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Let $\phi : \mathbb{R}^{1+d} \rightarrow \mathbb{R}^{m+1}$ be a classical solution to the nonlinear wave equation $$ \partial^2_{tt} \phi - \Delta \phi = ( - |\partial_t \phi|^2 + |\nabla \phi|^2 ) \phi \, . $$

Show that if the initial data $(\phi(0,x), \partial_t \phi (0,x) )$ obeys the conditions $$ \phi(0,x) \cdot \phi(0,x) = 1\, ; \phi(0,x) \cdot \partial_t \phi(0,x) = 0\, , $$ then we have $$ \phi(t,x) \cdot \phi(t,x) = 1\, ; \phi(t,x) \cdot \partial_t \phi(t,x) = 0\, , $$ for all times $t$.

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Yes. Let's define $\mu = |\phi|^2$. A direct computation reveals that $$ \partial_t^2 \mu - \Delta \mu = 2(1-\mu) ( |\partial_t u|^2 - |\nabla u|^2), $$ and so $\mu$ solves this wave equation with data $\mu(0) = 1$ and $\partial_t \mu(0) = 2\phi(0)\cdot \partial_t \phi(0)=0$.

The wave-type problem $$ \begin{cases} \partial_t^2 \xi -\Delta \xi + \rho \xi = f \\ \xi(0) = g, \partial_t \xi(0) = h \end{cases} $$ admits unique classical solutions as long as $\rho, f,g,h$ are "nice enough." For the above $\mu$ problem, since we assume $u$ is a classical solution, the corresponding $\rho, f, g, h$ are "nice enough," and so the problem for $\mu$ admits unique solutions.

However, we can easily check by hand that $\theta(x,t) =1$ identically solves the same PDE as $\mu$ and satisfies the same initial conditions. By uniqueness we then have that $\mu = \theta$ identically, i.e. $$ |\phi(x,t)|^2 = \mu(x,t) = \theta(x,t) =1 \text{ for all }x,t. $$

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    Thank you for your help. I got it ! By the way, I have anther question. Does the following equation preserve 0 solution ? \begin{align} \left\{ \begin{array}{l} \partial_t^2 \xi - \lambda_1 \partial_t \xi - \Delta \xi = \gamma \xi \\ \xi(0) = 0, \partial_t \xi (0) = 0 , \\ \end{array} \right. \end{align} where $\lambda_1$ is constant and $\gamma$ is "nice enough". – 2017-01-12