Answer to Question #94224 in Differential Equations for Jennylyn

Question

3y(dy/ dx)=-3x ; y(2)=-π unique or not?

Answer 1

We have the equation $\frac{dy}{dx}=-\frac{x}{y}$ with initial condition $y(2)=-\pi$. One of solutions for this equation is $x^2+y^2=\pi^2+4$ . It can be transformes into two functions: $y(x)=\sqrt{\pi^2+4-x^2}$ and $y(x)=-\sqrt{\pi^2+4-x^2}$ .Function $y(x)=\sqrt{\pi^2+4-x^2}$ does not satisfy the initial condition $y(2)=-\pi$ . The second function,

$y(x)=-\sqrt{\pi^2+4-x^2}$ , defined on $(-\sqrt{\pi^2+4},\sqrt{\pi^2+4})$, satisfies the initial condition.

From https://www.science.unitn.it/~bagagiol/noteODE.pdf, theorem 2.10 we have that for all $a\in (0,\pi)$ the equation has a unique local solution in $(-\sqrt{\pi^2+4},\sqrt{\pi^2+4})\times (-\infty,a)$ (which is $y=-\sqrt{\pi^2+4-x^2}$ ).

Hence the equation has a unique local solution on $(-\sqrt{\pi^2+4},\sqrt{\pi^2+4})\times (-\infty,0)$ .Indeed, assume that there is solution $y_1(x)$ defined on $I$ , where $y_1(u)\neq -\sqrt{\pi^2+4-u^2}$ for some $u\in (-\sqrt{\pi^2+4},\sqrt{\pi^2+4})\cap I$ . If$v=\max\{y_1(u), -\sqrt{\pi^2+4-u^2}\}$ , we have two different solutions in $\bigl(I\cap (-\sqrt{\pi^2+4},\sqrt{\pi^2+4})\bigr)\times (-\infty,\frac{v}{2})$ . Contradiction with uniqueness of $y(x)=-\sqrt{\pi^2+4-x^2}$.

Assume that there is solution $y_2(x)$, defined on $I_1$ such that $I_1\setminus [-\sqrt{\pi^2+4},\sqrt{\pi^2+4}]\neq\varnothing$ . If there is $c\in I_1\cap (-\infty,-\sqrt{\pi^2+4})$ , then $[c,2]\subset I_1$ and $y_2(x)=-\sqrt{\pi^2+4-x^2}$ on $[-\sqrt{\pi^2+4},2]$ . So $y_2(-\sqrt{\pi^2+4})=0$ and $\bigl|y_2'(-\sqrt{\pi^2+4})\bigr|=+\infty$.

Let $x\in (c,-\sqrt{\pi^2+4})$. Since $y_2'(x)=-\frac{x}{y_2(x)}$ and $x<0$ , we have $sgn (y_2'(x))=sgn(y_2(x))$ .

If $y_2'(-\sqrt{\pi^2+4}-0)=-\infty$ , then $y'(x)<0$ in $(d,-\sqrt{\pi^2+4})$ for some $d\in (c,-\sqrt{\pi^2+4})$ . But since $y_2$ is decreasing function in $(d,-\sqrt{\pi^2+4})$ and $y_2(-\sqrt{\pi^2+4})=0$ , we have $y_2>0$ in $(d,-\sqrt{\pi^2+4})$ . Contradiction with statement that $sgn (y_2')=sgn(y_2)$ in $(d,-\sqrt{\pi^2+4})$ . Similarly we obtain a contradiction while assuming that $y_2'(-\sqrt{\pi^2+4}-0)=+\infty$. So $I_1\cap (-\infty,-\sqrt{\pi^2+4})=\varnothing$ .

Likewise we can obtain that $I_1\cap (\sqrt{\pi^2+4},+\infty)=\varnothing$ .

Hence every solution of the equation is defined on a subset of $[-\sqrt{\pi^2+4},\sqrt{\pi^2+4}]$ . Uniqueness of the local solution of $y(x)=-\sqrt{\pi^2+4-x^2}$ in $[-\sqrt{\pi^2+4},\sqrt{\pi^2+4}]\times (-\infty,0]$ gives us uniqueness of the solution $y(x)=-\sqrt{\pi^2+4-x^2}$ in $\mathbb R^2$ .