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  1. MHB Master
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    #1
    Hey!!

    Let $\Omega$ a bounded space. Using the maximum principle I have to show that the following problem has an unique solution.

    $$u_t(x, t)-\Delta u(x, t)=f(x, t), x \in \Omega,t>0\\ u(x, t)=h(x, t), x\in \partial{\Omega}, t>0 \\ u(x, 0)=g(x), x \in \Omega$$

    I have done the following:

    We suppose that $u_1$,$u_2$ are two different solutions of the problem, so $w=u_1−u_2$ solves the following two problems:

    $$w_t(x,t)-\Delta w(x,t), x \in \Omega, t>0\\ w(x, t)=0, x \in \partial{\Omega}, t>0\\ w(x, 0)=0, x \in \Omega$$
    and
    $$-w_t(x,t)-\Delta (-w(x,t)), x \in \Omega, t>0\\ -w(x, t)=0, x \in \partial{\Omega}, t>0\\ -w(x, 0)=0, x \in \Omega$$

    Since $w_t−\Delta w \leq 0$ from the maximum principle for $w$ we have that
    $$\max_{x \in \Omega, t \in [0, T]}w(x, t)=\max_{(\Omega \times \{0\})\cup (\partial{\Omega} \times [0, T])}w(x, t)=0$$

    Since $−w_t−\Delta (−w) \leq 0$ from the maximum principle for $−w$ we have that
    $$\max_{x \in \Omega, t \in [0, T]}(-w(x, t))=\max_{(\Omega \times \{0\})\cup (\partial{\Omega} \times [0, T])}(-w(x, t))=0$$

    Since $\max (−w)=\min (w)$ we have that $w \equiv 0$. So, $u_1=u_2$.

    Is this correct??

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    #2
    Hii!!!

    Quote Originally Posted by mathmari View Post
    $$w_t(x,t)-\Delta w(x,t), x \in \Omega, t>0$$
    and
    $$-w_t(x,t)-\Delta (-w(x,t)), x \in \Omega, t>0$$
    I think that should be $w_t(x,t)-\Delta w(x,t) = 0$.


    Quote Originally Posted by mathmari View Post
    Since $w_t−\Delta w \leq 0$ from the maximum principle for $w$ we have that
    $$\max_{x \in \Omega, t \in [0, T]}w(x, t)=\max_{(\Omega \times \{0\})\cup (\partial{\Omega} \times [0, T])}w(x, t)=0$$
    I've looked up the maximum principle and found for instance on :

    Let $u = u(x), x = (x1, , xn)$ be a $C^2$ function which satisfies the differential inequality
    $$Lu = \sum_{ij} a_{ij}(x)\frac{\partial^2 u}{\partial x_i\partial x_j} +
    \sum_i b_i\frac{\partial u}{\partial x_i} \geq 0$$
    in an open domain $Ω$, where the symmetric matrix $a_{ij} = a_{ij}(x)$ is locally uniformly positive definite in $Ω$ and the coefficients $a_{ij}, b_i = b_i(x)$ are locally bounded. If $u$ takes a maximum value $M$ in $Ω$ then $u ≡ M$.

    If we would use this version, then I think all conditions of the proposition should be covered.

    Btw, which version of the maximum principle do you have?

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