Partial Differential Equations/Parallel Plate Flow: Realistic IC

Parallel Plate Flow: Realistic IC

The Steady State

The initial velocity profile chosen in the last problem agreed with intuition but honestly came out of thin air. A more realistic development follows.

The problem stated that (to come up with an IC) the fluid was under a pressure difference for some time, so that the flow became steady. "Steady" is another way of saying "not changing with time", and "not changing with time" is another way of saying that:

$\frac{\partial u}{\partial t}=0$

Putting this into the PDE from the previous section:

$\frac{\partial u}{\partial t}=\nu \frac{{\partial }^{2}u}{\partial y^2}-\frac{P_x}{\rho }$

$\Downarrow$

$0=\nu \frac{{\partial }^{2}u}{\partial y^2}-\frac{P_x}{\rho }\Rightarrow \frac{P_x}{\nu \rho }=\frac{d^{2}u}{d{y}^2}$

Independent of $t$, the PDE became an ODE. The no slip condition results in the following BCs: $u=0$ at $y=0$ and $y=1$.

$\frac{d^{2}u}{d{y}^2}=\frac{P_x}{\nu \rho }\Rightarrow u=\frac{P_x}{2\nu \rho }{y}^2+C_{1}y+C_0$

$0=\frac{P_x}{2\nu \rho }\cdot 0^2+C_1\cdot 0+C_0\Rightarrow C_0=0\text{(Bottom plate BC: u = 0, y = 0)}$

$0=\frac{P_x}{2\nu \rho }\cdot 1^2+C_1\cdot 1\Rightarrow C_1=-\frac{P_x}{2\nu \rho }\text{(Top plate BC: u = 0, y = 1)}$

$\Downarrow$

$u=\frac{P_x}{2\nu \rho }(y^2-y)$

For the sake of example, take $P_x/(2\nu \rho )=-4$ (recall that a negative pressure gradient causes left to right flow). This gives a parabola which starts at $0$, increases to a maximum of $1$ at $y=1/2$, and returns to $0$ at $y=1$.

This parabola looks pretty much identical to the sinusoid previously used (you must zoom in to see a difference). However, even on the narrow domain of interest, the two are very different functions (look at their taylor expansions, for example). Using the parabola instead of the sine function results in a much more involved solution.

So this derives the steady state flow, which we will use as an improved, realistic IC. Recall that the problem is about a fluid that's initially in motion that is coming to a stop due to the absence of a driving force. The IBVP is now subtly different:

$\frac{\partial u}{\partial t}=\nu \frac{{\partial }^{2}u}{\partial y^2}\text{(PDE)}$

$u(y,0)=4y-4y^2\text{(IC)}$

$u(0,t)=0$

$u(1,t)=0\text{(BCs)}$

Separation

Since the only difference from the problem in the last section is the IC, the variables may be separated and the BCs applied with no difference, giving:

$u(y,t)=e^{-(n\pi {)}^2\nu t}B\sin (n\pi y)$

But now we're stuck! The IC can't match this:

$u(y,0)=4y-4y^2$

$B\sin (n\pi y)=4y-4y^2$

What went wrong? It was the assumption that $u(y,t)=Y(y)T(t)$. The fact that the IC couldn't be fulfilled means that the assumption was wrong. It should be apparent now why the IC was chosen to be $\sin (\pi y)$ in the previous section.

We can proceed however, thanks to the linearity of the problem. It gets long.

Linearity (the superposition principle specifically) says that if $u_1$ is a solution to the BVP (not the whole IBVP, only the BVP) and so is $u_2$, then $C_1{u}_1+C_2{u}_2$, a linear combination, is also a solution.

Let's take a step back: suppose that the IC was $u(y,0)=\sin (\pi y)+1/5\sin (3\pi y)$ (no longer a realistic flow problem). This may be put into the half baked solution:

$u(y,0)=\sin (\pi y)+\frac{1}{5}\sin (3\pi y)$

$B\sin (n\pi y)=\sin (\pi y)+\frac{1}{5}\sin (3\pi y)$

And it still can't match. However, observe that the individual terms in the IC can:

$B_1\sin (n_1\pi y)=\sin (\pi y)\Rightarrow n_1=1\text{and}{B}_1=1$

$B_3\sin (n_3\pi y)=\frac{1}{5}\sin (3\pi y)\Rightarrow n_3=3\text{and}{B}_3=\frac{1}{5}$

Note the subscripts used to identify the terms: they reflect the integer $n$ from the separation constant. Solutions may be obtained for each individual term of the IC, identified with $n$:

$u_1(y,t)=e^{-(1\pi {)}^2\nu t}1\sin (1\pi y)=e^{-{\pi }^2\nu t}\sin (\pi y)$

$u_3(y,t)=e^{-(3\pi {)}^2\nu t}\frac{1}{5}\sin (3\pi y)=e^{-9{\pi }^2\nu t}\frac{1}{5}\sin (3\pi y)$

Linearity states that the sum of these two solutions is also a solution to the BVP (no need for new constants):

$u_{1+3}(y,t)=e^{-{\pi }^2\nu t}\sin (\pi y)+\frac{1}{5}{e}^{-9{\pi }^2\nu t}\sin (3\pi y)$

So we added the solutions and got a new solution... what is this good for? Try setting $t=0$:

$u_{1+3}(y,0)=e^{-{\pi }^2\nu \cdot 0}\sin (\pi y)+\frac{1}{5}{e}^{-9{\pi }^2\nu \cdot 0}\sin (3\pi y)=\sin (\pi y)+\frac{1}{5}\sin (3\pi y)$

Each component solution satisfied the BVP, and the sum of these just happened to satisfy the IC. The IBVP with IC $u(y,0)=\sin (\pi y)+1/5\sin (3\pi y)$ is now solved. It would work the same way for any linear combination of sine functions whose half frequencies vary are $n\pi$ ("linear combination" means a sum of terms, each multiplied by a constant), the sums assumed to converge and be term by term differentiable:

$u(y,0)=\sum _{n=1}^{\mathrm{\infty }}{B}_n\sin (n\pi y)\text{(IC: an arbitrary linear combination of sines)}$

$u_n(y,t)=e^{-(n\pi {)}^2\nu t}{B}_n\sin (n\pi y)\text{(solution of}{n}^{th}\text{term)}$

$u(y,t)=\sum _{n=1}^{\mathrm{\infty }}{u}_n(y,t)=\sum _{n=1}^{\mathrm{\infty }}{e}^{-(n\pi {)}^2\nu t}{B}_n\sin (n\pi y)\text{(sum of solutions)}$

$u(y,0)=\sum _{n=1}^{\mathrm{\infty }}{e}^{-(n\pi {)}^2\nu \cdot 0}{B}_n\sin (n\pi y)=\sum _{n=1}^{\mathrm{\infty }}{B}_n\sin (n\pi y)\text{(IC recovered)}$

So now we can solve the problem if the IC is a linear combination of sine functions. But the IC for this problem isn't such a sum, it's just a stupid parabola. Or is it?

Series Construction

In the 19th century, a man named Joseph Fourier took a break from helping Napoleon take over the world to ask an important question while studying this same BVP (concerning heat flow): can a function be expressed as a sum of sinusoids, similar to a taylor series? The short answer is yes, if a few reasonable conditions apply. The long answer follows, and the next chapter is a longer answer.

A function meeting certain criteria may indeed be expanded into a sum of sines, cosines, or both. In our case, all that is needed to accomplish this expansion is to find the coefficients $B_n$. A little trick involving an integral makes this possible.

The sine function has a very important property called orthogonality. There are many flavors of this, which will be served in the next chapter. Relevant to this problem is the following:

$\underset{0}{\overset{1}{\int }}2\sin (m\pi y)\sin (n\pi y)dy=\{\begin{array}{cc}1,& m=n\\ 0,& m\ne n\end{array}$

Let's call the IC $\varphi (y)$ to generalize it. We equate the IC with its expansion, and then apply some craftiness:

$\sum _{n=1}^{\mathrm{\infty }}{B}_n\sin (n\pi y)=\varphi (y)$

$2\sin (m\pi y)\cdot \sum _{n=1}^{\mathrm{\infty }}{B}_n\sin (n\pi y)=2\sin (m\pi y)\cdot \varphi (y)$

$\sum _{n=1}^{\mathrm{\infty }}{B}_n\cdot 2\sin (m\pi y)\sin (n\pi y)=2\sin (m\pi y)\varphi (y)$

$\underset{0}{\overset{1}{\int }}\sum _{n=1}^{\mathrm{\infty }}{B}_n\cdot 2\sin (m\pi y)\sin (n\pi y)dy=\underset{0}{\overset{1}{\int }}2\sin (m\pi y)\varphi (y)dy$

$\sum _{n=1}^{\mathrm{\infty }}{B}_n\underset{0}{\overset{1}{\int }}2\sin (m\pi y)\sin (n\pi y)dy=\underset{0}{\overset{1}{\int }}2\sin (m\pi y)\varphi (y)dy$

$B_m=\underset{0}{\overset{1}{\int }}2\sin (m\pi y)\varphi (y)dy$

In the last step, all of the terms in the sum became $0$ except for the $m^{th}$ term (the term where $m=n$), the only case where orthogonality gave $1$. This isolates and explicitly defines $A_m$. The expansion for $\varphi (y)$ is then:

$\varphi (y)=\sum _{m=1}^{\mathrm{\infty }}\underset{0}{\overset{1}{\int }}2\sin (m\pi y)\varphi (y)dy\cdot \sin (m\pi y)$

Or equivalently:

$\varphi (y)=\sum _{n=1}^{\mathrm{\infty }}{B}_n\sin (n\pi y);B_n=\underset{0}{\overset{1}{\int }}2\sin (n\pi y)\varphi (y)dy$

Many important details have been left out later in a devoted chapter; one noteworthy detail is that this expansion is what it's (very superficially) expected to be only on the interval $0\le y\le 1$.

This expansion may finally be combined with the sum of sines solution to the BVP developed previously. Note that the last equation looks very similar to $u(y,0)$. Following from this:

$u(y,0)=\sum _{n=1}^{\mathrm{\infty }}{B}_n\sin (n\pi y)$

$u(y,0)=\sum _{n=1}^{\mathrm{\infty }}\underset{0}{\overset{1}{\int }}2\sin (n\pi y)\varphi (y)dy\cdot \sin (n\pi y)$

So the expansion will satisfy the IC given as $\varphi (y)$ (surprised?). The full solution for the problem with arbitrary IC is then:

$u(y,t)=\sum _{n=1}^{\mathrm{\infty }}{e}^{-(n\pi {)}^2\nu t}{B}_n\sin (n\pi y)$

$u(y,t)=\sum _{n=1}^{\mathrm{\infty }}{e}^{-(n\pi {)}^2\nu t}\underset{0}{\overset{1}{\int }}2\sin (n\pi y)\varphi (y)dy\cdot \sin (n\pi y)$

In this problem specifically, the IC is $\varphi (y)=4y-4y^2$, so:

$B_n=\underset{0}{\overset{1}{\int }}2\sin (n\pi y)(4y-4y^2)dy=\frac{8}{n^3{\pi }^3}(2-2\cos (n\pi )-n\pi \sin (n\pi ))$

Sines and cosines appear from the integration dependent only on $n\pi$. Since $n$ is an integer, these can be made more aesthetic.

$\sin (n\pi )=0;n\text{ is an integer.}$

$\cos (n\pi )=(-1{)}^n;n\text{ is an integer.}$

$\Downarrow$

$B_n=\frac{16-16(-1{)}^n}{n^3{\pi }^3}$

Note that for even $n$, $B_n=0$. Putting everything together finally completes the solution to the IBVP:

$u(y,t)=\sum _{n=1}^{\mathrm{\infty }}{e}^{-(n\pi {)}^2\nu t}\frac{16-16(-1{)}^n}{n^3{\pi }^3}\sin (n\pi y)$

There are many interesting things to observe. To begin with, $u(y,t)$ is not a product of a function of $y$ and a function of $t$. Such a solution was assumed in the beginning, proved to be wrong, but eventually happened to yield a solution anyway thanks to linearity and what is called a Fourier sine expansion.

A careful look at the procedure reveals something that may be disturbing: this lengthy solution is strictly valid for the given BCs. Thanks to the definition of $\varphi (y)$, the solution is generic as far as the IC is concerned (the IC doesn't even need to match the BCs), however a slight change in either BC would mandate starting over almost from the beginning.

The parabolic IC, which looks very similar to the sine function used in the previous section, is wholly to blame (or thank once you understand the beauty of a Fourier series!) for the infinite sum. It is interesting to approximate the first several numeric values of the sequence $B_n$:

$B_1\approx 1.03205$

$B_3\approx 0.03822$

$B_5\approx 0.00826$

$B_7\approx 0.00301$

Recall that the even terms are all $0$. The first term by far dominates, this makes sense since the first term already looks very, very similar to the parabola. Recall that $n^2$ appears in an exponential, making the higher terms even smaller for time not too close to $0$.