Question #94224
3y(dy/ dx)=-3x ; y(2)=-π unique or not?
1
Expert's answer
2019-09-11T13:39:55-0400

We have the equation dydx=xy\frac{dy}{dx}=-\frac{x}{y} with initial condition y(2)=πy(2)=-\pi. One of solutions for this equation is x2+y2=π2+4x^2+y^2=\pi^2+4 . It can be transformes into two functions: y(x)=π2+4x2y(x)=\sqrt{\pi^2+4-x^2} and y(x)=π2+4x2y(x)=-\sqrt{\pi^2+4-x^2} .Function y(x)=π2+4x2y(x)=\sqrt{\pi^2+4-x^2} does not satisfy the initial condition y(2)=πy(2)=-\pi . The second function,

y(x)=π2+4x2y(x)=-\sqrt{\pi^2+4-x^2} , defined on (π2+4,π2+4)(-\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(0,π)a\in (0,\pi) the equation has a unique local solution in (π2+4,π2+4)×(,a)(-\sqrt{\pi^2+4},\sqrt{\pi^2+4})\times (-\infty,a) (which is y=π2+4x2y=-\sqrt{\pi^2+4-x^2} ).

Hence the equation has a unique local solution on (π2+4,π2+4)×(,0)(-\sqrt{\pi^2+4},\sqrt{\pi^2+4})\times (-\infty,0) .Indeed, assume that there is solution y1(x)y_1(x) defined on II , where y1(u)π2+4u2y_1(u)\neq -\sqrt{\pi^2+4-u^2} for some u(π2+4,π2+4)Iu\in (-\sqrt{\pi^2+4},\sqrt{\pi^2+4})\cap I . Ifv=max{y1(u),π2+4u2}v=\max\{y_1(u), -\sqrt{\pi^2+4-u^2}\} , we have two different solutions in (I(π2+4,π2+4))×(,v2)\bigl(I\cap (-\sqrt{\pi^2+4},\sqrt{\pi^2+4})\bigr)\times (-\infty,\frac{v}{2}) . Contradiction with uniqueness of y(x)=π2+4x2y(x)=-\sqrt{\pi^2+4-x^2}.

Assume that there is solution y2(x)y_2(x), defined on I1I_1 such that I1[π2+4,π2+4]I_1\setminus [-\sqrt{\pi^2+4},\sqrt{\pi^2+4}]\neq\varnothing . If there is cI1(,π2+4)c\in I_1\cap (-\infty,-\sqrt{\pi^2+4}) , then [c,2]I1[c,2]\subset I_1 and y2(x)=π2+4x2y_2(x)=-\sqrt{\pi^2+4-x^2} on [π2+4,2][-\sqrt{\pi^2+4},2] . So y2(π2+4)=0y_2(-\sqrt{\pi^2+4})=0 and y2(π2+4)=+\bigl|y_2'(-\sqrt{\pi^2+4})\bigr|=+\infty.

Let x(c,π2+4)x\in (c,-\sqrt{\pi^2+4}). Since y2(x)=xy2(x)y_2'(x)=-\frac{x}{y_2(x)} and x<0x<0 , we have sgn(y2(x))=sgn(y2(x))sgn (y_2'(x))=sgn(y_2(x)) .

If y2(π2+40)=y_2'(-\sqrt{\pi^2+4}-0)=-\infty , then y(x)<0y'(x)<0 in (d,π2+4)(d,-\sqrt{\pi^2+4}) for some d(c,π2+4)d\in (c,-\sqrt{\pi^2+4}) . But since y2y_2 is decreasing function in (d,π2+4)(d,-\sqrt{\pi^2+4}) and y2(π2+4)=0y_2(-\sqrt{\pi^2+4})=0 , we have y2>0y_2>0 in (d,π2+4)(d,-\sqrt{\pi^2+4}) . Contradiction with statement that sgn(y2)=sgn(y2)sgn (y_2')=sgn(y_2) in (d,π2+4)(d,-\sqrt{\pi^2+4}) . Similarly we obtain a contradiction while assuming that y2(π2+40)=+y_2'(-\sqrt{\pi^2+4}-0)=+\infty. So I1(,π2+4)=I_1\cap (-\infty,-\sqrt{\pi^2+4})=\varnothing .

Likewise we can obtain that I1(π2+4,+)=I_1\cap (\sqrt{\pi^2+4},+\infty)=\varnothing .

Hence every solution of the equation is defined on a subset of [π2+4,π2+4][-\sqrt{\pi^2+4},\sqrt{\pi^2+4}] . Uniqueness of the local solution of y(x)=π2+4x2y(x)=-\sqrt{\pi^2+4-x^2} in [π2+4,π2+4]×(,0][-\sqrt{\pi^2+4},\sqrt{\pi^2+4}]\times (-\infty,0] gives us uniqueness of the solution y(x)=π2+4x2y(x)=-\sqrt{\pi^2+4-x^2} in R2\mathbb R^2 .


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