Wave equation: $u_{tt}=au_{xx}, a>0$
We are looking for solutions of the wave equation of the form of a wave function.
We suppose that $u(x,t)=A \cos(kx- \omega t)$ is a solution of $u_{tt}=au_{xx}, a>0$.
We have:
$$u_x(x,t)=-Ak \sin(kx- \omega t)\\u_{xx}(x,t)=-Ak^2 \cos(kx-\omega t)\\u_t(x,t)=-\omega A \sin(kx- \omega t)\\ u_{tt}(x,t)=-\omega^2 A \cos(kx-\omega t)$$
Thus, it has to hold that:
$$-A \omega^2 \cos(kx-\omega t)=-a A k^2 \cos(kx \omega t)$$
or equivalently
$$A(\omega^2-ak^2) \cos(kx-\omega t)=0 (\forall x,t \in \mathbb{R})$$
The above holds if $\omega^2=ak^2$.
Thus $u(x,t)=Acos(k(x- \sqrt{a}t))$ is a solution of the wave equation of the form of a wave funcion, where $k>0$.
Remark:
At the above example we see that, if $k_1, k_2>0$ wave numbers with $k_1 \neq k_2$ then we have solutions of the form of a wavefunction
$$u_1(x,t)=A \cos(k_1(x- \sqrt{a}t))$$
$$u_2(x,t)=A \cos(k_2(x- \sqrt{a}t))$$
that "travel" with the same velocity $\sqrt{a}$.
Could you explain me why we can just change the wavenumber but let the circular frequency as it is? I.e. why can we write the following?
$$u_1(x,t)=A \cos(k_1(x- \sqrt{a}t))$$
$$u_2(x,t)=A \cos(k_2(x- \sqrt{a}t))$$