I am trying to understand whether there are any theorems for showing whether a system of non-linear equations has a unique solution? I have searched a bit online already, but not finding anything concrete.

Consider the following system to non-linear equations: $$y^3 = x + 2$$ $$y = x^3 + 1$$ So we have 2 equations and 2 unknowns. We can't analytically solve for $x,y.$ We can of course use numerical methods (such as newtons) from software like R or Matlab to determine a solution. Plotting the above equations on Desmos, I see there is a unique solutions of x = 0.736 and y = 1.39.

So the question is, suppose I just want to show above system has a unique solution, but don't care about solving for it. How can I go about doing that?


4 Answers 4


Substituting the second equation for $y$ into the first equation, we get $$(x^3+1)^3 = x+2 \iff (x^3+1)^3 - x - 2 = 0.$$ Now consider the polynomial $f(x) = (x^3+1)^3 - x - 2$, for which any root yields a solution to the original system. Thus we want to analyze $f$ in such a way that we can prove there is only one real root. We can see that for any $x$ with $|x| \leq \frac{1}{2}$ we have that $$ |(x^3+1)^3 - x| \leq (|x|^3+|1|)^3 + |x| \leq (|.5|^3+1)^3 + |.5| < 2,$$ and so $f$ is negative on the interval $\left[-\frac{1}{2},\frac{1}{2}\right]$. Additionally, if $-1 < x < -\frac{1}{2}$ then we have that $$(x^3+1)^3 - x < ((-.5)^3+1)^3 + 1 < 2$$ and if $-2 < x < -1$ we have that $$(x^3+1)^3 - x < ((-1)^3+1)^3 + 2 = 2.$$ Thus $f$ is negative on the interval $\left(-2,\frac{1}{2}\right]$.

Now take the derivative to get $$ f'(x) = 9x^2(x^3+1)^2 - 1,$$ and notice that $f'(x) > 0$ for every $x \in (-\infty,-2]\cup \left(\frac{1}{2},\infty\right)$. Therefore $f$ is monotonically increasing in this region. All together, we see that $f$ is negative for $x \leq \frac{1}{2}$ and monotonically increasing for all $x > \frac{1}{2}$; thus $f$ has exactly 1 real root.


So the question is, suppose I just want to show above system has a unique solution, but don't care about solving for it

Here we are trying to show that the system has only 1 real root.

Using the 2 equations, cube the 2nd equation and subtract to get:


Applying the Rule of Signs, we get the maximum number of real positive roots to be one, since there is only 1 sign change.

As the comment below, we have to take in consideration the number of negative roots that could be $0, 2 ,4$.

To investigate other roots, we divide by $(x - r)$ where $r$ is a real root we know.

Using Newton's method, we get a root for $P_9(x)$ to be: $$x=0.7359655$$

Let: $$P_8(x)=\frac{x^9+3x^6+3x^3-x-1}{x-0.7359655}$$

We need to see if $P_8$ has any real roots.

for $\delta$ almost equal zero we get: $$ P_8(x)=x^8+0.7359655x^2+0.54164x^6+3.39863x^5+2.50127x^4+1.84085x^3+4.35480x^2+3.20498x+1.35875-\delta$$

Using the rule of signs on $P_8(x)$, since there is no sign change, $P_8(x)$ has no real roots. This leaves you with the only real root for $P_9(x)$ to be $0.7359655$. There are 8 other complex roots.

Note: all values are rounded.

  • 1
    $\begingroup$ The rule of signs only tells you about the number of positive roots; while it is true that the number of positive roots is exactly 1, using this rule on $P_9(-x)$ we see that there are 4 sign changes and so there can be 0, 2, or 4 negative roots. Thus we have to exclude the possibility of negative roots before concluding that there is a single real solution. $\endgroup$
    – Andrew
    Commented Sep 10, 2019 at 4:23
  • $\begingroup$ You are correct, I forgot about counting the negative roots. According to your note, the division step is not optional as I mentioned. Corrected answer as per note. Thanks. $\endgroup$
    – NoChance
    Commented Sep 10, 2019 at 5:08

Think that

$$ 3y^2 \frac{dy}{dx} = 1\Rightarrow\frac{dy}{dx} = \frac{1}{3y^2} > 0\\ 3x^2 \frac{dx}{dy} = 1\Rightarrow\frac{dx}{dy} = \frac{1}{3x^2} > 0 $$

so $y=x^3+1$ is strictly increasing regarding the vertical axis and $x = y^3-2$ is strictly increasing regarding the horizontal axis. Also regarding $y = x^3+1$ it pass across the point $(0,1)$ and $x = y^3-2$ pass across the point $(0,\sqrt[3]{2})$ so they cross only at one point, because $\sqrt[3]{2} > 1$


Using @Andrew's notation let $f(x)=(x^3+1)^3-x-2.$ If expanded there is a single sign change so exactly one positive zero. Now put $x=-2-t$ and expand that, giving a degree $9$ polynomial with all negative coefficients, so $f$ has no zeroes less than $-2.$ [I can list the coefficients if wanted, first few $-1,-18,-144,-669$ via a symbolic calculator I use.] So what remains to look at is $x \in [-2,0].$

For this we go back to the original system and note we seek the intersection of $y_1=x^3+1$ and $y_2=(x+2)^{\frac{1}{3}}.$

On $(-2,-1)$ we have $y_1<0,y_2>0$ so no crossing there. And on $(-1,0)$ we have (fairly easily) $y_1<1$ along with $y_2>1$ so no crossing there either. Finally should check $x=-2,-1,0$ since those were omitted from the intervals used.


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