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Let $u \in C^{2,1}(\Omega_{T}) \cap C(\partial_{p} \Omega_{T})$ and $$ u_{t} - k\Delta u = f(x,t),$$ with $f(x,t)$ continuous.One can show that for all $(x_{0}, t_0) \in \Omega_{T}$, $$ -t_{0} \max_{\overline{\Omega_{t_{0}}}}|f| - \max_{\partial_p \Omega_{t_{0}}}|u| \le u(x_0 , t_0) \le t_0 \max_{\overline{\Omega_{t_0}}}|f| + \max_{\partial_p \Omega_{t_0}}|u|$$.

Let $v,w \in C^{2,1}(\Omega_{T}) \cap C(\partial_{p} \Omega_{T})$ and \begin{gather} v_{t} - k\Delta v = f_{1}(x,t) \quad \text{in} \ \ \Omega_{t}\\ w_{t} - k\Delta w = f_{2}(x,t) \quad \text{in} \ \ \Omega_{t} \end{gather} Prove that for any $(x_{0}, t_{0}) \in \Omega_{T}$, $$ \lvert v(x_{0}, t_{0}) - w(x_{0},t_{0})\rvert \le t \max_{\overline{\Omega_{t_{0}}}}\lvert f_{1} - f_{2} \rvert + \max_{\partial_{p}\Omega_{t_{0}}}\lvert v - w \rvert.$$

Attempt at solution

\begin{align} \lvert v(x_{0},t_{0}) - w(x_{0},t_{0})\rvert &\le \lvert v(x_{0},t_{0}) \rvert + \lvert w(x_{0},t_{0}) \rvert \\ &\le \lvert t_0 \max_{\overline{\Omega_{t_0}}}|f_{1}| + \max_{\partial_p \Omega_{t_0}}|v| \rvert + \lvert t_0 \max_{\overline{\Omega_{t_0}}}|f_{2}| + \max_{\partial_p \Omega_{t_0}}|w| \rvert \\ &\le t_0 \max_{\overline{\Omega_{t_0}}}|f_{1}| + t_{0}\max_{\overline{\Omega_{t_0}}}|f_{2}| +\max_{\partial_p \Omega_{t_0}}|v| + \max_{\partial_p \Omega_{t_0}}|w|\\ &\le t_{0}\max_{\overline{\Omega_{t_0}}}|f_{1}+f_{2}| + \max_{\partial_p \Omega_{t_0}}|v+w| \end{align}

I don't understand where I'm going wrong , I'm getting $f_{1}+f_{2}$ instead of $f_{1}-f_{2}$ and similarly for $v+w \to v-w$. Anyone knows where I'm going wrong ?

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    $\begingroup$ You can’t use the triangle inequality right away; in doing so, you are disregarding any cancellation caused by the subtraction. Instead just set $u=v-w$, wrote down the equation solved by $u$ and apply the previous result. $\endgroup$
    – User8128
    Sep 15, 2020 at 22:38
  • $\begingroup$ Thank you for your reply, but I'm not quite sure to understand the procedure you're suggesting to me. $u=v-w$ solves which equation? $\endgroup$
    – hexaquark
    Sep 15, 2020 at 23:08
  • $\begingroup$ I added an answer explaining the process. $\endgroup$
    – User8128
    Sep 16, 2020 at 1:41

1 Answer 1

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Set $u = v - w$. Then $$u_t - k\Delta u = v_t - w_t - k\Delta v + k\Delta w = (v_t - k\Delta v) - (w_t - k\Delta w) = f_1 - f_2.$$ That is $u$ solves the equation $$u_t - k\Delta u = f_1 - f_2. $$ Thus applying the previous result yields $$u(x_0,t_0) \le t_0 \max_{\overline \Omega_{t_0}} \lvert f_1 - f_2 \rvert + \max_{\overline{\partial_p \Omega_{t_0}}} \lvert u \rvert.$$ But putting in the definition of $u$, this gives $$v(x_0,t_0) - w(x_0,t_0) \le t_0 \max_{\overline \Omega_{t_0}} \lvert f_1 - f_2 \rvert + \max_{\overline{\partial_p \Omega_{t_0}}} \lvert v-w \rvert.$$

Now do the same thing with $w -v$ to get the reverse inequality.

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    $\begingroup$ Hey , looking back I wonder, do we assume $v-w$ solves $u_{t} -k\Delta u$ since the PDE is linear and hence a linear combination of solutions is also a solution ? $\endgroup$
    – hexaquark
    Sep 16, 2020 at 21:16
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    $\begingroup$ No, you don’t assume that: you show that using the computation above. $\endgroup$
    – User8128
    Sep 17, 2020 at 3:50

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