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I need to know what the dimensions are of things! It seems horrible!

Okay the equation models a thin channel with some pollution or something in it. The notes say: "$u(x,t)$ denotes the pollution per unit length", so far I'm okay. I accept this is what we want to find.

It then goes on to define mass over some length, it does this as: $M_{x,x+h}(t)=\int^{x+h}_xu(k,t)dk$ which I am also fine with.

Now it starts thinking about $\frac{d}{dt}M_{x,x+h}(t)$, this I am less sure about, I might have used partials instead. I'm okay with this though because I accept time is not a dependent variable, I'd like to be happy with this though, so help there please.

It then leads to $\frac{d}{dt}M_{x,x+h}(t)=q(x,t)-q(x+h,t)+\int^{x+h}_xs(k,t)dk$ Here $s(x,t)$ is a source/sink (how much is flowing into our channel at time t at position x) so I think it has units "pollution per meter per second" (say)

By integrating over a bit of length we now get "pollution per second" so kg coming in or out, per second. Now q(x,t) is the rightward flow of pollution at x, so over the boundaries of the channel we're looking at. The amount coming in per second on the left, take away the amount "coming in" on the right (as flow is measured to the right) gives us another pollution per second one.

So far I am quite happy! This is the confusing bit:

$\frac{d}{dt}\frac{1}{h}\int^{x+h}_xu(k,t)dk = -\frac{1}{h}(q(x+h,t)-q(x,t))+\frac{1}{h}\int^{x+h}_xs(k,t)dk$

It simplifies this to:

$u_t(x,t)=-q_x(x,t)+s(x,t)$

I have mixed feelings towards this, on the one hand I'm happy with: $\frac{1}{h}\int^b_af(x)dx=\frac{F(b)-F(a)}{h}$ So surely we should have $\frac{S(x+h,t)-S(x,t)}{h}$ which is one way of writing (limit $h\rightarrow 0$ of course) $\frac{\partial S}{\partial x}$

Yet the $u$ on the LHS of the $=$ becomes a partial?

I'd like to be able to construct these with confidence, can someone please clarify the dimensions of things, and the differences between $\partial$ and $d$ WRT dimensions.

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  • $\begingroup$ Regarding the derivative of $M$: It might be better to use the partial derivative symbol "$\partial$", since one might conjecture that $x$ must be made into a function of $t$ also. I don't think that there is any risk of ambiguity in this particular case, since there is no dependence of $x$ on $t$ in sight. But since you asked... $\endgroup$ – Giuseppe Negro Jun 12 '14 at 17:03
  • $\begingroup$ @GiuseppeNegro look at the $S$ please. $\endgroup$ – Alec Teal Jun 12 '14 at 17:04
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The derivative becomes partial, because it is brought in side the integral, I.e:

$\frac{d}{dt}\int_x^{x+h}u(k,t)dt=\int_x^{x+h}\frac{\partial}{\partial t}u(k,t)dt$.

With respect to dimension $\partial$ and $d$ are the same, all they do is "divide" by a dimension, for instance, if we have velocity $v$, which has dimension $LT^{-1}$, then we differentiate with respect to $t$ and we get $\frac{dv}{dt}=a$, which is accelaration, which has dimension $LT^{-2}$.

To address the $s$, let $S$ be the $x$ antiderivative of $s$, I.e $S_x=s$, then:

$\frac{1}{h}\int_x^{x+h}s(k,t)dk=\frac{1}{h}(S(x+h,t)-S(x,t))\to S_x=s(x,t)$ (as $h\to 0$).

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  • $\begingroup$ But what about S? That one looks like the definition of partial $\endgroup$ – Alec Teal Jun 12 '14 at 17:02
  • $\begingroup$ See my edit for the s part $\endgroup$ – Ellya Jun 12 '14 at 17:23

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