I am considering the following wave equation (for $\phi=\phi(x,t)$)
$$
\phi\_{tt} - \Delta \phi = a |\nabla \phi|^2, \quad (x,t) \in \mathbb{R}^3 \times \mathbb{R}
$$
where $\nabla$ is just spatial gradient, i.e., $\nabla \phi= (\partial\_{x\_1} \phi, \partial\_{x\_2} \phi, \partial\_{x\_3} \phi)$ and $a \in \mathbb{R}$. And assume that initial data $\phi(x,0)$ and $\partial\_t \phi(x,0)$ are regular enough and compactly supported.


This PDE in some sense is the complementary case of the PDE in John's blow-up theorem, where we have
$$
\phi\_{tt} - \Delta \phi = (\partial\_t\phi)^2.
$$


Now, I have the following questions: is this PDE well-posed? Or the solution will blow-up? Is there any paper on this PDE in 3+1 dimensions?



On a nonlinear wave equation
By replacing $\phi$ by $-\phi$ you can set $\alpha$ to be positive (or negative, if you wish). By replacing $\phi$ by $\lambda \phi$ you can rescale away $\alpha$. So you can set $\alpha$ to be either $+1$ or $-1$ as you wish.


So for now let's consider the situation you are solving
$$ \partial^2\_{tt}\phi - \Delta \phi = |\nabla \phi|^2 $$
First observation: suppose $\phi$ is a $C^3$ solution with compactly supported initial data, then $\phi$ is uniformly bounded from below on $\mathbb{R}^3\times [0,\infty)$. This is due to the fact that on $\mathbb{R}^3$ the fundamental solution to the wave equation is signed (a fact also used by John in his proof), so that the solution to the above equation is pointwise $\geq$ the solution to the linear equation with the same initial data. The latter is globally uniformly bounded, and hence $\phi$ is uniformly bounded below.


Now $v = e^\phi - 1$; you have that $v$ is also $C^3$, and there exists $\beta$ such that $v + 1 > \beta$ uniformly on $\mathbb{R}^3\times [0\infty)$, using the uniform bound on $\phi$.


Now we compute
$$ (\partial^2\_{tt} - \Delta) v = e^\phi (\partial^2\_{tt} - \Delta) \phi + e^\phi (\partial^2\_t \phi - |\nabla\phi|^2) = \frac{1}{1+v}( \partial\_t v)^2 $$


The lower bound above ensures that the denominator on the RHS is never zero.


Suppose now that $\phi$ is globally bounded *above* also: then $(1+v)$ is bounded above globally by some $M > 0$. Then the condition for Theorem 2 in John's 1981 paper is satisfied and the result follows.


So the remaining question is whether it is possible for $\phi$ to blow-up only at infinity and not at any finite time. I think there may be a way to rule this out, but that would require checking some more technical details so I'll have to report back later.


