Usbrth

Problem

Let $,OmegasubsetmathbbR^n,;Omega_textrmT=Omegatimesleft(0,Tright);$ and $,p>1.$

$$left(Pright);,left{
beginaligned
psi_t;-; Delta psi; &= ;psi^,p & &textrmon;;; Omega_textrmT &\
psi; &=; 0 & &textrmon;; partialOmegatimesleft(0,,Tright)&\
psi; &= ;g & &textrmon;; Omegatimesleftt;=;0right&
endalignedright.\$$

Assuming the existence of solution (the one that doesn't blow up in finite time), I want to prove a comparison principle to guarantee the uniqueness of solution.

My attempt

Given a sub-solution $u$ and super-solution $v$, I have defined the following function
$$w = v - u + varepsilon e^lambda tquad;left(forallvarepsilon>0right)$$

which verifies

$$w(x,t) geqvarepsilon e^lambda t>; 0, ;;(x,t)inpartialOmegatimesleft(0,,Tright) qquadqquad
w(x,0) geq varepsilon>0, ;; xinOmega$$

Since I want to prove that $,uleq v,$ on $,Omega_textrmT$, I suppose it exists $(x_0,t_0)$ such that
$$w(x_0,t_0)=0qquad w(x,t_0)geq0,;xin B(x_0,delta)subsetOmega $$
This implies $,u(x_0,t_0)>v(x_0,t_0),$ and also $,(x_0,t_0),$ is a minimum of $,w,$ which yields

$$w_t-Delta w-w^pleq0$$

Now, I try to prove the opposite.

Assuming $v$ attains a minimum in $,Omega_textrmT$ such that $,v=0,$ I get a contradiction, so I can state $u(x_0,t_0)>v(x_0,t_0)$ only if both $,u,$ and $,v,$ are positive at $,Utimesleftt=t_0right$. Then,

$$beginalign*w_t-Delta w-w^p =& ;left(v_t-Delta vright) - left(u_t-Delta uright) + lambdavarepsilon e^lambda t_0-left(v-u+varepsilon e^lambda t_0right)^p\
geq&;v^p-u^p + lambdavarepsilon e^lambda t_0-left(v-u+varepsilon e^lambda t_0right)^punderset?> 0endalign*$$

I think I am on the right way, but I don't know how to prove the last inequality. Any kind of help would be appreciated. Thank you in advance.

I referred to a professor, expert in PDE theory, and he gave me numerous hints to solve the problem. I'm going to post an aswer for those who may be interested.

The function $w$ I defined is a good choice to prove the Comparison principle, and the assumptions I made about $(x_0,t_0)inOmega_textrmT,$ are correct.
However, we have to fix some extra hypothesis:

• 
  $,t_0,$ must be the first $,tin(0,T)$ which verifies $w=0;$ and $,w(x,t_0)geq0 ,;xin B(x_0,delta)$.


• 
  $u, v,$ are bounded (that's why I said the solution shouldn't blow-up in finite time).


• Remove the "minimum principle" assumed about $v$.

Now, on the one hand, we have
$$left(w_t-Delta w- w^pright)(x_0,t_0)leq0$$

On the other hand, we have
$$left(w_t-Delta w-w^pright)(x_0,t_0)geq v^p(x_0,t_0)-u^p(x_0,t_0)+lambdavarepsilon e^lambda t_0$$

Since $,w(x_0,t_0)=0,$ we can state $,u(x_0,t_0)>v(x_0,t_0)$. Thus, we define the interval $I=left[v(x_0,t_0),u(x_0,t_0)right]$ and the function

beginalign*
f:I&;longrightarrow;mathbbR\
eta&;longmapsto;eta^p
endalign*

which verifies the Mean Value Theorem. So, it exists $xiintextrmint(I)$ such that
$$v^p(x_0,t_0)-u^p(x_0,t_0)+lambdavarepsilon e^lambda t_0=-pxi^p-1left(u-vright)(x_0,t_0)+lambdavarepsilon e^lambda t_0=varepsilon e^lambda t_0left(lambda-pxi^p-1right)$$

Supposing $,lambda>pxi^p-1$ we reach a contradiction. This implies $,w(x,t)>0;;forallvarepsilon>0$.

Making $,varepsilonsearrow0,$ we prove $,uleq v,$ on $,barOmega_textrmT$.