# Intuition behind the maximum principle for the heat equation

The maximum principle for the heat equation say that if $$u$$ solves the heat equation on $$\Omega_T=\Omega \times (0,T]$$, then it will take its maximum on the parabolic boundary $$\Gamma_T=\overline{\Omega_T}-\Omega_T$$, and if the maximum is attained at some $$(x_0,t_0)$$ point inside, then $$u$$ is constant on $$\bar{\Omega}$$ for all $$t\leq t_0$$.

I wonder what is the physical intuition behind this theorem. Any help would be greatly appreciated.

• What does "parabolic boundary" mean? What is $\Gamma_T$? Which initial and boundary conditions are you using when solving the heat equation? Mar 6 at 2:11
• I just revised the question. Mar 6 at 2:14
• If you think about a metal rod that is insulated at both ends, the initial heat distribution is just going to spread out evenly over the bar until all points on the bar are the same temperature. You can also imagine what would happen with other types of boundary conditions (rather than insulated endpoints). Mar 6 at 3:29

The heat operator is "smoothing" in the sense that the temperature at any point is the average of the temperature in a ball around that point at a previous time, i.e. $$u(x_0,t_0) = \text{avg} \; u(B_r(x_0), s_0) \;\; \text{ for } s_0 < t_0.$$
The maximum principle says that the maximum temperature occurs either at the initial condition (temporal boundary), or on $$\bar{\Omega} - \Omega$$ (spatial boundary). Intuitively, this says that any "spikes" in temperature diffuse out over time, and that if at some time $$t_0$$ there is a maximum at $$x_0$$ in the interior of $$\Omega$$, then it cannot be a (new) spike.
This second point makes sense because because of the property $$u(x_0,t_0) = \text{avg} \; u(B_r(x_0), s_0) \;\; \text{ for } s_0 < t_0.$$
Since we're assuming $$u(x_0,t_0) = \max_{\bar{\Omega}_T} u$$, this means that $$u(B_r(x_0),s_0) \equiv u(x_0,t_0)$$ (i.e. the temperature throughout the ball is the same as the max temperature, since any point in the ball is not allowed to be hotter than $$u(x_0,t_0)$$, and yet the average needs to be equal). If you extend this argument to previous times, expanding the ball, we can see that the temperature must have been constant over all $$\bar{\Omega}$$ up to time $$t_0$$!