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**Is the solution of differential equation be unique always?**

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- Thread starter moh salem
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Simon Bridge

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No.

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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

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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.

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

For

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.

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Is the solution of differential equation be unique always?

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.

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Thank you so much for all members.

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