I would like to find the steady states to the following system:
\begin{align*} \dfrac{dX}{dt}=k_{1}S \dfrac{K_{d}^p}{K_{d}^p + Y^p} - k_{dx}X \end{align*}
\begin{align*} \dfrac{dY}{dt}=k_{sy}X-k_{2}E_{T}\dfrac{Y}{K_{m}+Y} \end{align*}
With the proposed values:
\begin{align*} \dfrac{dX}{dt}=0.1*2 \dfrac{0.5^2}{0.5^2 + Y^2} - 0.1X=0 \end{align*}
\begin{align*} \dfrac{dY}{dt}=1X-0.5\dfrac{Y}{2+Y}=0 \end{align*}
Then I tried to find the x and y steady states for the first function, f(x,y), in matlab and finish with strange values and I would expect real values as steady states:
syms X Y
f = 0.1*2*(0.5^2)/(0.5^2 + Y^2) - 0.1*X;
f = simplify(f)
diff(f, X)
diff(f, Y)
But I finished with strange values and I don't think that's right:
>> diff(f, X)
ans =
-1/10
>> diff(f, Y)
ans =
-Y/(10*(Y^2 + 1/4)^2)
Apparently I am a little bit lost how to perform the partial derivatives to find the steady states. I would be grateful for any help to solve the problem, even without matlab.
If you extract $X$ from the first equation $$X=\frac{2}{1+4 Y^2}$$ and replace in the second, you arrive to $$\frac{-4 Y^3+3 Y+8}{8 Y^3+16 Y^2+2 Y+4}=0$$ So, the problem is to find the root of $$4 Y^3-3 Y-8=0$$ If you apply the standard method, you will notice that there is only one real root which is given by $$Y=\frac{1}{2} \left(\sqrt[3]{8-3 \sqrt{7}}+\sqrt[3]{8+3 \sqrt{7}}\right)\approx 1.45696$$ to which corresponds $$X=\frac{2}{3+\left(8-3 \sqrt{7}\right)^{2/3}+\left(8+3 \sqrt{7}\right)^{2/3}}\approx 0.210728$$
If you apply the method of hyperbolic solution for one real root, you should find that $$Y=\cosh \left(\frac{1}{3} \cosh ^{-1}(8)\right)\qquad X=\frac{2}{3+2 \cosh \left(\frac{2}{3} \cosh ^{-1}(8)\right)}$$