# Proving Unique Solution to The Following System of Two Non-Linear Equations and 2 Unknown

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?

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:

$$P_9(x)=\:\left(\left(x^3+1\right)^3-x-2\right)=x^9+3x^6+3x^3-x-1=0$$

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.

• 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. Sep 10, 2019 at 4:23
• 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. 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.