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Maybe the question can simply be answered by the observation that a wave like

$$\Psi(x,t)=A \cos(x)-A \cos(x+\omega\ t),$$

where the two cosines cancel at periodic times $$t_n=\frac{2\pi}{\omega}n\ \ \longrightarrow\ \ \Psi(x,t_n)=0,$$ still has nonvanishing kinetic energy, if it looks something like $$E=\sum_\mu\left(\frac{\partial \Psi}{\partial x^\mu} \right)^2+\ ...$$

You would really have to construct an example.

Since non-dissipative waves whose equations of motions can be formulated by a Lagrangian will have an energy associated to them, as you say, you'd have to find a sitation/theory without an energy quantity. The energy is related to the wave by its relation to the equation of motion. So if the energy is defined as that which is constant because of time symmetriy and you don't have such a thing, then there is no question.

Also don't make the mistake and talk about about two different waves with different energy. If you have a linear problem, the wave will be "one wave" in the energy expression, whereever its parts may wander around.

edit: See also the other answer(s) for a discussion of a more physical reading of the question.

Maybe the question can simply be answered by the observation that a wave like

$$\Psi(x,t)=A \cos(x)-A \cos(x+\omega\ t),$$

where the two cosines cancel at periodic times $$t_n=\frac{2\pi}{\omega}n\ \ \longrightarrow\ \ \Psi(x,t_n)=0,$$ still has nonvanishing kinetic energy, if it looks something like $$E=\sum_\mu\left(\frac{\partial \Psi}{\partial x^\mu} \right)^2+\ ...$$

You really would have to construct an example.

Since non-dissipative waves whose equations of motions can be formulated by a Lagrangian will have an energy associated to them, as you say, you'd have to find a situation/theory without an energy quantity. The energy is related to the wave by its relation to the equation of motion. So if the energy is defined as that which is constant because of time symmetry and you don't have such a thing, then there is no question.

Also don't make the mistake and talk about about two different waves with different energy. If you have a linear problem, the wave will be "one wave" in the energy expression, wherever its parts may wander around.

edit: See also the other answer(s) for a discussion of a more physical reading of the question.

Maybe the question can simply be answered by the observation that a "wave" like

$$\Psi(x,t)=\cos(x)+\cos(x+\omega\ t),$$

where the two cosines cancel at times $$t_n=\frac{2\pi}{\omega}n\ \ \longrightarrow\ \ \Psi(x,t_n)=0,$$ still has nonvanishing kinetic energy, if it looks something like $$E=\sum_\mu\left(\frac{\partial \Psi}{\partial x^\mu} \right)^2+\ ...$$

You would really have to construct an example.

Since non-dissipative waves whose equations of motions can be formulated by a Lagrangian will have an energy associated to them, as you say, you'd have to find a sitation/theory without an energy quantity. The energy is related to the wave by its relation to the equation of motion. So if the energy is defined as that which is constant because of time symmetriy and you don't have such a thing, then there is no question.

Also don't make the mistake and talk about about two different waves with different energy. If you have a linear problem, the wave will be "one wave" in the energy expression, whereever its parts may wander around.

edit: See also the other answer(s) for a discussion of a more physical reading of the question.

Maybe the question can simply be answered by the observation that a wave like

$$\Psi(x,t)=A \cos(x)-A \cos(x+\omega\ t),$$

where the two cosines cancel at periodic times $$t_n=\frac{2\pi}{\omega}n\ \ \longrightarrow\ \ \Psi(x,t_n)=0,$$ still has nonvanishing kinetic energy, if it looks something like $$E=\sum_\mu\left(\frac{\partial \Psi}{\partial x^\mu} \right)^2+\ ...$$

You would really have to construct an example.

Since non-dissipative waves whose equations of motions can be formulated by a Lagrangian will have an energy associated to them, as you say, you'd have to find a sitation/theory without an energy quantity. The energy is related to the wave by its relation to the equation of motion. So if the energy is defined as that which is constant because of time symmetriy and you don't have such a thing, then there is no question.

Also don't make the mistake and talk about about two different waves with different energy. If you have a linear problem, the wave will be "one wave" in the energy expression, whereever its parts may wander around.

edit: See also the other answer(s) for a discussion of a more physical reading of the question.

Maybe the question can simply be answered by the observation that a "wave" like

$$\Psi(x,t)=\cos(x)+\cos(x+\omega\ t),$$

where the two cosines cancel at times $$t_n=\frac{2\pi}{\omega}n\ \ \longrightarrow\ \ \Psi(x,t_n)=0,$$ still has nonvanishing kinetic energy, if it looks something like $$E=\sum_\mu\left(\frac{\partial \Psi}{\partial x^\mu} \right)^2+\ ...$$

You would really have to construct an example.

Since non-dissipative waves whose equations of motions can be formulated by a Lagrangian will have an energy associated to them, as you say, you'd have to find a sitation/theory without an energy quantity. The energy is related to the wave by its relation to the equation of motion. So if the energy is defined as that which is constant because of time symmetriy and you don't have such a thing, then there is no question.

Also don't make the mistake and talk about about two different waves with different energy. If you have a linear problem, the wave will be "one wave" in the energy expression, whereever its party may wander around.

See also the other answers for a discussion of a more physical interpretation of the question.

Maybe the question can simply be answered by the observation that a "wave" like

$$\Psi(x,t)=\cos(x)+\cos(x+\omega\ t),$$

where the two cosines cancel at times $$t_n=\frac{2\pi}{\omega}n\ \ \longrightarrow\ \ \Psi(x,t_n)=0,$$ still has nonvanishing kinetic energy, if it looks something like $$E=\sum_\mu\left(\frac{\partial \Psi}{\partial x^\mu} \right)^2+\ ...$$

You would really have to construct an example.

Since non-dissipative waves whose equations of motions can be formulated by a Lagrangian will have an energy associated to them, as you say, you'd have to find a sitation/theory without an energy quantity. The energy is related to the wave by its relation to the equation of motion. So if the energy is defined as that which is constant because of time symmetriy and you don't have such a thing, then there is no question.

Also don't make the mistake and talk about about two different waves with different energy. If you have a linear problem, the wave will be "one wave" in the energy expression, whereever its parts may wander around.

edit: See also the other answer(s) for a discussion of a more physical reading of the question.