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The solution of ODE

  1. Dec 18, 2014 #1
    Is the solution of differential equation be unique always?
     
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  3. Dec 19, 2014 #2

    Simon Bridge

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  4. Dec 23, 2014 #3
    Nah, just because there is a solution doesn't always mean it's unique.
    Take something like:
    ##\frac{dy}{dx} = 3y^{2/3}##, where y(0) = 0.
    ##f(x,y)## is continuous about (0 , 0), but ##\frac{\partial f}{\partial y}## is not continuous about (0 , 0); this violates the conditions for ``uniqueness".
    So, you won't always get a unique solution. When you get a solution, you have to consider the fact that existence doesn't warrant uniqueness.
     
  5. Dec 24, 2014 #4

    HallsofIvy

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    First, the solution to a "differential equation" is never unique. Solving a differential equation is equivalent to integrating so there is always an arbitrary "constant of integration". There is an "existence and uniqueness" theorem for initial value problems: if the function f(x, y) is continuous in both variables and "Lipshitz" in y for some region about [itex](x_0, y_0)[/itex] then there exist a unique function, in some region of [itex](x_0, y_0)[/itex], satisfying dy/dx= f(x, y), with the "initial condition" [itex]y(x_0)= y_0[/itex].

    Generally, if the function is continuous but not Lipschitz, you may have existence without uniqueness. For example the initial value problem [itex]dy/dx= \sqrt{y}[/itex], y(0)= 0, has both [itex]y= x^2/4[/itex] and [itex]y= 0[/itex], for all x, as solutions.

    (A function of a single variable, f(x), is "Lipschitz" in set A of real numbers if and only if [itex]|f(x)- f(y)|\le C|x- y|[/itex] for some constant C, for x and y in A. A function, f(x, y), of two variables, is "Lipschitz in y" in set A of pairs of numbers, (x, y), if and only if [itex]f(x_0, y)[/itex] is Lipschitz in y for every [itex]x_0[/itex]. "lipschitz" lies between "continuous" and "differentiable"- every Lipshitz function is continuous but there exist continuous functions that are not Lispchitz. Similarly, every differentiable function is Lipshitz but there exist Lipschitz functions that are not differentiable. Some textbooks give the condition ""differentiable in y" rather than "Lipschitz". That is "sufficient" but not "necessary".)

    For boundary value problems, where we are given values of y at two or more points rather than values of y and its derivatives at a single point, there is no such theorem. Both "existence" and "uniqueness" depend on the equation and the boundary values.

    Obvious examples are:
    [itex]d^2y/dt^2= -y[/itex], [itex]y(0)= 0[/itex], [itex]y(\pi)= 1[/itex] has NO solution while
    [itex]d^2y/dt^2= -y[/itex], [itex]y(0)= 0[/itex], [itex]y(\pi)= 0[/itex] has infinitely many solutions.
     
    Last edited: Jan 1, 2015
  6. Dec 28, 2014 #5
    The solution to a differential equation is never unique, as others have pointed out, unless conditions are provided. The solution to a differential equation is always a "family" of solutions, that is, they are related in that they have the same derivative. And because it's always possible to add an arbitrary constant, this means there will be an infinite number of functions that have the same derivative.
     
  7. Dec 29, 2014 #6
    Thank you so much for all members.
     
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