Show at most one solution exists for Cauchy Problem Consider the Cauchy problem in $\mathbb{R^n},$
$$ \begin{cases} u_t - \Delta u + | u_{x_1} | = 0 & t > 0, x \in \mathbb{R^n}  \\ u(x,0) = g(x) & t = 0 \end{cases}$$
where $g$ is continuous with compact support.  Show at most one solution exists that solves the above problem and decays to zero as $|x| \to \infty.$

I feel like we ultimately want to use the fact that if $u_t - \Delta u \leq 0,$ the max of $u$ occurs on the parabolic boundary, which in this case would be at $t = 0.$  But if we let $w = u - v,$ then we get a zero initial condition but $w_t - \Delta w = |v_{x_1}| - |u_{x_1}|$ and I'm not sure if we can say anything about this RHS.  Does anyone see a better way to go about this?
Thanks in advance.
 A: My proposed solution:
Let $w = u-v.$ Then $w_t - \Delta w = |v_{x_1}| - |u_{x_1}|.$
By the reverse triangle inequality, we have,
$$|w_{x_1}| = |u_{x_1}-v_{x_1}| \geq \big| |u_{x_1}| - |v_{x_1}| \big| \geq  |u_{x_1}| - |v_{x_1}| = -( |v_{x_1}| - |u_{x_1}|),$$
meaning
$$ -|w_{x_1}| \leq w_t - \Delta w \;\text{ or } \; w_t - \Delta w + |w_{x_1}| \geq 0.$$
My claim, which I'll prove after this solution, is that $\min_{U_T} w = \min_{\partial U_T} w,$ where $U_T = \{ (x,t) \, | \, t \geq 0, x \in \mathbb{R}^n\}.$  We see that if this is true, then $min_{U_T} u-v = 0,$ since $w$ decays at infinity and $w(x,0) = 0.$
Similarly, we could go through the same exact steps for $w = v-u,$ and conclude that $min_{U_T} v-u = -max_{U_T} u-v = 0,$ meaning that $u-v=0$ and therefore we have uniqueness.

Proof of claim: Let $z = w + \mu e^{t}$ where in the end, we will let $\mu$ go to zero. Differentiating, $$z_t = w_t + \mu e^{t} \geq \Delta z - |z_{x_1}| + \mu e^{t}.$$
Assume $z$ takes a minimum inside the domain.  Then $z_t \leq 0,\; -\Delta z \leq 0, \;z_{x_1} = 0,$ so $z_t - \Delta z - \mu e^{t} < 0$ since $-\mu e^{t} < 0,$ thus a contradiction with the fact that $z_t - \Delta z + |z_{x_1}|  - \mu e^{t} \geq 0$ for all $(x,t).$ Letting $\mu \to 0,$ we see that $w$ must take its minimum on the boundary of the domain, meaning at $t = 0.$

If anyone spots any errors, please comment!
