An ordinary Differential Equation which has no solution?

Can we think of an Ordinary Differential Equation which has no solution?

How can we think of this ODE?

Like what $f$ should I use such that $\frac{dy}{dx} = f(x,y)$ with $y(x_{0}) = f(x_{0},y_{0})$.

From existence theorem I think I have to use a $f$ which is not Lipschitz continuous?

• – mattos May 30 '18 at 4:28

The Lipschitz condition is needed for uniqueness, not existence. As long as $f(x,y)$ is continuous in a neighbourhood of $(x_0, y_0)$ there is a solution.
For a simple example of non-existence, consider the d.e. $$\dfrac{dy}{dx} = \cases{1 & if xy < 0\cr -1 & if xy \ge 0\cr}$$ with initial condition $y(0)=0$.
• Nice, but I was thinking why a product of $x$ and $y$ was taken while defining $f$, cannot I define $\frac{dy}{dx} = 1$ if $x <0$ and $-1$ if $x \geq 0$? – BAYMAX May 30 '18 at 5:06
• You could do that too, but $y = -|x|$ would be a solution everywhere except at $x=0$ itself. In my example, the nonexistence is a bit more extreme. There is no solution on an interval $(0,\epsilon)$ or $(-\epsilon, 0)$ with $\epsilon > 0$ that has one-sided limit $0$ at $x=0$. – Robert Israel May 30 '18 at 6:59
• Illustrative, but I was thinking why your argument will work if you have the product $xy$ that is $xy \geq 0$ and $xy \leq 0$, like I could visualize why the above mentioned solution by me will not work when i think of $y =|x|$ graph but how could I visualize in your case when $xy\geq 0$and $xy \leq 0$ is involved? – BAYMAX May 30 '18 at 7:04
• Also, I think why $y = -|x|$ will not be a solution to the ODE you said, as I observe $y=-|x|$ satisfies the conditions of the function $y$? – BAYMAX May 30 '18 at 7:13
Derivatives have the Darboux property, that is, if $g:I\to\mathbb{R}$ is differentiable in the interval $I$ and $g'$ takes two distinct values, then it takes all the values in between. So in order to have an ODE $y'=f(x,y)$ with no solutions, we can take $f(x,y)=g(x)$ where $g$ is any function with jump discontinuities like the Dirichlet Function.