I'm trying to solve question 4.6 using maximum principle
Suppose that $u(x, t)>0$ solves $$ \frac{\partial u}{\partial t}=\frac{\partial}{\partial x}\left(u \frac{\partial u}{\partial x}\right)+u $$ in a region $D$ bounded by the lines $t=0$ and $t=\tau$ and two non-intersecting smooth curves $C_1$ and $C_2$.
Prove that if a solution exists then $u(x, t)$ attains its minimum value on $t=0$ or on one of the curves $C_1$ and $C_2$.
My working
Let
$$
∂_P D=(\left[a_1, a_2\right] ×\{0\})∪C_1∪C_2.
$$
Since $D$ is compact, a continuous function $u$ has a minimum point in $D$.
First suppose $u>0$ and $∂_tu>∂_x(u∂_xu)+u=(∂_xu)^2+u∂_{xx}u+u$
If $u(x,t)$ has an internal minimum, then at this point
$$∂_xu=∂_tu=0 ∂_{xx}u≥0$$
If $u(x,t)$ has a minimum at a point on $t=τ$, then at this point
$$∂_xu=0 ∂_tu≤0 ∂_{xx}u≥0$$
Both contradicts $∂_tu>∂_x(u∂_xu)+u$ and it follows that minimum point of $u$ must be on $∂_PD$.
Suppose now that $∂_tu=∂_x(u∂_xu)+u$, then define $v(x,t)=u(x,t)-ϵ$ for $ϵ>0$.
Then $v$ satisfies $$\tag1 ∂_tv-∂_x(v∂_xv)-v=[∂_tu-∂_x(u∂_xu)-u]+(1+∂_{xx}u)ϵ>0$$ By the first part, the minimum point of $v$ is on $∂_PD$. $$u>v≥\min_{∂_PD}u-ϵ$$ Let $ϵ→0$ we conclude that $u$ takes its minimum on $∂_PD$.
But my solution has a flaw: in $(1)$ why do we have $1+∂_{xx}u\ge0$ ?
I find a similar question: Maximum principle for a nonlinear heat equation
Below is Theorem 4.3. in $\S$4.4.2 The maximum principle for heat equation
Suppose that $u(x, t)$ satisfies $$ u_t-u_{x x} \leq 0 $$ in a region $D_\tau$ bounded by the lines $t=0, t=\tau>0$, and two non-intersecting smooth curves $C_1$ and $C_2$ that are nowhere parallel to the x-axis. Suppose also that $f \leq 0$ in $D_\tau$. Then $u$ takes its maximum value either on $t=0$ or on one of the curves $C_1$ or $C_2$.
Proof: We first observe that since $u$ is a continuous function on a compact set $\bar{D}_\tau$ it will achieve its maximum on $\bar{D}_\tau$
Suppose first that $u_t-u_{x x}<0$ in $D_\tau$. At an internal maximum inside $D_\tau, u$ must satisfy $$ u_x=u_t=0 u_{x x} \leq 0,\left(u_{t t} \leq 0\right) $$ On the other hand, if $u$ has a maximum at a point on $t=\tau$, then there it must satisfy $$ u_x=0, u_t \geq 0, u_{x x} \leq 0 $$ With $u_t-u_{x x}$ assumed to be strictly negative, both of these lead to contradictions, and it follows that $u$ must take its maximum value somewhere on $\partial D_\tau$ but not on $t=\tau$. We are done. Suppose now that $u_t-u_{x x} \leq 0$, then define $$ v(x, t)=u(x, t)+\frac{\epsilon}{2} x^2 $$ where $\epsilon$ is a positive constant. Then $v$ satisfies $$ v_t-v_{x x}=u_t-u_{x x}-\epsilon<0 $$ in $D_\tau$. So by the earlier step $v$ takes its maximum value on $\partial D_\tau$ but not on $t=\tau$. Now if the maximum value of $u$ over these three portions of $\partial D_\tau$ is $M$, and the maximum value of $|x|$ on $C_1$ and $C_2$ is $L$, then $$ u \leq v \leq \frac{L^2 \epsilon}{2}+M $$ Now we let $\epsilon \rightarrow 0$ and conclude that $u \leq M$, i.e. $\mathrm{u}$ takes its maximum value on $\partial D_\tau$ but not on $t=\tau.\quad\square$
If $u_t-u_{x x} \geq 0$ in $D_\tau$, then a similar argument shows that $u$ attains its minimum value on $\partial D_\tau$ but not on $t=\tau$. Thus, for the homogeneous equation (the heat equation) $u$ attains both its maximum and its minimum values on $\partial D_\tau$ but not on $t=\tau$