Let $p(x,t)$ denote the number density of bacteria growing in a straight pipe (straight along the x-axis) through which water is streaming with a velocity $v(x)$ to the right. The growth may be modelled by a term linear in density. The growth rate $r$ depends on the illumination, hence $r = r(x)$. The initial density of bacteria is $p(x,0) = p_0 (x) $.

(a) Write down the continuity equation for the density $p(x,t) $ and a general current $j = j(p)$ with the growth term. Solve this PDE for vanishing velocity, $v = 0$, and constant rate $r = r_0$.

I have a final examination on PDE next month, was solving past exam papers and

tried the following:

Since bacteria growing in a pipe line can be represented as a flux flowing.

Rate that $q$ is flowing through the imaginary surface $S$ can be written as

$$\int \int_s j \space ds$$

which in a differential form can be written as

$$ \partial_t p + \nabla j = \delta$$

$\delta$ is the generation of $q$ per unit volume per unit time. In this question $\delta$ is a growth term

hence $$\partial_t p + \nabla j = r$$

which is just the one-way wave equation, I am guessing the solution should look like

something like $p(x,t) = p_0 (x-v_0 t)$......

I am having some trouble finding the correct continuity equation and its solution, any help will be appreciated.

  • $\begingroup$ It's the inhomogeneous heat equation, not wave equation. $\endgroup$ – Mattos Nov 20 '16 at 11:32

Consider a point $z$ along the axis and let $h >0$. Let's examine the interval $(z, z+h)$ and see what's happening to the concentration $p$. First we have a velocity $v$ pushing the bacteria along. This means we'll have an outward flux term $$ F_{out} = (v(z+h) p(z+h,t) - v(z)p(z,t)) $$ with the signs chosen by the fact that $v$ is the velocity to the right. By the fundamental theorem of calculus we can rewrite $$ F_{out} = \int_z^{z+h} \partial_x(v(x)p(x,t))dx. $$ Next we consider the growth of bacteria. They reproduce at a rate $r$ that depends on where we are in the pipe. At each point $x$ the bacterial creation is then $r(x) p(x,t)$. When we add this up over the length of pipe we have that the creation of bacteria, $C$, is given by $$ C = \int_{z}^{z+h} r(x) p(x,t) dx. $$

Finally, we combine these by asserting that $$ \frac{d}{dt} \int_z^{z+h} p(x,t) dx = - F_{out} + C, $$ i.e. the change in the amount of bacteria equals the creation term minus the outward flux (or plus the inward flux, if you prefer). Then $$ \int_{z}^{z+h} \partial_t p(x,t) dx = \int_{z}^{z+h} [-\partial_x(v(x)p(x,t)) + r(x) p(x,t)]dx. $$ Dividing both sides by $h$, sending $h \to 0$, and using the fact that $z$ was arbitrary then shows that $$ \partial_t p(,t) + \partial_x(v(x) p(x,t)) = r(x) p(x,t), $$ which is a transport-type equation for the evolution of $p$. This can be expanded as $$ \partial_t p + v \partial_x p = (r-\partial_x v) p $$ to make it look more like the standard transport equation.

I will trail off here and leave the rest of the problem to you now that you know the proper PDE for the model.


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