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Suppose that $u = u(x, t)$ and $v = v(x, t)$ have partial derivatives as follows: $$u_t=-v_x \quad\text{and}\quad v_t=-u_x.$$ Show that $u$ and $v$ are solutions of the wave equation: $$u_{tt}=u_{xx}.$$

My solution effort: taking derivative of $v_t=-u_x$ w.r.t $x$, $v_{xt}=-u_{xx}$, we get $u_{xx}=-v_{xt}$. Similiarly; $u_{tt}=-v_{xt}$. Then $u_{tt}=u_{xx}$.

Is this right?

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If $v$ is a $C^2$ function, then $v_{xt}=v_{tx}$ (Schwaz's theorem) and you are right.

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  • $\begingroup$ We don' t know that $v$ is a $C^2$ function. It is not given in the question. Then, how can we show that $u$ and $v$ are solutions of the wave equation? $\endgroup$
    – user383978
    Mar 24 '17 at 9:15
  • $\begingroup$ I suspect that the $C^2$ assumption was implicit in the question. $\endgroup$ Mar 24 '17 at 9:20
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This reduces to show that $$v_{tx} = v_{xt}$$ This is true (Schwarz's theorem) as soon as your function $v$ admits a second derivative (you don't need $C^2$ assumption).

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