Lecture 15 - Nonlinear Differential Equations

#math

Notations:
Let $u = u (x; t), x \in R^{n}, n \geq 2, t \in R^{+}$ and $\nabla u = (u_{x_{1}}, . . .,_{x_{n}})$ where $x = (x_{1}, . . . x_{n})$

Porous medium Equation:

\begin{aligned} u_{t} & = Δ u^{m}, m > 0 \\ u_{t} & = Δ (| u |^{m - 1} u) \end{aligned}

If $m = 1$ , we have the classical heat equation. With the heat equation, we have the maximum principle.
If $u_{0} > 0$ for some small open set on $R$ then $u > 0$ for $t > 0$ . We notice that heat in the heat equation has an infinite speed of propagation.

The physics behind the porous medium equation.

For $m = 2$ we have $u_{t} = (u^{2})_{x x}$ which is known as the Boussinesq equation. It models flow of a gas in a porous medium where $ρ$ is the density of the gas and $v$ is the velocity.

Conservation of mass $f ρ_{t} + div (ρ v) = 0 f \in (0, 1)$
Darcy's law $v = - \frac{ϵ}{μ} \nabla p, ϵ : permeability μ : viscosity p : pressure$
Equation of state $ρ = ρ_{0} p^{γ}, 0 < γ < 1$

Thus putting everything together we obtain:

\begin{aligned} f ρ_{t} & = - div (p (- \frac{ϵ}{μ} \nabla p)) \\ = \frac{ϵ}{μ} div (p \nabla {\frac{ρ}{ρ_{0}}}^{\frac{1}{γ}}) \\ = \frac{ϵ}{μ} \cdot \frac{1}{ρ_{0}^{\frac{1}{γ}}} div (ρ ρ^{\frac{1}{γ} - 1} \nabla ρ) \\ = \frac{ϵ}{μ} \frac{1}{ρ_{0}^{\frac{1}{γ}}} \frac{1}{\frac{1}{γ} + 1} \underset{Δ ρ^{1 \frac{+ 1}{γ}}}{\underset{⏟}{div (\nabla ρ^{1 \frac{+ 1}{γ}})}} \\ = \end{aligned}

If we scale the time

t \to c t

and choose $c$ appropriately we can scale away the constants thus:

ρ_{t} = Δ ρ^{m}

Instantaneous point source solution case of the heat solution

u_{t} = Δ u, x \in R^{n} t > 0

such that $\int_{R^{n}} u (x; t) = 1$ , i.e, constant energy and when $t = 0$ we have $u (x, 0) = δ (x)$ .
Using Fourier transform, we obtain:

φ (x, t) = (4 π t)^{- \frac{n}{2}} e^{\frac{- | | x | |^{2}}{4 t}}

$φ$ is the fundamental solution of the heat equation.

Now if $Δ u = u_{t}$ :

u (x, 0) = u_{0} (x)

u (x, t) = \int_{R^{n}} φ (x - y, t) u_{0} (y) d y

Notice that if $u_{0} > 0$ in some $(- ϵ, ϵ)^{n}$ and 0 elsewhere then $u > 0$ for $t > 0$ and notice that if $u_{0} (x) = δ (x)$ then $u (x, t) = φ (x, t)$

Self-similar solutions

\begin{aligned} u_{t} = div & (u^{σ} \nabla u), x \in R^{n}, t > 0 \\ u & (x, 0) = δ (x) \end{aligned}

We look for a particular solution:

u (x, t) = t^{α} θ (\frac{x}{t^{β}})

set $ξ = \frac{x}{t^{β}} \in R^{n}$ . Let us compute $α, β$ and $θ$ :

\begin{aligned} u_{t} & = α t^{α - 1} θ + t^{α} \sum_{i = 1}^{n} - β x_{i} t^{- β - 1} θ_{ξ_{i}} \\ = α t^{α - 1} θ - β t^{α - 1} \underset{ξ \cdot \nabla θ}{\underset{⏟}{\sum_{i = 1}^{n} ξ_{i} θ_{ξ_{i}}}} \end{aligned}

Thus

div (u^{σ} \nabla u) = t^{α (σ + 1) - 2 β} d i v_{ξ} (θ^{σ} \nabla θ)

Now we have an equation:

\begin{matrix} (*) & α t^{α - 1} θ - β t^{α - 1} ξ \cdot \nabla θ = t^{α (σ + 1) - 2 β} {div}_{ξ} (θ^{σ} \nabla θ) \end{matrix}

If we require that:

α - 1 = α (σ + 1) - 2 β

then we can cancel out the $t$ 's on both side of the equation. We also have:

1 = \int_{R^{n}} u (x, t) d x = \int_{R^{n}} t^{α} θ (\frac{x}{t^{β}}) d x

With a change of variable we obtain:

1 = t^{α + n β} \int_{R^{n}} θ (ξ) d ξ

Now we our second requirement is:

\begin{matrix} (* *) & α + n β = 0 \end{matrix}

Finally we have equations $(*)$ and $(* *)$ to solve for $α$ and $β$ :

\begin{aligned} α & = \frac{- n}{σ n + 2} \\ β & = \frac{1}{σ n + 2} \end{aligned}

To obtain $θ$ , I have to solve an Elliptic equation:

div (θ^{σ} \nabla θ) + \frac{n}{σ n + 2} θ + \frac{1}{σ n + 2} ξ \cdot \nabla θ = 0

We look for a radial solution:

θ = θ (η), η = | | ξ | |

In this case, our elliptic equation is reduced to an ODE:

\frac{1}{η^{n - 1}} (η^{n - 1} θ^{σ} θ^{'})^{'} \frac{+ n}{σ n + 2} θ + \frac{1}{σ n + 2} θ^{'} = 0

Which is equivalent to:

(η^{n - 1} θ^{σ} θ^{'})^{'} + \frac{1}{σ n + 2} (η^{n} θ)^{'} = 0

Integrating once we obtain:

η^{n - 1} θ^{σ} θ^{'} + \frac{1}{σ n + 2} η^{n} θ = const = 0

Reducing this equation to:

θ^{σ - 1} θ^{'} = \frac{- 1}{σ n + 2} η

Integrating both sides we obtain our solution:

θ (η) = {(\frac{σ}{2 (σ n + 2)})}^{\frac{1}{σ}} ({η_{0}}^{2} - η^{2})_{+}^{\frac{1}{σ}}

Remember that $\int_{R^{n}} θ (ξ) d ξ = 1$ which fixes $η_{0}$ . Thus:

\begin{aligned} u (x, t) & = t^{α} θ (\frac{x}{t^{β}}) \\ = t^{\frac{- n}{σ n + 2}} {(\frac{σ}{2 (σ n + 2)} (η_{0}^{2} - \frac{| | x | |^{2}}{t^{\frac{2}{σ n + 2}}}))}_{+}^{\frac{1}{σ}} \end{aligned}