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Consider a differential separable equation $$y^{\prime}=f(y,t)$$ with initial solution $y(t_0)=y_0$. Suppose that $f(y_0,t_0)$ is not defined. Is there a theorem which can be used to prove the existence and the uniqueness of the solution of this Differential Equation?

The trouble is because $f(y_0,t_0)$ is not defined (much worse than discontinuous where we can still use Carathéodory's existence theorem)

For example a separable differential equation $y^{\prime}=\frac{1}{y-1}+2$ with initial solution $y(0)=1$.

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  • $\begingroup$ Another example would be $y'=\frac{1}{2\sqrt y},\quad y(0)=0.$ It illustrates the existence much better=) $\endgroup$ Jul 12 '13 at 13:28
  • $\begingroup$ If $f(y_0,t_0)$ is not defined, how do you interpret the equality $y'(t)=f(y(t),t)$ in the point $t=t_0$? $\endgroup$
    – Tomás
    Jul 12 '13 at 13:29
  • $\begingroup$ @Tomás we don't need to. We pose the differential equation on the open set $\mathbb R\setminus \{t_0\}$ and initial condition in $t_0$. $\endgroup$ Jul 12 '13 at 13:30
  • $\begingroup$ @TZakrevskiy, ok I understood, thank you. $\endgroup$
    – Tomás
    Jul 12 '13 at 13:31
  • $\begingroup$ I don't think there exists a general result in that direction. Not sure at all. In your example, I believe there are two solutions. $\endgroup$ Sep 6 '13 at 23:23
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I think what you are after is the concept of weak solution, that is precisely used to give a sense to differential equations without having to worry about single points in the domain. You can check intuition behind weak solution , https://en.wikipedia.org/wiki/Weak_solution The idea is to use the integral formulation of the equation, that in your case is: \begin{equation} y(t)-y(t_0)=\int^t_{t_0}f(s,y)d s \end{equation} if you are willing to adopt this as a definition of your solution then you can dispense with the C1 assumption and even the "definition" of the function on a set of measure zero.

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The answer is no. At least there is no such definable solution, because, according to the definition, a solution of the IVP $$ y'=f(t,y), \quad y(t_0)=y_0, $$ is a $C^1$ function $\psi$ which is defined in an open interval $I$, with $t_0\in I$, and satisfying the both the initial condition and the ODE, i.e., $$ \psi(t_0)=y_0\,\,\,\text{and}\,\,\,\psi'(t)=f\big(t,\psi(t)\big), $$ for all $t\in I$. Satisfaction of the ODE implies that $$ \big(t,\psi(t)\big) \in D, \quad \text{for all $t\in I$}, $$ where $D$ is the domain of definition (and continuity) of $f$. Hence $(t_0,y_0)\in D$.

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