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Let the initial value problem : $$y'=(x-y)^{1/2}, y(2)=2$$ Study the existence of solutions in a domain around the initial value and study its uniqueness too.

Attempt :

In order to show the uniqueness, I'll work around the function $f(x,y) = (x-y)^{1/2}$ and it's Lipschitz condition with respect to $y$ :

$$|f(x,y_2) - f(x,y_1)| = |(x-y_2)^{1/2} - (x-y_1)^{1/2}|$$

For the function $g(y) = (x-y)^{1/2}$, we can use the Mean Value Theorem since it's continuous and differentiable for $x-y \geq 0 \Leftrightarrow x \geq y$, for some $ξ \in (y_1,y_2) $ :

$$\big((x-y)^{1/2}\big)'|_{y=ξ} = \frac{(x-y_2)^{1/2} - (x-y_1)^{1/2}}{y_2 - y_1}$$

$$\Leftrightarrow$$

$$-\frac{1}{2\sqrt{x-ξ}}(y_2-y_1) = (x-y_2)^{1/2}-(x-y_1)^{1/2}$$

$$\Leftrightarrow$$

$$\bigg|-\frac{1}{2\sqrt{x-ξ}}\bigg|\bigg|y_2-y_1\bigg| = \bigg| (x-y_2)^{1/2}-(x-y_1)^{1/2}\bigg|$$

Now, I am not sure if I can make an assumption about a constant $L$ with respect to $y$. I observe that $k=\bigg|-\frac{1}{2\sqrt{x-ξ}}\bigg|$ is independent of $y_2,y_1$ but I cannot see how to make an inequality out of this, since it's value would change for any different $x$. Can you give me some help regarding how to show that there exists a Lipschitz constant or not ?

Also, regarding the solution, how would I study the existence of solutions in a domain around $y(2) = 2$ ? Can you please explain thoroughly with a complete solution regarding this, because it's an issue I have in a lot of similar problems.

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    $\begingroup$ At $y=x$ you have a vertical tangent in direction $y$ and thus no Lipschitz constant. $\endgroup$ – Dr. Lutz Lehmann Oct 23 '17 at 7:46
  • $\begingroup$ @LutzL ty for the reply. What about the existence of solution though around the domain of the initial value ? $\endgroup$ – Rebellos Oct 23 '17 at 9:31
  • $\begingroup$ @LutzL By the way, I suppose you mean a vertical tangent at $x=ξ$ ? $\endgroup$ – Rebellos Oct 23 '17 at 9:45
  • $\begingroup$ At any point $(x,y)$ where $y=x$, so also at the given initial point. However, contrary to $y'=\sqrt{y}$ the zero curve of the radicant is not a solution itself, so that the possibilities for multiple solutions are reduced. $\endgroup$ – Dr. Lutz Lehmann Oct 23 '17 at 13:39
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Let $z(x)=x-y(x)$, then $z'(x)=1-y'(x)$ hence $$z'(x)=F(z(x))\qquad z(2)=0$$ with $$F:u\mapsto1-\sqrt{u}$$ This is an autonomous differential equation with $F(z(2))\ne0$ hence, for every $x\geqslant2$, $$\int_0^{z(x)}\frac{du}{1-\sqrt{u}}=x-2$$ that is, using the change of variable $v=\sqrt u$, $$x-2=\int_0^{\sqrt{z(x)}}\frac{2vdv}{1-v}=\left.-2\log(1-v)-2v\right|_0^{\sqrt{z(x)}}=-2\log(1-\sqrt{z(x)})-2\sqrt{z(x)}$$ or, equivalently, $$(1-\sqrt{z(x)})\,e^{\sqrt{z(x)}}=e^{1-x/2}$$ One sees that $z(x)$ increases from $z(2)=0$ to $z(+\infty)=1$, and that, for every $x\geqslant2$, $y(x)$ is the unique solution of the identity $$(1-\sqrt{x-y(x)})\,e^{\sqrt{x-y(x)}}=e^{1-x/2}$$ On the other hand, again using $F(z(2))>0$, one sees that the solution is not defined on $x<2$.

Nota: The fact that $F(z(2))>0$ implies that every solution $z(x)$ is instantaneously positive, that is, that $z(x)>0$ for every $x>2$, and then Cauchy-Lipschitz uniqueness and existence criterion applies. However, a rigorous argument showing uniqueness and existence starting at $z(2)=0$ is lacking (Cauchy-Peano-Arzelà provides existence, confirmed by the explicit computations above, but the hypotheses of Osgood's criterion, for example, are not met).

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