I'm modeling a pressure relief valve in order to find the limits of stability using the Jacobian matrix at an equilibrium point. The equations are the folowing:
$$ \ddot{y} = \frac{1}{m}(A\cdot p\space - k\cdot y \space -d_a \cdot \dot{y}) $$ $$ \dot{p} = \frac{E}{V}(Q_{IN}\space - y\cdot \gamma \cdot \sqrt{p}) $$
For more clarity I introduced the following variables and functions:
$$ X = \begin{bmatrix} y \\ \dot{y} \\ p \end{bmatrix} = \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} $$
$$ \begin{cases} \dot{x_{1}} = f_{1}(x_{1}, x_{2}, x_{3}) = x_{2}\\ \dot{x_{2}} = f_{2}(x_{1}, x_{2}, x_{3}) = \frac{1}{m}(A\cdot x_{3}\space - k\cdot x_{1} \space -d_a \cdot x_{2})\\ \dot{x_{3}} = f_{3}(x_{1}, x_{2}, x_{3}) = \frac{E}{V}(Q_{IN}\space - x_{1}\cdot \gamma \cdot \sqrt{x_{3}}) \end{cases}\, $$
And I find this Jacobian matrix: $$ J = \begin{bmatrix} \frac{\partial f_{1}(x_{1},x_{2},x_{3})}{\partial x_{1} } & \frac{\partial f_{1}(x_{1},x_{2},x_{3})}{\partial x_{2} } & \frac{\partial f_{1}(x_{1},x_{2},x_{3})}{\partial x_{3} }\\[1ex] % <-- 1ex more space between rows of matrix \frac{\partial f_{2}(x_{1},x_{2},x_{3})}{\partial x_{1} } & \frac{\partial f_{2}(x_{1},x_{2},x_{3})}{\partial x_{2} } & \frac{\partial f_{2}(x_{1},x_{2},x_{3})}{\partial x_{3} } \\[1ex] \frac{\partial f_{3}(x_{1},x_{2},x_{3})}{\partial x_{1} } & \frac{\partial f_{3}(x_{1},x_{2},x_{3})}{\partial x_{2} } & \frac{\partial f_{3}(x_{1},x_{2},x_{3})}{\partial x_{3} } \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\[1ex] % <-- 1ex more space between rows of matrix -\frac{k}{m} & -\frac{d_{a}}{m} & -\frac{A}{m} \\[1ex] -\frac{E}{V}\cdot \gamma \cdot \sqrt{x_{3}}& 0 & -\frac{E}{V}\cdot \gamma \cdot \frac{x_{1}}{2\sqrt{x_{3}}} \end{bmatrix} $$
but I'm not sure for the derivatives $\frac{\partial x_{3}}{\partial x_{1}}$ and $\frac{\partial x_{1}}{\partial x_{3}}$, as $\dot{x_{3}}$ depends on $x_{1}$ the derivative of $x_{3}$ and $x_{1}$ aren't null. How can I find these terms? Am I missing something?
