Is this function quasi convex

Question

I have a function $f(x,y) = y(k_1x^2 + k_2x + k_3)$ which describes chemical potential of a species ($y$ is mole fraction and $x$ is temperature)

I only want to check quasi convexity over a limited range. k1,k2 and k3 are coefficients of a polynomial function to calculate Gibbs energy of formation. They differ widely between different chemical species. In my case they have the following restrictions.

$0<k1<0.001$

$-0.1<k2<0.1$

$-400<k3<0$

Temperature, $100<=T<=3000$

Mole fraction, $0<=y<=1$

From the hessian matrix I know it is not convex or concave. But can I check whether it is quasi convex.

Thanks for the answer. So to check quasi convexity I have to find the roots of the polynomial. If I get equal or complex roots then the function will be convex.

Thanks for the update. However I am a confused about how the domain for $x_1,x_2$ was determined in the last step. Also would there be any change in the domain if the function were divided by $x$ so that,

$f(x,y)=y(k_1x+k_2+k_3/x)$

Answer 1

Let $k_{1}=1, k_{2}=k_{3}=0$ and consider the level set $$ \{(x,y)\in R^{2}: f(x,y)\leq 1\} $$ This condition is equivalent to $y\leq\tfrac{1}{x^{2}}$ and the set $$\{(x,y)\in R^{2}: y\leq\tfrac{1}{x^{2}}\}$$ is convex in $R^{2}$.

This can be generealized. If the function $k_{1}x^{2}+k_{2}x+k_{3}$ has two different real roots, it is not quasi convex, if it has two identical real roots it is quasi convex, if it has no real roots it is also quasi convex, but this seems only be right for a special selection of $k_{1}, k_{2},k_{3}$ :-(...

If $k_{1}=0$ and $k_{2}\neq 0$ the function is not quasi convex. ...if I did no mistake.

So an update: We have summarized the following situation: $$ f:[100,3000]\times[0,1]\rightarrow R,\quad (x,y)\mapsto y(k_{1}x^2+k_{2}x+k_{3})$$ where $k_{1},k_{2},k_{3}$ parameters in $R$ with restriction $$ k_{1}\in (0,0.001),\quad k_{2}\in [-0.1,0.1], \quad k_{3}\in [-400,0].$$ So for given $\alpha\in R$ we have to research if the set $$\{(x,y)\in[100,3000]\times[0,1]: f(x,y)\leq \alpha\}$$ is convex $\forall \alpha\in R$. This can be transformed into $$y\leq \frac{\alpha}{k_{1}x^{2}+k_{2}x+k_{3}}.$$ Seeing this as the graph of a one dimensional function we have a rational function. Computing the asymptotes we obtain $$x_{1,2}=\frac{-k_{2}\pm\sqrt{k_{2}^{2}-4k_{1}k_{3}}}{2k_{1}}.$$ So there are always two different roots for every selection of parameters an we have $$ y\leq\frac{\alpha}{k_{1}(x-x_{1})(x-x_{2})}. $$ So if $x_{1}$ or $x_{2}\in[100,3000]$ the function will not be quasi convex. Otherwise it will be quasi convex. The restriction on $y$ seems to be irrelevant since we have it always compensated by $\alpha$.

That should answer your question. Maybe some constants are not quite right, but in general this idea can be used rigorously...