PDEs: Deriving a Simple Advection Equation

Suppose there’s a pipe with water flowing through it at a constant rate v, and that a chemical of some kind is suspended in and moving with the water. This process is called advection. Suppose also that the chemical does not diffuse (i.e. spread out) as it moves. In other words, each particle of the chemical moves parallel to the pipe.

Now let u(x,t) describe the concentration of this chemical at position x and at time t. We get the following picture, where the chemical is moving in the positive x direction (to see this, just note how the graph changes as time passes).

lineartransport1.png

If there’s no diffusion, then this PDE should be a piece of cake to find. Why? Because if the chemical particles only move parallel to the pipe, and if the water flow is constant, then the orientation or the “shape” of the chemical particles as they move doesn’t change. This means that the curve above is the same shape for any point x. Well, if the curve is the same for any point x, then clearly the area under the curve is also the same, too.

So fix an x=s_1 and an x=s_2. Then by the logic above, we get the following:

\int\limits_0^{s_1 } {u(x,0) \cdot dx = \int\limits_{s_2 }^{s_1  + s_2 } {u(x,t) \cdot dx} }

Let’s mess with units a little bit to try and understand this. Inside the integral is u, which represents the concentration of the pollutant. Concentration (also called density) is basically the amount of stuff (or mass, if you want to be fancy) per unit area. So if we want to look at the units of the u, we can write (in a very non-mathematical way) that u = \frac{mass}{area}. Remember that the integral is a kind of summation. What are we summing? In the first integral, we’re summing u(x,0)\cdot dx. In other words, we’re summing the concentration u times little changes in distance. Putting that into units, we get u\cdot dx = \frac{(mass)(area)}{area}, which is just mass. So let’s call the above integral equation M, for mass.

Eventually we want to get a PDE, right? So first we need to get rid of those integrals. Assuming s_1 varies, let’s differentiate M with respect to s_1. By the fundamental theorem of calculus, we get:

\frac{{\partial M}}{{\partial s_1 }} = u(s_1,0) = u(s_1+s_2,t)

This is closer to what we want, but not quite. Notice that each function u has an s_1 term. So let’s look at s_2 more closely, keeping in mind our assumptions. This term represents a distance. Well, we know that the water is moving at a constant speed v, and we know that speed is distance over time. Thus we can represent s_2 as v\cdot t (which is distance over time times time, or just distance). Then M_{s_1} becomes:

u(s_1,0) = u(s_1+v\cdot t, t)

This looks promising. Now we have an equation involving one distance parameter and one time parameter instead of two distance parameters and a time parameter. Clearly if we want a PDE we need to differentiate again. If we decided to differentiate with respect to s_1, it doesn’t look like we’d get far. We’d just end up with the same equation, except with derivatives. So let’s differentiate with respect to the other variable, t. The left-hand side is easy, it’s just zero. The right-hand side requires the use of the chain rule for multivariable functions. Grinding that out, and replacing the dummy variable s_1 with x we get:

v\cdot u_x  + u_t  = 0

We’ll learn how to solve this PDE, and others like it, in the next post.

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2 responses to this post.

  1. […] 23, 2008 Last time we derived a simple PDE describing the process of non-diffusive advection in a constantly flowing […]

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  2. […] in this post, we need to differentiate to get rid of the integrals. Differentiating with respect to , we […]

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