Problem Sheet Question:
A PDE initial value problem has three associated characteristics equations to solve $x,y,u$. The first, $y$, is a simple ODE and can be solved by integration with a constant.
The other two have the characteristics system as follows:
\begin{align*} \frac{dx}{dt} &= u\\ \frac{du}{dt} &= x \end{align*}
Where $x,u$ are both functions of the parametrised variables $s$ and $t$.
Solution:
Quoting from the mark scheme:
"The above form a system which can be reduced to one equation by eliminating one of the functions. For example, if we eliminate $u$, we will have \begin{align*} \frac{d^2x}{dt^2} &= x\\ \end{align*} Which can be integrated using the methods relevant to ODE's. Thus, the general solution is given by: $$\ x = C_{1}e^{t} + C_{2}e^{-t} , u= C_{1}e^{t} - C_{2}e^{-t}$$ Where $C_{1}$ and $C_{2}$ are constants of integration."
My Question:
what is the method in full to obtain $x$ and $u$ using the solution above?
First think of it like this:
\begin{align*} \frac{dx}{dt} &= u\\ \frac{du}{dt} &= x \end{align*}
can be written as
\begin{align*} x\frac{d}{dt} &= u\\ u\frac{d}{dt} &= x \end{align*}
Substituting $u$ in both systems seperatly will give:
\begin{align*} \frac{d^2x}{dt^2} &= u\\ \frac{d^2u}{dt^2} &= x \end{align*}
On rearranging we have:
\begin{align*} \frac{d^2x}{dt^2} & - u= 0\\ \frac{d^2u}{dt^2} & - x= 0 \end{align*}
Which should remind you of a general method for solving 2nd order homogeneous linear ODEs with constant coefficients. Hence we need to solve the quadratic characterisc equation:
$$m^2-1 = 0$$
Which has the roots:
$$m_{1,2}=1,-1$$
Since the roots are real and distinct we have for x:
$$x=C_1e^{M_1t}+C_2e^{M_2t}$$
Which gives you the solution your looking for $x$. However, i don't know why $u$ is negative, i could only find one solution.