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I'm having a difficulty figuring a way to proving that the following system has a unique solution :

$$\begin{cases} -y+x\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2} =0 \\ \space \space\space x+y\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2}=0\end{cases}$$

I can obviously see that $O(0,0)$ is a solution, but I need to figure out if it's the only one.

One thing that look catchy to me, is that if we interchange variables on the first equation, $x$ and $y$ (which means $x:=y$ and $y:=x$) we get exactly the same equation but with a minus sign in front of the "free" variable that is not part of the product expression of the equation.

Also, since everything involves variables, bringing the system to a form to calculate determinants to decide if the solution is unique, would also be complicated if not impossible.

I would really appreciate any help or tip towards this, as I can't find a way to prove that the solution is unique.

Here's a link to Wolfram Alpha for this particular system.

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3 Answers 3

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Assume $x,y\ne0$. Multiply the first equation by $-y$ and the second by $x$ and add. You get $y^2+x^2=0$, which is not compatible with the hypothesis.

Now assume $x\ne0,y=0$. You have

$$\begin{cases} x^3\cdot \sin\sqrt{x^2}=0\\x=0\end{cases}$$

which is also incompatible. Same with $x=0,y\ne0$.

Finally, $x=y=0$ is a solution.

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    $\begingroup$ Was just wondering about the theoritical part of the multiplication ! Perfect lead through, thanks a lot ! $\endgroup$
    – Rebellos
    Nov 16, 2017 at 0:45
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You are right that swapping x and y in one equation gives the other. Another way of saying that is that if we multiply the first equation by y, we get $-y^2+ xy(x^2+ y^2)sin\sqrt{x^2+ y^2}= 0$ and if we multiply the second equation by x we get $x^2+ xy(x^2+ y^2)sin\sqrt{x^2+ y^2}= 0$. Subtracting the first of those from the second, $x^2+ y^2= 0$. The only (real) x and y that satisfy that are x= y= 0.

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  • $\begingroup$ Was close ! Thanks a lot for hinting the rest out ! Isn't there any theoritical "error" involving the multiplications (not knowing if $x,y$ take the values zero in case of not observing that $O(0,0)$ satisfies the equation) ? (was just answered by Yves Daoust!) $\endgroup$
    – Rebellos
    Nov 16, 2017 at 0:45
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let $a = (x^2 + y^2) \sin \sqrt {x^2 + y^2}$

$ax - y = 0\\ x + ay = 0$

has a non-trival solution if $a^2 + 1 = 0$

Which requires $a$ be complex.

there are no non-trivial real solutions.

$(x,y) = 0$ is the unique solution

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  • $\begingroup$ One question, is it theoritically acceptable to letting a variable $a$ be expressed in the form of the unknowns that we are initially letting intact to form out the discriminant hypothesis ? $\endgroup$
    – Rebellos
    Nov 16, 2017 at 0:47
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    $\begingroup$ Why not? If our solution led us somewhere interesting, we may need to convert a back to where it was before we could proceed. $\endgroup$
    – Doug M
    Nov 16, 2017 at 0:50
  • $\begingroup$ True that ! Was just wondering if the discriminant test works even if the constants $a_n$ are dependent from the variables of the system. $\endgroup$
    – Rebellos
    Nov 16, 2017 at 0:52
  • $\begingroup$ @DougM: the OP is right to ask. This works "by chance" because the determinant is guaranteed to be nonzero. It might not work in other cases. The theory of linear equations does not apply. $\endgroup$
    – user65203
    Nov 16, 2017 at 0:55
  • $\begingroup$ @YvesDaoust What I have done above shows that no other solutions exist. But, suppose there exists some value of $a$ such that the determinant of that matrix equals $0$ Then this would but some $(x,y)$ pair in the kernel. and we would be able to say $x = ky$ and plug it into the equation for $a$ and then we can solve for $x,y$ $\endgroup$
    – Doug M
    Nov 16, 2017 at 1:21

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