# Heat Equation - Energy (Neumann Condition)

Consider the heat equation

$$$$u_t - u_{xx} = 0, \ \ \ x \in [0, L],\ t > 0.$$$$ and $$$$E(t) = \int^L_0 (u(x,t))^2 dx.$$$$ If $$u$$ satisfies the neumann condition $$u_x(0, t) = u_x(L, t) = 0$$, show that $$E(t)$$ is constant.

Attempt: Integrating by parts, we have $$$$\frac{1}{2}E'(t) = \int^L_0 uu_t dx = \int^L_0 uu_{xx} dx = uu_x|^L_0 - \int^L_0 u_x^2 dx.$$$$

Integrating the term $$\int^L_0 u_x^2 dx$$ by parts, i arrive at the same place. Help!

• Hint: Multiply the equation by $u$ and then integrate. Commented Dec 6, 2021 at 23:54

$$E(t)$$ is indeed non-constant: e.g.
$$u(x, t) = \exp \left(-\frac{\pi^2t}{L^2} \right) \cos \left( \frac{\pi x}{L}\right)$$
satisfies the heat equation $$u_t = u_{xx}$$ and $$u_x ( 0, t) = u_x (L, t) = 0,$$ but
$$E(t) = \int_0^L u^2(x, t)dx = \exp \left(-\frac{2\pi^2t}{L^2} \right)\int_0^L \cos ^2\left( \frac{\pi x}{L}\right) dx$$
is not constant in $$t$$. What you can show is only that $$E(t)$$ is non-increasing (as you did).