How do I demonstrate that a non-linear system has solutions?

Question

I've a question about non-linear systems: I've started to do some exercises on the implicit function, when I've found one that asks me to discuss the existence of solutions $x,y,z\in\mathbb{R}$ of the following system:

$$\begin{cases} x+e^z+yz\sin(x)=1 \\ ze^z+\sin(xyz)+y^2x=0 \end{cases}$$

How do I proceed? I can't see how implicit functions could help me here...

Answer 1

While there is no "recipe" to get a solution for arbitrary non-linear problems, here the situation is fairly nice.

Setting $x=0$ in:

$$ \begin{cases} x+e^z+yz\sin(x)=1 \\ ze^z+\sin(xyz)+y^2x=0 \end{cases} $$

gives the simplified problem (since also $\sin x = 0$):

$$ \begin{cases} e^z=1 \\ ze^z=0 \end{cases} $$

This is clearly satisfied when $z=0$ and for any $y$ whatsoever. Thus we have a line of solutions (the $y$-axis) to which the implicit function theorem may be applied.

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With marty cohen's further observations all the solutions that lie in any coordinate plane are simply the points in the $y$-axis. We may then ask if any solutions with only non-zero coordinates exist.

Note that generically the two equations in three unknowns would produce a one-dimensional family of solutions. This is no doubt a part of the discussion called for in the OP's motivating exercise, that by specifying a nonzero value of one variable we can in principle find how varying that value affects the other variables.

To be more explicit about solving the given non-linear system, the first equation can be "solved" for $y$ in terms of $x$ and $z$:

$$ y = \frac{1 - x - e^z}{z \sin(x)} $$

whenever $z \sin(x) \neq 0$ of course. Given that $x,z\neq 0$, the only such unaccounted possibilities are when $x$ is an integer multiple of $\pi$.

This expression for $y$ can then be used to eliminate $y$ in the second equation, so that the problem will mostly reduce to solving one equation in the two unknowns $x,z$.

Answer 2

Continuing in hardmath's path, if we set $y=0$,

$\begin{cases} x+e^z+yz\sin(x)=1 \\ ze^z+\sin(xyz)+y^2x=0 \end{cases} $

becomes $\begin{cases} x+e^z=1 \\ ze^z=0 \end{cases} $

for which the only solution is $x=z=0$.

Similarly, if we set $z=0$, the equations become

$\begin{cases} x+1=1 \\ y^2x=0 \end{cases} $

for which the only solution is $x=0$ and $y$ can be any value.