## Diffusion ##
## Diffusion ## ![utlogo](http://www.math.toronto.edu/colliand/images/utlogo.jpeg)
### Diffusion vs. Turbulence * Diffusive Transport: perfume, cigarette smoke * [Fick's Law](http://en.wikipedia.org/wiki/Fick's_laws_of_diffusion): Diffusion flux = $\kappa \times$ concentration gradient. * Turbulent Transport: eddies carry particles ![turbulence](http://upload.wikimedia.org/wikipedia/commons/thumb/f/fe/Airplane_vortex_edit.jpg/220px-Airplane_vortex_edit.jpg)
### Example: Dye in a pipe * A circular pipe is filled with a fluid at rest. * Dye in the fluid varying along length $x$. * Density (mass per unit length) of dye: $u(t,x)$. The amount of dye between $x\_1$ and $x\_2$ is $$M(t) = \int\_{x\_1}^{x\_2} u(t,x) dx.$$ How does $M(t)$ change as dye diffuses?
### How does $M(t)$ change? $$\frac{dM}{dt} = \int\_{x\_1}^{x\_2} u_t (t,x) dx.$$ Dye is neither destroyed nor created. Mass changes as dye flows. Fick's diffusion law shows: $$\frac{dM}{dt} = dye~ flows ~in - dye ~flows~ out = \kappa [u\_x (t, x\_2) - u\_x (t, x\_1)].$$ Setting expressions for $\frac{dM}{dt}$ equal to each other $$\int\_{x\_1}^{x\_2} u_t (t,x) dx = \kappa [u\_x (t, x\_2) - u\_x (t, x\_1)].$$
### Deriving Diffusion Equation $$\frac{1}{x\_2 - x\_1} \int\_{x\_1}^{x\_2} u_t (t,x) dx = \kappa \frac{ u\_x (t, x\_2) - u\_x (t, x\_1)}{x\_2 - x\_1}$$ Taking the limit $x\_2 \rightarrow x\_1$ $$\implies u\_t = u\_{xx}.$$ **Initial Value Problem for the Diffusion Equation:** $$\begin{cases} \partial\_t u = \partial_x^2 u \\\\ u(0, x) = u\_0(x) \end{cases}$$
### Maximum Principle If $u(t,x)$ solves the diffusion equation on a spacetime rectangle $[(t,x): t \in [0,T], x \in [0,L] ]$ then the maximum value of $u$ in the rectangle is either located when $t=0$ or on the lateral sides $x=0$ or $x =L$. * *Miniumum Principle* also holds true. * Initial hot spots cool off.
### Uniqueness Consider the *Dirichlet Problem* for the diffusion equation: $$\begin{cases} \partial\_t u - \kappa \partial_x^2 u = f(t,x), ~ x \in [0,L], t>0 \\\\ u(0, x) = u\_0 (x) \\\\ u(0,t) = g(t), ~ u(L,t) = h(t). \end{cases}$$ Here $f, u\_0, g$ and $h$ are given functions. Maximum Principle $\implies \exists$ at most one solution.
### Proof of Uniqueness * Suppose $u\_1$ and $u\_2$ are two distinct solutions. * Form $w(t,x) = u\_1 (t,x) - u\_2 (t,x)$. Then $w$ satisfies the diffusion equation $$\begin{cases} \partial\_t w - \kappa \partial_x^2 w = 0, ~ x \in [0,L], t>0 \\\\ u(0, x) = 0 \\\\ w(0,t) = 0, ~ w(L,t) = 0. \end{cases}$$ By the Maximum Principle, $w$ has maximum along $t=0$ or $x=0$ or $x=L$, so $w \leq 0$. By the Minimum Principle, $w$ has minimum along $t=0$ or $x=0$ or $x=L$, so $w \geq 0$. We conclude that $w = 0$ and $u\_1 = u\_2.$
### Deriving the Fundamental Solution We want to solve the initial value problem: $$\begin{cases} \partial\_t u = \partial_x^2 u \\\\ u(0, x) = u\_0(x) \end{cases}$$ **Strategy:** * Find a *special* solution. * Build *general* solution using the special one.
## Invariance Properties 1. Any **translate** $u(t, x - y)$ solves if $u(t,x)$ solves. 2. Any **derivative** ($u\_x, u\_t, u\_{xx}, ...$) solves if $u$ solves. 3. **Linear** combos of solutions are also solutions. * $\implies$ An **integral** of solutions is a solution. 4. **Parabolic dilations** of a solution are solutions: $$u_a (t,x) = u (at, \sqrt{a} x), ~\forall a > 0.$$ (Check it with the chain rule.)
### We seek a special solution $Q(t,x)$ **Boundary Conditions for $Q$** * $Q(0,x) =1$ for $x>0$ * $Q(0,x) = 0$ for $x<0$. **Step 1:** Seek $Q$ in special form. $$Q(t,x) = g(p), ~p = \frac{x}{\sqrt{t}}$$ Why is this a good idea?
**Step 2: Diffusion Equation converts into an ODE!** $$Q\_t = \frac{dg}{dp} \frac{\partial p}{\partial t} = - \frac{1}{2t} \frac{x}{\sqrt{4kt}} g' (p).$$ $$Q\_x = \frac{dg}{dp} \frac{\partial p}{\partial x} = \frac{1}{\sqrt{4kt}} g' (p).$$ $$Q\_{xx} = \frac{d {Q_x}}{dp} \frac{\partial p}{\partial x} = \frac{1}{4kt} g'' (p).$$ Using $Q\_t = Q\_{xx}$, we find an ODE for $g$: $$g'' + 2p g' = 0.$$
**Step 3: Solve the ODE** Multiply by the integrating factor $e^{\int 2p dp}$ to recast as $$g'(p) = c\_1 e^{-p^2}.$$ Integrating again yields $$Q(t,x) = g(p) = c\_1 \int e^{-p^2} dp + c\_2.$$ Recalling $p$, we write more explicitly $\forall ~ t>0$, $$Q(t,x) = c\_1 \int\_0^{\frac{x}{\sqrt{4kt}}} e^{-p^2} dp + c\_2.$$
**Step 4: Determine $c\_1, c\_2$ using boundary conditions.** If $x>0$, we want $$1 = \lim_{t \searrow 0} Q(t,x) = c\_1 \int\_0^\infty e^{-p^2} dp + c\_2 = c\_1 \frac{\pi}{2} + c\_2.$$ If $x<0$, we want $$1 = \lim_{t \searrow 0} Q(t,x) = c\_1 \int\_0^{-\infty} e^{-p^2} dp + c\_2 = - c\_1 \frac{\pi}{2} + c\_2.$$ These conditions lead us to choose: * $c\_1 = \frac{1}{\sqrt{\pi}}$ * $c\_2 = \frac{1}{2}$
$$Q(t,x) = \frac{1}{2} + \frac{1}{\sqrt{\pi}} \int\_0^{\frac{x}{\sqrt{4kt}}} e^{-p^2} dp .$$ **Step 5: Define $S = \frac{\partial Q}{\partial x}$.** $$S(t,x) = \frac{1}{2 \sqrt{\pi k t}} e^{-{\frac{x^2}{4kt}}}, ~ t>0.$$ *** $$u(t,x) = \int\_{-\infty}^{+\infty} S(t, x-y) u\_0 (y) dy, ~t>0$$ solves the initial value problem!
## Solution of the Initial Value Problem $$\begin{cases} \partial\_t u - \partial_x^2 u = 0 \\\\ u(0, x) = u\_0(x) \end{cases}$$ $\implies$ $$u(t,x) = \int\_{-\infty}^{+\infty} S(t, x-y) u\_0 (y) dy, ~t>0$$
## Solution of the inhomogeneous IVP $$\begin{cases} \partial\_t u - \partial_x^2 u = f(t,x) \\\\ u(0, x) = u\_0(x) \end{cases}$$ $\implies$ $$u(t,x) = \int\_{-\infty}^{+\infty} S(t, x-y) u\_0 (y) dy + \int\_0^t \int\_{-\infty}^{+\infty} S(t-s, x-y) f(y,s) dy ds.$$