I'm having a problem trying to compute the total heat in a fully insulated rod. Here is the important parts of the problem statement I've been given:

The one-dimensional heat conduction equation, which governs the spatial and temporal evolution of the temperature in a bar, is given by the following partial differential equation (PDE):

$$\frac{∂^2u(x,t)}{∂x^2}=D\frac{∂u(x,t)}{∂t} , D>0.$$

for$$0 < x <L$$

The bar is fully insulated. So that would be that the boundaries are insulated, which means $$\frac{∂u}{∂x}=0$$ at $$ x=0$$ and $$ x =L$$

Consider the initial temperature: $$u(x,0)=f(x)=−x(x−L)$$

Use Vector Calculus to compute the total heat in the bar for all time t.

I know that the integral to solve for the total heat is: $$∫^{L}_{0}u(x,t)dx, t\ge0$$

I've solved for the Sum of the Series for $u(x,t)$ but I'm not sure how to use that or anything else to solve for the total heat.

Any help would be greatly appreciated!


Not sure just how much vector calculus I'm using here, but consider:

Denote the total heat in the bar at time $t \ge 0$ by $Q(t)$; thus

$Q(t) = \int_0^L u(x, t) dx. \tag{1}$

Then we have

$\dfrac{dQ(t)}{dt} =\dfrac{d}{dt} \int_0^L u(x, t) dx$ $=\int_0^L \dfrac{\partial u(x, t)}{\partial t} dx; \tag{2}$

now since

$\dfrac{\partial^2 u(x,t)}{\partial x^2}=D\dfrac{\partial u(x,t)}{\partial t} \tag{3}$

with $D > 0$, (2) becomes

$\dfrac{dQ(t)}{dt} = D^{-1} \int_0^L \dfrac{\partial^2 u(x, t)}{\partial x^2} dx. \tag{4}$

The integral on the right of (4) may be evaluated thusly:

$\int_0^L \dfrac{\partial^2 u(x, t)}{\partial x^2} dx = \dfrac{\partial u(L, t)}{\partial x} - \dfrac{\partial u(0, t)}{\partial x} = 0 \tag{5}$

by virtue of the boundary conditions

$\dfrac{\partial u(L, t)}{\partial x} = \dfrac{\partial u(0, t)}{\partial x} = 0. \tag{6}$

Thus by (4) we see that

$\dfrac{dQ(t)}{dt} = 0, t \ge 0; \tag{7}$

i.e. $Q(t) = Q(0)$ is constant for all $t \ge 0$. This result is of course in perfect accord with our physical intuition that a completely thermally insulated body should suffer no net gain or loss in total heat energy over the course of time. Finally, we may compute the (constant) total heat $Q(t)$ by evaluating

$Q(0) = \int_0^L u(x, 0)dx = \int_0^L (Lx - x^2)dx$ $= (\dfrac{Lx^2}{2} - \dfrac{x^3}{3} \mid_0^L = \dfrac{L^3}{2} - \dfrac{L^3}{3} = \dfrac{L^3}{6}; \tag{8}$

the total heat in the bar is $L^3 / 6$ for all $t \ge 0$.

  • 2
    $\begingroup$ Thanks a lot for the explanation. $\endgroup$ – user281476 Oct 19 '15 at 7:57
  • $\begingroup$ You're welcome! You might consider "accepting" my answer if you don't get any others you like better! Cheers! $\endgroup$ – Robert Lewis Oct 19 '15 at 7:59

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