Solving a non linear system

Question

I have the 4 following equations.

\begin{align}x^{0} &= (c + \xi^1)\sinh(g\xi^0) \\ x^{1} &= (c + \xi^1)\cosh(g\xi^0) \\ x^2 &= \xi^2 \\ x^3 &= \xi^3 \end{align}

I am given that $x^1 = x^3 = 0$ and $x^2 = x^0$, where $x^2$ and $x^0$ evolve with time. One can think of $x^0$ as the parameter.

Using this information, is there a consistent solution of these non linear systems of equations, given that the $\xi$ co-ordinates also change (with time)$(\dagger)$. I can't seem to find that this is the case.

$(\dagger)$ One can think that the $x^0$ coordinate is time in $x-$frame and the $\xi^0$ coordinate is the time in the $\xi-$frame.

Answer 1

Use

$$(x^1)^2-(x^0)^2=(c+\xi_1)^2$$ and

$$\frac{x^0}{x^1}=\tanh g\xi^0.$$

Answer 2

From the above set one has $$ \frac{x^0}{x^1}=\tanh(g\xi_0) $$ that can be inverted using the logarithm or just formally writing $$ \xi_0=g^{-1}\operatorname{arctanh}\left(\frac{x^0}{x^1}\right). $$ being $\operatorname{arctanh} (x) = \frac{1}{2}\ln \left( \frac{1 + x}{1 - x} \right); \left| x \right| < 1$. Then, we just note that $$ \sinh(x)=\frac{\tanh(x)}{\sqrt{1-\tanh^2(x)}} $$ and so, $$ \xi_1=-c+x_1\sqrt{1-\frac{x_0^2}{x_1^2}}. $$

Answer 3

With a system of nonlinear equations, any explicit solution is rare and valuable. The possibility of blowup is a small price to pay.

Answer 4

Others have correct answers, but let me try to use what you know to make it even simpler.

1. $x^1 = 0$. Hence $(c + \xi^1) \cosh(g\xi^0) = 0$. Since $\cosh$ is always positive, that means that $c + \xi^1 = 0$, so $\xi^1 = -c$.

2. The first equation now implies that $x^0 = 0$.

3. Your assumption that $x^2 = x^0$ tells you $x^2 = 0$.

4. $x^3$ and $\xi^3$ can be anything. (Except that since you changed the problem, this last line no longer holds.)

Oops! I just saw that it now says $x_1 = x_3 = 0$, so the system has a solution only when $x^0 = x^1 = x^2 = x^3 = 0$. The solution is

$$ \xi^1 = -c\\ \xi^2 = \xi^3 = 0\\ $$ while $\xi_0$ can have any value.