I have the following system of partial differential equations:
$$ \frac{\partial\Lambda^{\mu}}{\partial x^{\nu}}= $$ $$ \small \begin{pmatrix} \cosh\phi_{0}&\sinh\phi_{1}&\sinh\phi_{2}&\sinh\phi_{3}\\ \sin\theta_{0}\cos\omega_{0}\sinh\phi_{0}&\sin\theta_{1}\cos\omega_{1}\cosh\phi_{1}&\sin\theta_{2}\cos\omega_{2}\cosh\phi_{2}&\sin\theta_{3}\cos\omega_{3}\cosh\phi_{3}\\ \sin\theta_{0}\sin\omega_{0}\sinh\phi_{0}&\sin\theta_{1}\sinh\omega_{1}\cosh\phi_{1}&\sin\theta_{2}\sin\omega_{2}\cosh\phi_{2}&\sin\theta_{3}\sinh\omega_{3}\cosh\phi_{3}\\ \cos\theta_{0}\sinh\phi_{0}&\cos\theta_{1}\cosh\phi_{1}&\cos\theta_{2}\cosh\phi_{2}&\cos\theta_{3}\cosh\phi_{3} \end{pmatrix} $$ Where $\Lambda:\Bbb{R}^{4}\to\Bbb{R}^{4}$ is written as $\Lambda(x^{0},x^{1},x^{2},x^{3})=(\Lambda^{0},\Lambda^{1},\Lambda^{2},\Lambda^{3})$ and $\mu$ and $\nu$ denote respectively the rows and columns of the jacobian matrix. Is there any way by which I can say a priori that the hyperbolic and trigonometric angles do not depend themselves on the coordinates $(x^{0},x^{1},x^{2},x^{3})$ without, for example, computing the Schwartz conditions on the second derivatives for each and every one of them? $$ \\ $$ Should it be necessary, the system itself is accompanied by the equations $$ \cosh\phi_{0}\sinh\phi_{1}=\cosh\phi_{1}\sinh\phi_{0}\left[\sin\theta_{0}\sin\theta_{1}\cos(\omega_{0}-\omega_{1})+\cos\theta_{0}\cos\theta_{1}\right]\\ \\ \cosh\phi_{0}\sinh\phi_{2}=\cosh\phi_{2}\sinh\phi_{0}\left[\sin\theta_{0}\sin\theta_{2}\cos(\omega_{0}-\omega_{2})+\cos\theta_{0}\cos\theta_{2}\right]\\ \\ \cosh\phi_{0}\sinh\phi_{3}=\cosh\phi_{3}\sinh\phi_{0}\left[\sin\theta_{0}\sin\theta_{3}\cos(\omega_{0}-\omega_{3})+\cos\theta_{0}\cos\theta_{3}\right]\\ \\ \sinh\phi_{1}\sinh\phi_{2}=\cosh\phi_{1}\cosh\phi_{2}\left[\sin\theta_{1}\sin\theta_{2}\cos(\omega_{1}-\omega_{2})+\cos\theta_{1}\cos\theta_{2}\right]\\ \\ \sinh\phi_{1}\sinh\phi_{3}=\cosh\phi_{1}\cosh\phi_{3}\left[\sin\theta_{1}\sin\theta_{3}\cos(\omega_{1}-\omega_{3})+\cos\theta_{1}\cos\theta_{3}\right]\\ \\ \sinh\phi_{2}\sinh\phi_{3}=\cosh\phi_{2}\cosh\phi_{3}\left[\sin\theta_{2}\sin\theta_{3}\cos(\omega_{2}-\omega_{3})+\cos\theta_{2}\cos\theta_{3}\right] $$
$$ \\ $$ P.S.: The solution to the equations represent the general form of the symmetry transformations of Minkowski spacetime.
$$\\$$ EDIT: The matrix and the equations that I wrote above are really a solution to the PDEs
$$ \left(\frac{\partial\Lambda^{0}}{\partial x^{0}}\right)^{2}-\left[\left(\frac{\partial\Lambda^{1}}{\partial x^{0}}\right)^{2}+\left(\frac{\partial\Lambda^{2}}{\partial x^{0}}\right)^{2}+\left(\frac{\partial\Lambda^{3}}{\partial x^{0}}\right)^{2}\right]=1\\ \\ \left(\frac{\partial\Lambda^{0}}{\partial x^{1}}\right)^{2}-\left[\left(\frac{\partial\Lambda^{1}}{\partial x^{1}}\right)^{2}+\left(\frac{\partial\Lambda^{2}}{\partial x^{1}}\right)^{2}+\left(\frac{\partial\Lambda^{3}}{\partial x^{1}}\right)^{2}\right]=-1\\ \\ \left(\frac{\partial\Lambda^{0}}{\partial x^{2}}\right)^{2}-\left[\left(\frac{\partial\Lambda^{1}}{\partial x^{2}}\right)^{2}+\left(\frac{\partial\Lambda^{2}}{\partial x^{2}}\right)^{2}+\left(\frac{\partial\Lambda^{3}}{\partial x^{2}}\right)^{2}\right]=-1\\ \\ \left(\frac{\partial\Lambda^{0}}{\partial x^{1}}\right)^{3}-\left[\left(\frac{\partial\Lambda^{1}}{\partial x^{3}}\right)^{2}+\left(\frac{\partial\Lambda^{2}}{\partial x^{3}}\right)^{3}+\left(\frac{\partial\Lambda^{3}}{\partial x^{3}}\right)^{2}\right]=-1\\ \\ \\ \frac{\partial \Lambda^{0}}{\partial x^{0}}\frac{\partial \Lambda^{0}}{\partial x^{1}}=\frac{\partial \Lambda^{1}}{\partial x^{0}}\frac{\partial \Lambda^{1}}{\partial x^{1}}+\frac{\partial \Lambda^{2}}{\partial x^{0}}\frac{\partial \Lambda^{2}}{\partial x^{1}}+\frac{\partial \Lambda^{3}}{\partial x^{0}}\frac{\partial \Lambda^{3}}{\partial x^{1}}\\ \\ \frac{\partial \Lambda^{0}}{\partial x^{0}}\frac{\partial \Lambda^{0}}{\partial x^{2}}=\frac{\partial \Lambda^{1}}{\partial x^{0}}\frac{\partial \Lambda^{1}}{\partial x^{2}}+\frac{\partial \Lambda^{2}}{\partial x^{0}}\frac{\partial \Lambda^{2}}{\partial x^{2}}+\frac{\partial \Lambda^{3}}{\partial x^{0}}\frac{\partial \Lambda^{3}}{\partial x^{2}}\\ \\ \frac{\partial \Lambda^{0}}{\partial x^{0}}\frac{\partial \Lambda^{0}}{\partial x^{3}}=\frac{\partial \Lambda^{1}}{\partial x^{0}}\frac{\partial \Lambda^{1}}{\partial x^{3}}+\frac{\partial \Lambda^{2}}{\partial x^{0}}\frac{\partial \Lambda^{2}}{\partial x^{3}}+\frac{\partial \Lambda^{3}}{\partial x^{0}}\frac{\partial \Lambda^{3}}{\partial x^{3}}\\ \\ \frac{\partial \Lambda^{0}}{\partial x^{1}}\frac{\partial \Lambda^{0}}{\partial x^{2}}=\frac{\partial \Lambda^{1}}{\partial x^{1}}\frac{\partial \Lambda^{1}}{\partial x^{2}}+\frac{\partial \Lambda^{2}}{\partial x^{1}}\frac{\partial \Lambda^{2}}{\partial x^{2}}+\frac{\partial \Lambda^{3}}{\partial x^{1}}\frac{\partial \Lambda^{3}}{\partial x^{2}}\\ \\ \frac{\partial \Lambda^{0}}{\partial x^{1}}\frac{\partial \Lambda^{0}}{\partial x^{3}}=\frac{\partial \Lambda^{1}}{\partial x^{1}}\frac{\partial \Lambda^{1}}{\partial x^{3}}+\frac{\partial \Lambda^{2}}{\partial x^{1}}\frac{\partial \Lambda^{2}}{\partial x^{3}}+\frac{\partial \Lambda^{3}}{\partial x^{1}}\frac{\partial \Lambda^{3}}{\partial x^{3}}\\ \\ \frac{\partial \Lambda^{0}}{\partial x^{2}}\frac{\partial \Lambda^{0}}{\partial x^{3}}=\frac{\partial \Lambda^{1}}{\partial x^{2}}\frac{\partial \Lambda^{1}}{\partial x^{3}}+\frac{\partial \Lambda^{2}}{\partial x^{2}}\frac{\partial \Lambda^{2}}{\partial x^{3}}+\frac{\partial \Lambda^{3}}{\partial x^{2}}\frac{\partial \Lambda^{3}}{\partial x^{3}} $$
I'm thinking that the actual system could actually be more significative than the solution I gave.