1
$\begingroup$

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?

$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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, 2017 at 9:15
  • $\begingroup$ I suspect that the $C^2$ assumption was implicit in the question. $\endgroup$
    – Giuseppe Negro
    Mar 24, 2017 at 9:20
1
$\begingroup$

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).

$\endgroup$

You must log in to answer this question.