I am given the following:
Consider the reaction-diffusion equation
$$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} + ku\left( 1-\frac{u}{K} \right) - du, \: \: -\infty<x<\infty, \: t>0,$$
which describes the density of bacteria in a liquid suspension with $u(x,0)=u_0$. Here $u = u(x, t)$ and all constants are assumed positive.
- Determine the dimensions of the equation parameters in terms of $T$ (time), $L$ (length) and $M$ (mass).
- By employing $\overline{x}$ to represent the non-dimensional spatrical coordinate and $\tau$ non-dimensional time, the above can be non-dimensionalised such that $$ \phi_\tau=\phi_{\overline{x}\overline{x}}+\phi(1-\phi)-\overline{d}\phi, \: \: -\infty<\overline{x}<\infty, \: \tau>0 $$ where $\phi=\phi(\overline{x},\tau)$ is the non-dimensional bacterial concentration with $\phi(\overline{x},0)=\phi_0$ and $\phi_0=u_0/K$. Given that $u(x, t)$ has been scaled according to $u(x,t)=K\phi(\overline{x},\tau)$ to yield this non-dimensionalisation, determine the scalings for $x$ and $t$ in terms of $\overline{x}$ and $\tau$ respectively.
Clearly $[K]=[u]=M/L^3$. Also, $$\left[\frac{\partial u}{\partial t}\right] = \left[D \frac{\partial^2 u}{\partial x^2} \right] \implies [D]=L^2/T.$$ Finally, $$[ku]=\left[ \frac{\partial u}{\partial t} \right] \implies [k]=1/T=[d].$$
For this question, let $x=X\overline{x}$ and $t=Y\tau$, where $X$ and $Y$ are parameters to be determined. Substitute these values into the governing equation: $$\begin{align} \frac{K}{Y}\phi_\tau &= \frac{DK}{X^2}\phi_{\overline{x}\overline{x}}+kK\phi(1-\phi)-dK\phi \\ \implies \phi_\tau &= \frac{DY}{X^2}\phi_{\overline{x}\overline{x}}+kY\phi(1-\phi)-dY\phi. \end{align} $$ Let $Y=1/k$ so that we have $$\phi_\tau = \frac{D}{kX^2}\phi_{\overline{x}\overline{x}}+\phi(1-\phi)-\frac{d}{k}\phi.$$ Now let $X^2=D/k \implies X=\sqrt{D/k}$ so that $$\phi_\tau=\phi_{\overline{x}\overline{x}}+\phi(1-\phi)-\overline{d}\phi,$$ where $\overline{d}=d/k$ is a non-dimensional parameter.
Let me know what you think!