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While studying the heat equation, I ran into the following exercise:

Consider conservation of thermal energy $(2)$ for any segment of a one-dimensional rod $a\leq x\leq b$. By using the fundamental theorem of calculus,

$$ \frac{\partial}{\partial b}\int_a^bf(x)\,dx=f(b),\tag{1} $$

derive the heat equation $(3)$.

For this exercise, we have that

$$\begin{align} \frac d{dt}\int_a^be(x,t)\,dx&=\varphi(a,t)-\varphi(b,t)+\int_a^bQ(x,t)\,dx,\tag{2}\\ c\rho\frac{\partial u}{\partial t}&=\frac\partial{\partial x}\left(K_0\frac{\partial u}{\partial x}\right)+Q.\tag{3} \end{align}$$

I do not know how to tackle this problem. Any hint would be greatly appreciated. Also, I do not understand why $(1)$ has a partial derivative when $f$ is a function of a single variable.

Edit 1: Following celtschk's advice, I managed to rewrite $(2)$ as

$$ \frac d{dt}\int_a^be(x,t)\,dx=\varphi(a,t)-\frac\partial{\partial b}\int_a^b\varphi(x,t)\,dx+\int_a^bQ(x,t)\,dx. $$

Now, I suppose that I must get rid of the $\partial/\partial b$ and subtract $\varphi(a,t)$?

Edit 2 One could also rewrite $(2)$ as

$$ \frac d{dt}\int_a^be(x,t)\,dx= - \frac{\partial}{\partial a} \int_a^b \varphi(x,t) dx -\frac\partial{\partial b}\int_a^b\varphi(x,t)\,dx+\int_a^bQ(x,t)\,dx. $$

using the fact that $ \frac{\partial}{\partial a} \int_b^a \varphi(x,t) dx$. Is there a way to collect together the two terms with partial derivatives in order to get the desired $\int_a^b \frac{d \varphi}{dx} dx $ term in the heat equation?

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2  
Note that the whole expression on the left of (1) has two free variables (and none of them is $x$). –  celtschk Sep 22 '12 at 17:50
    
Yes, get rid of the derivative of $b$ using the fundamental theorem of calculus. Then you will have the difference is values of $\varphi$ at the endpoints a,b. Divide by $b-a$ and pass to the limit (or use the Mean Value theorem for differences and for integrals). –  user31373 Sep 23 '12 at 15:12
    
I presume $e(x,t)$ represents a heat density, $\varphi$ a flux, and $Q$ a rate of heat source term? –  JohnD Dec 19 '12 at 4:32

2 Answers 2

up vote 1 down vote accepted

Seems to me that you are seeking the derivation of the heat equation based on a conservation law type of argument. Let $u(x,t)$ denote the density of the quantity of interest (heat, diffusing ink, etc.), $\varphi(x,t)$ the flux (which in 1D is positive for rightward flow in 1D), and $q(x,t)$ the rate at which the quantity of interest is generated inside the domain (source/sink term).

Let $[a,b]$ be an arbitrary subinterval of the overall domain $[\alpha,\beta]$. The conservation law in force here is that the time rate of change of the quantity in $[a,b]$ equals the rate of inflow at $x=a$ minus the rate of outflow at $x=b$ plus the amount of the quantity generated inside $[a,b]$. That is,

$$ {d\over dt}\int_a^b u(x,t)\,dx=\varphi(a,t)-\varphi(b,t)+\int_a^b q(x,t)\,dx. $$ Pushing the time derivative inside the integral (this is justified here) and applying the Fundamental Theorem of Calculus to the difference on the right-hand side,

\begin{align} \int_a^b {\partial u\over \partial t}(x,t)\,dx&=-\int_a^b {\partial \varphi\over \partial x}(x,t)\,dx+\int_a^b q(x,t)\,dx\\ \int_a^b \left[{\partial u\over \partial t}(x,t)+{\partial \varphi\over \partial x}(x,t)\right]\,dx&=\int_a^b q(x,t)\,dx. \end{align}

Since these integrals are equal for an arbitrary subinterval $[a,b]$ of $[\alpha,\beta]$, the integrands must be equal: $$ {\partial u\over \partial t}(x,t)+{\partial \varphi\over \partial x}(x,t)=q(x,t).\tag{1} $$ This is called the Fundamental Conservation Law because it applies to any scenario which obeys the conservation principle italicized above.

Of course this is not the heat equation---yet---because we need to impose one more assumption. Since $u(x,t)$ is the unknown we seek and $q(x,t)$ is the (known) source term, in order for solve for $u(x,t)$ we must know something about the flux $\varphi(x,t)$. In the context of heat diffusion, Fourier's Law of heat conduction says that the flux of heat energy is proportional to, but in the opposite direction as, the gradient of the density: $\varphi=-k{\partial u\over \partial x}$ in this 1D context. Substituting this into (1), $$ {\partial u\over\partial t}+{\partial\over \partial x}(-k\,u_x)=q(x,t), $$ or the much more familiar form $$ u_t=ku_{xx}+q(x,t) $$ which is the heat equation in 1D (with a source term).

Hope that helps.

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From the way the problem is posed, it seems as if a good first step is to use the Fundamental theorem of Calculus to write $\varphi(b,t)$ as $ \displaystyle\frac{\partial}{\partial b} \displaystyle\int_{a}^b \phi(x,t) dx$, but, for me, that made things much more complicated.

Instead, we can take $(2)$ and differentiate both sides with respect to $b$, getting $$ \displaystyle\frac{\partial}{\partial b} \left( \displaystyle\frac{\partial}{\partial t} \displaystyle\int_a^b e(x,t) dx \right) = \displaystyle\frac{\partial}{\partial b} \left( \varphi(a,t) \right) - \displaystyle\frac{\partial}{\partial b} \left(\varphi(b,t) \right) + \displaystyle\frac{\partial}{\partial b} \left( \displaystyle\int_a^b Q(x,t) dx \right) $$ On the lefthand side we can switch the order of the partial derivatives, and on the righthand side we note that $\displaystyle\frac{\partial}{\partial b} \left( \varphi(a,t) \right) = 0$ because $a$ and $b$ are independent by assumption. This leaves us with $$ \displaystyle\frac{\partial}{\partial t} \left( \displaystyle\frac{\partial}{\partial b} \displaystyle \int_a^b e(x,t) dx \right) = \displaystyle\frac{\partial}{\partial b} \left(\phi(b,t) \right) + \displaystyle\frac{\partial}{\partial b} \left( \displaystyle\int_a^b Q(x,t) dx \right) $$ Applying the Fundamental theorem of Calculus to both sides gives $$\displaystyle\frac{\partial}{\partial t} \left(e(b,t) \right) =- \displaystyle\frac{\partial}{\partial b} \left( \varphi(b,t) \right) + Q(b,t) $$ At this point $b$ is just an arbitrary position variable, and therefore we can replace $b$ by $x$ to obtain $$\displaystyle\frac{\partial}{\partial t} \left(e(x,t) \right) = - \displaystyle\frac{\partial}{\partial x} \left( \varphi(x,t) \right) + Q(x,t) $$

As shown in JohnD's answer, we can use Fourier's law of heat conduction to write $\displaystyle\frac{\partial}{\partial x} \varphi(x,t)$ as $- K_0(x) \displaystyle\frac{\partial u}{\partial x}$. This gives us the heat equation $$\displaystyle\frac{\partial}{\partial t} \left(e(x,t) \right) = \displaystyle\frac{\partial}{\partial x} \left( K_0(x) \displaystyle\frac{\partial u}{\partial x} \right) + Q(x,t) $$

Please note that this solution is based on techniques use on page 279 of Richard Haberman's book $\it{Mathematical \: Models}$, which is a great resource from which to learn a variety of tools for deriving ODEs and PDEs to model physical and biological situations.

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