Suppose $u$ is a solution of the heat equation with the property that $|\int\limits_{-\infty}^{\infty}u(x,0)dx| < \infty$, and $u_{x}(x,t) \rightarrow 0$ as $x \rightarrow \pm \infty$. Then integrating the PDE, we find

$$\frac{d}{dt}\int\limits_{-\infty}^{\infty}u(x,t)dx = 0 \quad (1)$$

so that thet total heat energy is conserved:

$$\int\limits_{-\infty}^{\infty}u(x,t)dx = \text{ constant.} \quad (2)$$

Could someone explain how to get from the initial conditions to equation $(1)$?

I was thinking to use Leibniz integral rule, but got stuck at the following stage:

$$\frac{d}{dt}\int\limits_{-\infty}^{\infty}u(x,t)dx = \int\limits_{-\infty}^{\infty}u_t(x,t)dx \quad + \quad 0$$

as I see no mention on how the above should behave at the boundaries.

The notes can be found online at IITD notes on The Heat Equation.

  • $\begingroup$ There's no need to differentiate the integrand: Defining $f(t)\equiv \int_{-\infty}^\infty u(x,t)\,dx$, we have $f'(t)=0\implies f(t)=\text{constant}$. The initial conditions only enter if one wants to determine the total energy, not in showing that it's constant. $\endgroup$ – Semiclassical May 3 '16 at 23:37
  • $\begingroup$ Why is $f^{\prime}(t) = 0$ ? Could $u(x,t)$ not be a more complex function where after integrating out $x$ one would be left with a function which is not a constant in terms of $t$? $\endgroup$ – Gabor Bakos May 3 '16 at 23:44
  • $\begingroup$ Your equation (1) says precisely that $\dfrac{d}{dt}f(t)=f'(t)=0$. $\endgroup$ – Semiclassical May 4 '16 at 0:06
  • $\begingroup$ Why does one need the finiteness assumption? $\endgroup$ – mavavilj Oct 30 '17 at 12:51

$\int_{-\infty}^\infty u_t(x,t) dx = \int_{-\infty}^\infty u_{xx}(x,t) dx = u_x(\infty,t)-u_x(-\infty,t)$. Now you use your decay condition. (The more interesting question is to prove that decay condition a priori.)

  • $\begingroup$ First of all, thank you for the answer, I should have realised that I need to use the fact that it is a heat equation... Secondly, by decay condition do you mean: $u_{x}(x,t) \rightarrow 0$ as $x \rightarrow \pm \infty$? $\endgroup$ – Gabor Bakos May 3 '16 at 23:51
  • $\begingroup$ @GaborBakos Yes. $\endgroup$ – Ian May 3 '16 at 23:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.