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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!

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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 $t \ge 0$.

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    $\begingroup$ 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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