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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.

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

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  • $\begingroup$ 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

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