MHB Why can we only change the wavenumber?

[evinda]

Hello! (Wave)

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 frequencyas it is?
I.e. why can we write the following? (Thinking)$$u_1(x,t)=A \cos(k_1(x- \sqrt{a}t))$$$$u_2(x,t)=A \cos(k_2(x- \sqrt{a}t))$$

 

[chisigma]

Hello! (Wave)

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 frequencyas it is?
I.e. why can we write the following? (Thinking)$$u_1(x,t)=A \cos(k_1(x- \sqrt{a}t))$$$$u_2(x,t)=A \cos(k_2(x- \sqrt{a}t))$$

All becomes 'all right' if You write the equation in the more usual form...

$\displaystyle u_{t t} = a^{2}\ u_{x x}\ (1)$

... so that the two independent solution are...

$\displaystyle u_{1} = A\ \cos \{k_{1}\ (x - a\ t) \}$

$\displaystyle u_{2} = A\ \cos \{k_{2}\ (x + a\ t)\}\ (2)$

The first is a wave traveling from left to right and the second a wave traveling from right to left...

Kind regards

$\chi$ $\sigma$

 

[evinda]

I think that we can also write the wave equation as $u_{tt}=au_{xx}$.

The solution is of the form $u(x,t)=A \cos(kx- \omega t)$ and it has to hold that $\omega^2=ak^2$.

If we suppose that $k=k_1$ then $\omega^2=ak_1^2 \Rightarrow \omega= \sqrt{a} k_1$.
And if $k=k_2$ then $\omega= \sqrt{a} k_2$.

So, for the first case $u(x,t)=A \cos(k_1x- \sqrt{a} k_1 t)=A \cos(k_1(x- \sqrt{a}t))$ and for the second case $u(x,t)=A \cos(k_2(x- \sqrt{a}t))$. So, we see that these two solutions have the same velocity, despite of the fact that the wave numbers are different.