}{)}^{2}}{+}\frac{1-a^2}{{(}{1-ax}{f'}{(}{z}{)}{)}^{2}}{=}{1}[/tex]
    [tex]{a}^{2}{(}{f}{(}{z}{)}{)}^{2}{+}{1}{-}{a}^{2}{=}{(}{1}{-}{ax}{f'}{(}{z}{)}{)}^{2}[/tex]

    If the above differential equation is solvable we should get a broader range of general values.Are we missing some solutions if the conventional general solution is considered?
    Now , the above equation contains x.
    We write,
    z=F(x,y)
    For any particular value of z say, z=k, we have,
    k=F(x,y)
    Keeping x fixed at some arbitrary value we can obtain k be changing y only[I believe that this may be possible in most situations or in many situations]. [We could have achieved the same effect,i.e, getting z=k, by changing both x and y in some manner]
    So we may try to solve our differential equation[the last one] by keeping x fixed at some arbitrary value. This will go in favor of my suggestion.

    Suppose we have only one local maximum for our function z=F(x,y) at
    the point (x0,y0) . For this point,
    k=F(x0,y0)
    The value of k in this situation, will not be accessible for any arbitrary x. One has to use x=x0. To surmount this difficulty one may think of dividing the the domain of definition of the function F,into sub-domains so that in any particular sub-domain, the value of z may be accessible for an arbitrary x in that sub-domain.
    [Rather we would look for a function,z=F(x,y), of this type]

    On Boundary Conditions


    The general solution[relation (2)]seems to be too restrictive with simple boundary conditions. We may consider a square domain: x=0,x=k,y=0,y=k [k: some constant]
    Let us take the line:x=k
    It is perpendicular to the x-axis
    If we use the value x=k in (2) ,then z changes linearly wrt y

    The general solution[conventional one] talks of plane surfaces given by (2). I can always take small pieces of such surfaces and sew them into a large curved surface ,z=F(x,y).Along the boundary, z may be a non-linear function of x or y.[One may consider the square:x=0,y=0,x=k,y=k]

    This can change the whole picture of the problem.

    [Equations of the form f(p,q)=0 do not have singular solutions."p" and "q" are the partial derivatives wrt to x and y respectively]
     
    Last edited by a moderator: Oct 8, 2011
  2. jcsd
  3. Sep 11, 2011 #2
    I'm pretty sure that's not the general solution but rather only a "special solution" which means we can't find the general solution but can find an expression representing a particular class of solutions. Often in non-linear equations, the general solution can't or is not easily found.
     
  4. Oct 7, 2011 #3

    JDoolin

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    If this helps, I plugged the given solution into mathematica
    [tex]{z}{=}{ax}{+}\sqrt{1-a^2}{y}{+}{B}[/tex]

    letting B = 0, while a changes from 0 to 1.

    attachment.gif

    Would there be any solutions for the equation that aren't planar?
     

    Attached Files:

  5. Oct 7, 2011 #4
    You may consider infinitesimal portions of several moving surfaces["a" changing at different adjustable rates from zero to one or in some different way--say from one to zero] at different instants of time and sew them into a new surface which is not a plane.Each infinitesimal portion should satisfy the differential equation.

    [Guest audience will get the picture of a moving surface in the attachment of the last post]
     
    Last edited: Oct 7, 2011
  6. Oct 8, 2011 #5

    JDoolin

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    Okay; what I am picturing now is a ball rolling around on the flat surface above as it shifts back and forth. At any given instant, the ball is rolling on a flat surface.

    More generally, just have a flat screen with an animation of a moving dot. Then shift the flat surface back and forth as the dot moves around the screen, (and raise it or lowering by changing your parameter B so that you are always rotating the surface around a fulcrum at the location of the dot) keeping track of the motion of the dot in three dimensional space.

    So then you can look at the path of the dot in three dimensional space and ask yourself if the path actually satisfies the original differential equation. I would say definitely not.

    So you lower your criteria and ask whether each "infinitesimal portion" of the path satisfies the differential equation. To which I have to ask, what do you mean by infiniitesimal?

    Do you mean (1) very very small? If you do, all I need to do is jerk the platform fast enough that the differential equation is not satisfied in your very very small infinitesimal portion.

    Or perhaps infinitesimal just means (2) "small enough an interval that the jerking of the platform will be almost unnoticeable."

    Or a third possibility of what you mean by infinitesimal, is to (3) treat the infinitesimal as a hyper-real number (see http://en.wikipedia.org/wiki/Non-standard_analysis" [Broken].) Once we go down to this level, perhaps we could say the jerking of the platform would be entirely gone. But I disagree with doing it this way; once we get into hyper-real numbers and "true infinitesimal paths", I don't think we're dealing with the differential equation. We're just committing a division-by-zero error on the left-hand-side, and making a meaningless claim about linearity on the right-hand-side.

    I think the best you can do is use definition (2), but with that word "almost" in there, you're not really satisfying the differential equation with a curved path. You're almost satisfying the differential equation because a the curved path is almost straight, over a small enough interval.
     
    Last edited by a moderator: May 5, 2017
  7. Oct 8, 2011 #6

    JDoolin

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    On a related topic,

    On pages 7-8 of http://www.blatword.co.uk/space-time/Carrol_GR_lectures.pdf" [Broken]

    There is a very confusing paragraph

    To me this whole paragraph appears to be utter nonsense. The idea of having a vector at a point I can kind of acknowledge if you are discussing forces and vector fields, although one still needs a continuous space around the point for any field or force to really cause any acceleration.

    On the other hand, Carroll's equation (1.23) [itex]A = A^\mu \hat u_{(\mu)}[/itex], uses a unit vector, which does not exist at a point in spacetime, but requires at least a differential displacement to define, unless he's thinking of employing non-standard analysis, and treating "infinitesimal" and "infinity" as though they have the same mathematical integrity as real numbers.


    Carroll is claiming that the idea of a displacement vector "is not a useful concept in relativity." I have to disagree with Carroll.

    There are times when you need to think about displacement vectors. The fact that Force vectors defined at a point is a useful concept does not mean that displacement vectors, defined over a finite space is no longer a useful concept.
     
    Last edited by a moderator: May 5, 2017
  8. Oct 9, 2011 #7

    We consider a ball[point] rolling over the moving flat surface which at any instant of time is a solution of the PDE given in Post#1[equation 1]

    Let the trajectory on the x-y plane[projected trajectory] be represented by:
    x=x(t); y=y(t)

    The motion is represented by

    z(x,y,t)=a(t)x+sqrt[1-[a(t)]^2]y+B(t)

    Partial differentiating z wrt to x and y we get the same PDe given in post #1[Equation 1]

    When we are partial differentiating z[ wrt x and y] at a particular location/point [x,y,t] we
    are not considering any explicit dependence of z on t .
    [x=x(t);y=y(t) and z=z(x,y,t) are assumed to be continuous, differentiable functions]
     
    Last edited by a moderator: May 5, 2017
  9. Oct 9, 2011 #8

    JDoolin

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    I went back to a differential equations text to help jog my memory on this;

    What we've been focusing on is the "particular" solution for the equation; not the "general" solution. The general solution to a differential equation is equal to the general solution of the homogenous differential equation plus the particular solution.

    The homogeneous equation, in this case is,

    [tex]\left (\frac{\partial z}{\partial x} \right )^2+\left (\frac{\partial z}{\partial y} \right )^2 = 0[/tex]

    and its solution is:

    [tex]z(x,y)=\sum_{n}e^{nx/ \lambda}e^{i n y / \lambda}[/tex]

    You can swap out the real-valued factor with sinh(___)+cosh(___), and swap out the imaginary-valued factor with sin(___)+cos(___) as needed, to get a real-valued solution.

    The general solution and the particular solution have to be added together

    If you are changing a and B on the fly, you aren't satisfying the differential equation. If a and B are functions of t and z is a function of a and B, I would say, to the contrary, that we are considering explicit dependence of z on t.
     
  10. Oct 9, 2011 #9
    Z is a function of t. There is no doubt about it
    But during the process of partial differentiation wrt x or y we take it[ie,t] as a constant.

    The relevant portion has been quoted below.
     
    Last edited: Oct 9, 2011
  11. Oct 9, 2011 #10
    Complete Integral:
    [tex]{z}{=}\Sigma{e}^{{nx}{/}{\lambda}}{e}^{{iny}{/}{\lambda}}{+}{.5}{x}{+}\sqrt{.75}{y}[/tex]
    [tex]\frac{\partial z}{\partial x}{=}\Sigma\frac{n}{\lambda}{e}^{{nx}{/}{\lambda}}{e}^{{iny}{/}{\lambda}}{+}{.5}[/tex]
    [tex]\frac{\partial z}{\partial y}{=}\Sigma\frac{in}{\lambda}{e}^{{nx}{/}{\lambda}}{e}^{{iny}{/}{\lambda}}{+}\sqrt{.75}[/tex]
    Therefore,
    [tex]{(}\frac{\partial z}{\partial x}{)}^{2}{+}{(}\frac{\partial z}{\partial y}{)}^{2}[/tex]
    [tex]{=} {(}\Sigma\frac{n}{\lambda}{e}^{{nx}{/}{\lambda}}{e}^{{iny}{/}
    {\lambda}}{+}{.5}{)}^{2}[/tex]
    [tex]{+} {(}\Sigma\frac{in}{\lambda}{e}^{{nx}{/}{\lambda}}{e}^{{iny}{/}{\lambda}}{+}\sqrt{.75}{)}^{2}[/tex]
    [tex] {=}{(}\Sigma\frac{n}{\lambda}{e}^{{nx}{/}{\lambda}}{e}^{{iny}{/}{\lambda}}{)}^{2}[/tex]
    [tex]{+} {(}\Sigma\frac{in}{\lambda}{e}^{{nx}{/}{\lambda}}{e}^{{iny}{/}{\lambda}}{)}^{2}{+}{.25}{+}{.75}[/tex]
    [tex]{+}{(}\Sigma\frac{n}{\lambda}{e}^{{nx}{/}{\lambda}}{e}^{{iny}{/}{\lambda}}{)}[/tex]
    [tex]{+}{2}\sqrt{.75}{(}\Sigma\frac{in}{\lambda}{e}^{{nx}{/}{\lambda}}{e}^{{iny}{/}{\lambda}}{)}[/tex]
    Now,
    [tex] {(}\Sigma\frac{n}{\lambda}{e}^{{nx}{/}{\lambda}}{e}^{{iny}{/}{\lambda}}{)}^{2}{+}{(}\Sigma\frac{in}{\lambda}{e}^{{nx}{/}{\lambda}}{e}^{{iny}{/}{\lambda}}{)}^{2}{=}{0}[/tex]
    [Just think of the Complementary function:#8.You may take i^2 outside the second summation sign in the above step]
    Therefore,
    [tex]{(}\frac{\partial z}{\partial x}{)}^{2}{+}{(}\frac{\partial z}{\partial y}{)}^{2}[/tex]
    Works out to 1+something
    [The non-linear nature of the PDE seems to upset the rule: Complete Integral= Complementary function + Particular Integral]
     
    Last edited: Oct 10, 2011
  12. Oct 10, 2011 #11

    JDoolin

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    Yep, I goofed. I gave the solution to



    [tex]\frac{\partial^2 z}{\partial x^2} + \frac{\partial ^2 z}{\partial y^2}= 0[/tex]


    rather than

    [tex]\left (\frac{\partial z}{\partial x} \right )^2 + \left (\frac{\partial z}{\partial y} \right )^2= 0[/tex]
     
  13. Oct 10, 2011 #12

    JDoolin

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    What if we get rid of the summation though and just try

    [tex]\begin{matrix} z=e^{i k x} e^{k y}\\ \left (\frac{\partial z}{\partial x} \right )^2 = -k^2 z\\ \left (\frac{\partial z}{\partial y} \right )^2 = +k^2 z\\ \left (\frac{\partial z}{\partial x} \right )^2 + \left (\frac{\partial z}{\partial y} \right )^2=0 \end{matrix}[/tex]

    I think it would work if you just use any individual wave-number. The summation would give difficulty though.
     
  14. Oct 11, 2011 #13

    JDoolin

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    Just for fun, I plotted a couple of the homogeneous solutions.

    attachment.php?attachmentid=39895&d=1318370732.png

    They are pretty.
     

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  15. Oct 11, 2011 #14
    What about the complete integral of the PDE:

    [tex]{(}\frac{\partial z}{\partial x}{)}^{2}{+}{(}\frac{\partial z}{\partial y}{)}^{2}{=}{1}[/tex]
    The rule[suggested in #8]: Complete Integral=Complementary Function+Particular Integral
    is not working!
    You may take the PI as [tex]{0.5}{x}{+}\sqrt{.75}{y}[/tex]

    [The solution you have suggested pertains to the homogeneous part only]
    [My comments are in consideration of the typos in the second and the third relations of #12: z should be replaced by z^2 on RHS]
     
    Last edited: Oct 11, 2011
  16. Oct 11, 2011 #15
    [tex]z(x,y)=\sum_{n}e^{nx/ \lambda}e^{i n y / \lambda}[/tex] is definitely a solution to the homogeneous part of the first order ,non-linear PDE
     
  17. Oct 11, 2011 #16

    JDoolin

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    Good point. Good point.

    Maybe that rule about General Solution = homogeneous solution + particular solution only applies to second order linear differential equations.

    I'm slowly coming around to understand the complexity of the question. My first inclination was to believe that those flat planes were the only solutions. But seeing that the homogeneous equation has other non-flat solutions, I would suspect that there are surely other solutions for the nonhomogeneous equation as well.

    As for whether these can be solved analytically, I have no idea, but of course that would be more satisfying if they could.

    But out of curiosity, have you already used numerical methods to solve this?

    (If so, how would you go about solving a 3D differential equation numerically? My differential equations skills, if you can't tell already, are rusty (or perhaps absent entirely). But to start a numerical solution in three dimensions, what do you need to start with as boundary values? Is it enough to define with three or four (x,y,z) points? )

    If not, it might be interesting to try, just so we could see what such a solution would look like; even if we don't have a closed-form equation for it.
     
    Last edited: Oct 11, 2011
  18. Oct 11, 2011 #17
    I have not tried it out for this case[But I would definitely do it]. I have tried numerical integration elsewhere in relation to simple integrations.
    https://www.physicsforums.com/showpost.php?p=3547857&postcount=2
    In the above link I have calculated some constants by writing programs in C. Then I verified the results by using a software.They are in good agreement.
     
  19. Oct 13, 2011 #18
    Information for the Audience:
    This is to draw the attention of the audience towards the editing on post #1 of this thread. The caption reads--- "Reason:removed spurious".Incidentally Astronuc had removed a third bracket so that the Latex code could be properly interpreted.I was informed of this when I requested information from him.

    But the word "Spurious " could have had a negative impact[or misleading impact] on the audience-- viewers could have interpreted that the author had posted something of false nature. It was the case of an extra third bracket that caused the problem--it had to be removed.

    This is a clarification from my side so that I am not misunderstood by audience. Incidentally the thread has already gathered more than a thousand views.
     
    Last edited: Oct 13, 2011
  20. Oct 14, 2011 #19
    PDE:
    [tex]{(}\frac{\partial z}{\partial x}{)}^{2}{+}{(}\frac{\partial z}{\partial y}{)}^{2}{=}{1}[/tex]
    We look for a solution of the form:
    [tex]{z}{=}{a}{(}{x,y}{)}{x}{+}\sqrt{1-a{(}{x,y}{)}^{2}}{y}{+}{B}[/tex]
    [tex]\frac{\partial z}{\partial x}{=}\frac{\partial a}{\partial x}{(}{x}{+}{y}\frac{a}{\sqrt{1-a^2}}{)}{+}{a}[/tex]
    [tex]\frac{\partial z}{\partial y}{=}\frac{\partial a}{\partial y}{(}{x}{+}{y}\frac{a}{\sqrt{1-a^2}}{)}{+}\sqrt{1-a^2}[/tex]
    [tex]{(}\frac{\partial z}{\partial x}{)}^{2}{+}{(}\frac{\partial z}{\partial y}{)}^{2}{=}{1}{+}{(}\frac{\partial a}{\partial x}{)}^{2}{(}{x}{+}{y}\frac{a}{\sqrt{1-a^2}}{)}^{2}{+}{(}\frac{\partial a}{\partial y}{)}^{2}{(}{x}{+}{y}\frac{a}{\sqrt{1-a^2}}{)}^{2}[/tex]
    [tex]{+}{2a}\frac{\partial a}{\partial x}{(}{x}{+}{y}\frac{a}{\sqrt{1-a^2}}{)} {+}{2}\sqrt{1-a^2}\frac{\partial a}{\partial y}{(}{x}{+}{y}\frac{a}{\sqrt{1-a^2}}{)} [/tex]
    The original Equation is satisfied if,
    [tex]{(}\frac{\partial a}{\partial x}{)}^{2}{(}{x}{+}{y}\frac{a}{\sqrt{1-a^2}}{)}^{2}{+}{(}\frac{\partial a}{\partial y}{)}^{2}{(}{x}{+}{y}\frac{a}{\sqrt{1-a^2}}{)}^{2}[/tex]
    [tex]{+}{2a}\frac{\partial a}{\partial x}{(}{x}{+}{y}\frac{a}{\sqrt{1-a^2}}{)} {+}{2}\sqrt{1-a^2}\frac{\partial a}{\partial y}{(}{x}{+}{y}\frac{a}{\sqrt{1-a^2}}{)} {=}{0}[/tex]
    This is a closer analytical representation[ rather the analytical background] of sewing/stitching small , flat surfaces into a large curved one. The solution will be more liberal towards boundary conditions.
     
    Last edited: Oct 14, 2011
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