# Question regarding Existence and Uniqueness of a Differential Equation

Here is the theorem

(a) If $$f$$ is continuous on an open rectangle $$R : \{ a < x < b , c < y < d \}$$ that contains $$(x_0, y_0)$$ then the initial value problem $$y ^ { \prime } = f ( x , y ) , \quad y \left( x _ { 0 } \right) = y _ { 0 }$$ has at least one solution on some open subinterval of $$(a, b)$$ that contains $$x_0$$.

(b) If both $$f$$ and $$f_y$$ are continuous on $$R$$ then the equation has a unique solution on some open subinterval of $$(a, b)$$ that contains $$x_0$$.

My question is
If the conditions of the Existence and Uniqueness theorem are met, does there exists a unique solution for all $$x\in(a,b)$$?

• No. Consider $f(x,y) = y^2$. – copper.hat Feb 24 at 23:33

## 2 Answers

Take $$f(x,y) = y^2$$ with $$(x_0,y_0) = (0,1)$$, then the solution is $$y(x) = {1 \over 1-x}$$ for $$x < 1$$.

Note that $$f$$ is smooth and defined everywhere.

In particular, if we choose the rectangle$$(-1,2) \times (0,100)$$ then it is impossible to find a solution starting from $$(0,1)$$ that is defined on $$(-1,2)$$.

Quick answer: No.

The reason is the function $$f$$ might met the conditions, but be nonlinear.

• I may not be understanding the theorem correctly but if you know that $f$ and $f_y$ are continuous on $(a,b)$ and you pick some arbitrary point in $(a,b)$ then why does it seem like the theorem doesn't apply to this point? – Jac Frall Feb 25 at 0:05
• @JacFrall: I have given an example that illustrates the issue. A solution exists in $R$ but, loosely, may 'escape out of the top or bottom of the box'. – copper.hat Feb 25 at 0:15
• @copper.hat yes it took some contemplation but I get it now. Thank you – Jac Frall Feb 25 at 0:24