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A necessary condition for uniqueness

  1. Jun 17, 2005 #1


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    Hello guys. I've been looking at uniqueness requirements for the differential equation:

    [tex]\frac{dy}{dx}=f(x,y);\qquad y(a)=b[/tex]

    And the extension of this to higher-ordered equations.

    I'd like to understand the sufficient and necessary conditions for uniqueness.

    Most proofs require that f(x,y) and its partial with respect to y be continuous. However, Ince shows that continuity of f(x,y) is not a necessary condition for uniqueness. As I read his analysis, it appears that the one necessary condition for uniqueness is that f(x,y) MUST satisfy a Lipschitz condition (or a condition of a similar nature) about the initial point [itex]x_0[/itex]:

    If [itex](x,y_1)[/itex] and [itex](x,y_2) [/itex] be any points about some defined region containing [itex]x_0[/itex], then:


    for some constant K.

    It's still a bit unclear to me.

    May I ask this: If f(x,y) does NOT satisfy a Lipschitz condition in a neighborhood of [itex]x_0[/itex], then no unique solution exists for the differential equation and if a unique solution exists, then f(x,y) must necessarilly satisfy a Lipschitz condition?
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  3. Jun 17, 2005 #2


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    I'd like to look at this one first. I know it's easy. I'm not concerned in this post about "how to solve ODEs", but rather with issues regarding existence and uniqueness:

    [tex]\frac{dy}{dx}=\sqrt{y};\qquad y(0)=0[/tex]

    Solving yields:


    Substituting the initial conditions yields:


    But y(x)=0 is another solution and thus uniqueness is not obtained.

    Looking at the Lipschitz condition in the neighborhood of the origin with [itex]y_1=0[/itex]:

    [tex]|\sqrt{y_2}|\leq K|(y_2)|[/tex]


    [tex] y_2=\frac{1}{K^4}[/tex]

    We obtain:

    [tex]\frac{1}{K^2}\leq \frac{1}{K3} \quad\text{obsurd}[/tex]

    Thus the Lipschitz condition is not met. However, the partial of f(x,y) is unbounded at y=0. Generally in the proofs of existence,

    [tex]\frac{\partial f}{\partial y}\leq K[/tex]

    Thus a bounded partial seems crucial to the Lipschitz condition.
  4. Jun 17, 2005 #3


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    As I look into this, I'm starting to suspect "containment" perhaps is the essential pre-requisite for uniqueness. If that is so, why is it? A differential equation, in the form specified above (and its higher ordered analogs I suspect), appears to admit uniqueness only if the function f(x,y) and it's partial are bounded or "contained" and once that containment is lost, uniqueness seems to be lost. Or is that not the case? Can there be a differential equation, unbounded at a point in its domain which still admits a unique solution at that point?

    What is the relationship between uniqueness and containment? Is containment the necessary condition for uniqueness and why must it be so if so?
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