Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is a differential equation still having a general solution even if the differential equation have a singular solution?

for example:

\begin{aligned} \frac {dy}{dx} = x y^{1/2} \end{aligned}

The Solution: \begin{aligned} y= \left(\frac {1}{4}x^2+c \right)^2 \end{aligned}

But also this singular solution (there is not a constant to obtain it from the above, but is a solution) \begin{aligned} y= 0 \end{aligned}

Is this function correct named as a general solution?: \begin{aligned} y= \left(\frac {1}{4}x^2+c \right)^2 \end{aligned}

share|improve this question
1  
And what then about, for example, $y=\cases{0&for $x\le 0$\\x^4/16&for $x\ge 0$}$? –  Henning Makholm Feb 27 '12 at 1:22
2  
The "general solution" to a differential equation is simply the set of all of its solutions. The set $S = \{y = (1/4 x^2 + c)^2 : c \text{ is a constant}\}$ is not the general solution to this differential equation, because $y = 0$ is a solution that is not in $S$. –  Tanner Swett Feb 27 '12 at 1:48

2 Answers 2

Your example has uniqueness problems because $x y^{1/2}$ is not locally Lipschitz as a function of $y$. But consider a differential equation $\frac{dy}{dx} = f(x,y)$ where $f(x,y)$ is continuous in a rectangle $(a,b) \times (c,d)$ and Lipschitz in the $y$ variable there. Then for any $x_0 \in (a,b)$, we can use $y(x_0)$ to parametrize all solutions with $y(x_0) \in (c,d)$. That is, for each $z \in (c,d)$, there is a unique solution $Y(x)$ with $Y(x_0) = z$, defined for $x$ in some maximal interval $(x_1, x_2)\subseteq (a,b)$ such that $Y(x) \in (c,d)$ for $x \in (x_1, x_2)$.

share|improve this answer

What you're hoping for is to name the "general solution" with a multi-parameter family, but that is not always possible for non-linear equations. The general solution of an ODE is the set of all solutions, parametrized by constants or not. For linear equations, since the solutions form an affine space (a linear space translated by some function), they can be parametrized by the form $y_p + c_1 y_1 + \dots + c_n y_n$ where $n$ is the order of the linear equation, $y_p$ a particular solution, $y_1, \dots, y_n$ linearly independent solutions of the homogeneous equation and $c_1, \dots, c_n$ the parameters which generate the general solution.

In non-linear equations you are not that lucky all the time. Sometimes solutions must be picked one by one and some other solutions can be "regrouped" in a parameter family (because you integrated somewhere).

Hope that helps,

share|improve this answer
    
Yes, helped alot, thank you, but as @Tanner L. Swett says, that's why im wondering about, but you say that the general solution of an ODE is the set of all solutions, parametrized by constants or not. How can be expresed that solution with linear equations? or there are no problems with linear equations? –  nEAnnam Feb 27 '12 at 2:09
    
I mean How can be expresed that solution if that was a linear equation? or there are not other solutions than the general solution within linear equations? –  nEAnnam Feb 27 '12 at 2:18
    
There are no problems with linear equations : in that case the general solution is just $\{ y_p + c_1 y_1 + \dots + c_n y_n \}$. In non-linear equations the set can have very various forms. –  Patrick Da Silva Feb 27 '12 at 2:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.