Equation with non constant coefficients with the method of characteristics?

Question

$$\dfrac{\partial u}{\partial t} +c\dfrac{\partial u}{\partial x} = 0$$

where $c$ is a constant, subject to $u(x,0) =\cosh(x)$

With the normal method I wrote down ($s$ is the parameter I'm using):

$\dfrac{du}{ds} = 0$ which gives me $u(s) = constant$

$\dfrac{dt}{ds} = 1$ which gives $t = s + d$ but at $s = 0$ this gives me $t = s$

$\dfrac{dx}{ds} = e^x$ which results in $x(s) = -\ln(-s-f)$, where $f$ is a constant. But then I don't really get anything when I set my initial condition $s = 0$

Any help would be appreciated :)

Answer 1

$$\dfrac{du}{dt} +c\dfrac{du}{dx} = 0$$ A characteristic equation comes from $\quad \dfrac{dt}{1}=\dfrac{dx}{c} \quad\to\quad x-ct=C_1$

Thus, the general solution of the PDE is $$u=F(x-ct)$$ where $F$ is any differentiable function.

With the initial condition

$$u(x,0)=F(x-c\:0)=F(x)=\cosh(x)$$ The function $F$ is determined :$\quad F(X)=\cosh(X)\quad$

With $X=x-ct$ the particular solution according to the initial condition is : $$u(x,t)=\cosh(x-ct)$$