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Let's suppose we know the following equality: $$\vec{v_1}e^{i(k_1x\omega_1t)}+\vec{v_2}e^{i(k_2x-\omega_2t)}=\vec{v_3}e^{i(k_3x-\omega_3t)}$$ is true for every real value of $x$ and for every positive real value of $t$. I think it is very trivial to show that under this assumption $$\omega_1=\omega_2=\omega_3$$ and: $$k_1=k_2=k_3,$$ but I don't seem to be able to prove it in a fairly formal way. Sorry if this is a dumb question.

davise
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1 Answers1

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It's not entirely true, but it is true under a few basic conditions.

I won't produce an entire proof, but let's look at a reduction of the problem by focusing on the behaviour at $x = 0$. That leaves us with:

$$\vec{v}_1 e^{-i \omega_1 t} + \vec{v}_2 e^{-i \omega_2 t} = \vec{v}_3 e^{-i \omega_3 t}$$

Now, taking an approach similar to using the Wronskian, let's differentiate this identity twice with respect to $t$, giving us:

$$\begin{eqnarray} -i \omega_1 \vec{v}_1 e^{-i \omega_1 t} -i \omega_2 \vec{v}_2 e^{-i \omega_2 t} & = & -i \omega_3 \vec{v}_3 e^{-i \omega_3 t} \\ - \omega_1^2 \vec{v}_1 e^{-i \omega t} - \omega_2^2 \vec{v}_2 e^{-i \omega_2 t} & = & - \omega_3^2 \vec{v}_3 e^{-i \omega_3 t} \end{eqnarray}$$

This gives us three linear equations, and if we pick a few "good" values of $t$ we can show that there are only a few ways to make everything consistent, those being:

  • $\omega_1 = \omega_2 = \omega_3$

  • $\vec{v}_1 = \vec{v}_2 = \vec{v}_3 = \vec{0}$

  • One of the $\vec{v}_i = \vec{0}$, and the other two terms have equal phase and matching coefficients

You can then extend this to look at the $x$ part similarly.

ConMan
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