The one dimensional linear transport equation $$\frac{\partial u}{\partial t}+c\frac{\partial u}{\partial x}=0$$ has solution $u(t,x)=v(\xi)$ where $\xi=x-ct$ is a characteristic variable. Now, the characteristic line $x=ct+\xi$ can be interpreted as the path an observer must follow through one dimensional space and time in order to observe the above wave as stationary. Additionally, the absolute value of the characteristic variable $\xi$ can be interpreted as the 'distance' the observer is from the wave, since we can choose any distance to be from the wave, it is a parametric variable. Suppose we choose the distance $\xi=k$ for some nonzero $k$. Then the path the observer must follow (in order to view the wave as stationary) is the curve $x(t)=ct+k$. Parametrizing the solution to follow this curve gives: $$u(t,ct+k)=v((ct+k)-ct)=v(k)$$ What has me thinking in circles is the following question. Let $u(t,x)=e^{-(x-ct)^2}$, then $u(t,ct+k)=e^{-k^2}$. As $k\rightarrow \infty$, $u(t,ct+k)\rightarrow 0$. In other words, the position of the observer changes the amplitude of the solution while it is on the characteristic curve. Why? What is the physical intuition behind this? Why does the wave 'decay' as the obeserver moves further away from the wave in this particular example.
What really has me boiled is this same question applied to d'Alembert's solution of the one dimensional wave equation $$\frac{\partial^2 u}{\partial t}=c^2\frac{\partial^2 u}{\partial x^2}$$ where $c$ is a constant. The solution takes the form $$u(t,x)=p(\xi)+q(\eta)=\frac{f(\xi)+f(\eta)}{2}-\frac{1}{2c}\int_0^\xi{g(z)}dz+\frac{1}{2c}\int_0^\eta {g(z)}dz$$ The first integral term of the equation above can be interpreted as the additional displacement that the velocity of the wave adds to $f(\xi)/2$. Notice however, that it is an integral from $0$ to $\xi$, so the same problem arises. I don't understand why the position of the observer, namely $\xi$, affects the amplitude of the wave. Thank you for reading this long post, and any clarification is helpful!