Uniqueness of a parabolic-like PDE

Question

Let $\Omega$ be a bounded open set in $\mathbb{R}^n$ and $0 < T < \infty$. Let $\Omega_T = \Omega \times (0, T]$. Given any functions $f, g, h$ show that \begin{equation} u_t - \Delta u + |Du|^2 + \sin(u) = f(x,t), \, (x,t) \in \Omega_T \\ u(x,0)=g(x), \, x \in \Omega \\ u(x,t)=h(x,t), \, x \in \partial\Omega, \, t \in [0, T] \end{equation} has at most one classical solution.

So far I have tried to do the following: Assume there are two solutions, say $u_1$ and $u_2$, and let $v = u_1 - u_2$. Then we can write the equation governing $v$ as \begin{equation} v_t - \Delta v + |Du_1|^2 - |Du_2|^2 + \sin(u_1) - \sin (u_2) = 0, \, (x,t) \in \Omega_T \\ v(x,0)=0, \, x \in \Omega \\ v(x,t)=0, \, x \in \partial\Omega, \, t \in [0, T] \end{equation} If I notice that $|\sin(u_1) - \sin(u_2)| \leq |u_1 - u_2| = |v|$, then I have $$0 \leq v_t - \Delta v + |Du_1|^2 - |Du_2|^2 + \sin(u_1) - \sin (u_2) \leq v_t - \Delta v + |Du_1|^2 - |Du_2|^2 + |v|$$ However, this leads to a dead end for me. I wanted to use the weak maximum principle for parabolic operators to finish my proof, but I don't see how I could apply it in this case.

Answer 1

The maximum principle is a little simpler: rewrite the PDE as $$ v_t-\Delta v+\alpha Dv=-\beta v. $$ Applying the weak maximum principle** and zero initial conditions yields $$ \sup_{[0,s]\times\Omega} |v|\le s\,\sup_{[0,s]\times\Omega}|\beta|\,\sup_{[0,s]\times\Omega}|v|. $$ Choosing $s$ small enough yields the result.

**Here we will recall why the right hand side depends on $s$. Putting $v=w+Kt$ in the PDE's left hand side, we get $$ Lw:=w_t-\Delta w+\alpha Dw=-\beta v-K. $$ Fixing $s>0$ and choosing $K=\sup_{[0,s]\times\Omega}|\beta|\sup_{[0,s]\times\Omega}|v|$ yields $$L(v-Kt)\le 0,\quad 0\le t\le s.$$ Replacing $K$ with $-K$ and arguing as before gives $$ L(v+Kt)\ge 0,\quad 0\le t\le s. $$ Applying the maximum principle to the zero initial condition yields $$ v-Kt\le 0,\quad 0\le t\le s, $$ and $$ v+Kt\ge 0,\quad 0\le t\le s. $$ This implies $|v|\le Kt$ if $0\le t\le s$.

Answer 2

First, assume there are two solutions, so $u_1$ and $u_2$. Let $v = u_1-u_2$. Then our PDE for $v$ is \begin{equation} v_t−\Delta v +|Du_1|^2−|Du_2|^2+ \sin(u_1)− \sin(u_2)=0 \, (x,t) \in \Omega \\ v(x,0)=0, x \in \Omega \\ v(x,t)=0, x \in \partial \Omega, t \in [0,T] \end{equation} Notice that we can find a function $B$ such that $$|Du_1|^2 - |Du_2|^2 = \int_{0}^1 \frac{d}{ds} B(sDu_1(x,t)+(1-s)Du_2(x,t))ds$$ $$=\int_{0}^1 D\{sDu_1(x,t)+(1-s)Du_2(x,t)\}ds \cdot (Du_1-Du_2) = \alpha(u_1,u_2) \cdot Dw$$ Similarly, $$\sin(u_1)-\sin(u_2)=\int_{0}^{1}\frac{d}{ds}\{\sin(su_1(x,t)+(1-s)u_2(x,t))\}ds \cdot (u - v) = \beta(u_1,u_2) w$$ Hence, we have \begin{equation} v_t−\Delta v + \alpha(u_1,u_2)Dv + \beta(u_1,u_2)v=0 \, (x,t) \in \Omega \\ v(x,0)=0, x \in \Omega \\ v(x,t)=0, x \in \partial \Omega, t \in [0,T] \end{equation} Define $$E(t) = \int_{\Omega}v^2 dx$$ Then $$E'(t) = 2\bigg(\int_{\Omega}vv_t \,dx = 2\int_{\Omega} v\Delta v \,dx - \int_{\Omega}\beta(u_1,u_2)v^2 \,dx -\int_{\Omega}\alpha(u_1,u_2)vDv \,dx\bigg)$$ By Green's Identity, $$2\int_{\Omega} v\Delta v \, dx = -2\int_{\Omega}|Dv|^2 \, dx$$ The Cauchy Inequality gives $$-\int_{\Omega}\alpha(u_1,u_2)vDv \,dx \leq \int_{\Omega}\alpha^2(u_1,u_2)v^2 \, dx + \int_{\Omega}|Dv|^2 \, dx$$ Combining all of this gives $$E'(t) \leq -2\int_{\Omega}|Dv|^2 \, dx + 2\int_{\Omega}\alpha^2(u_1,u_2)v^2 \, dx + 2\int_{\Omega}|Dv|^2 \, dx - \int_{\Omega}\beta(u_1,u_2)v^2 \,dx$$ $$= 2\int_{\Omega}(\alpha(u_1,u_2)^2 - \beta(u_1,u_2))v^2 \, dx$$ Continuity implies that $\alpha(u_1,u_2)^2 - \beta(u_1,u_2)$ is bounded on $\overline{\Omega}$. It follows that $E'(t) \leq M E(t)$ for some $M > 0$. Furthermore, $E(0) = 0$ since $v(0, x) = 0$ for all $x \in \Omega$. We apply Gronwall's Inequality to get $v^2 \equiv 0$. Hence, $v \equiv 0$, so $u_1 = u_2$.