Lecture 16 - Weak Solutions and Weak Derivatives

#math

L# Mathematical Reasons

u_{t} = Δ u^{m} = div (u^{σ} u)

degenerate equation \to finite support

We can't expect a classical solution.

We take note of the following: $Q_{T} = Ω \times [0, T]$ and $S_{T} = \partial (Ω \times [0, T])$ and also note that $Ω$ is bounded and regular henceforth.

Definition

Consider the differential equation:

L u = f

A classical solution is a function $u \in C (Ω)$ such that $L u = f$ on $Ω$ in the usual sense.

Thus its reasonable to ask what type of solution did we find in the previous class?

Formulation

$u$ is said to be a weak solution of the Porous Medium Equation iff:

\underset{Q_{T}}{\int \int} (\nabla u^{m} \nabla ϕ - u ϕ_{t}) d x d t = \int_{Ω} u_{0} (x) ϕ (x, 0) d x

for any $ϕ \in C^{1} ({\overset{―}{Q}}_{T})$

ϕ |_{\partial S_{T}}, ϕ (x, T) = 0

Definition (Weak Derivative)

Let $u$ and $v$ be Lesbesgue measurable and local integrable in $Ω$ . Let:

\begin{matrix} (*) & \int_{Ω} v ϕ d x = - \int_{Ω} u (\partial_{x_{i}} ϕ) d x \end{matrix}

for any $ϕ \in C_{c}^{\infty} (Ω)$ then we say $\partial_{x_{i}} u = v$ .

Example

$f (x) = | x |$ we have $| x |^{'} = {\begin{cases} - 1 & x < 0 \\ 1 & x > 0 \end{cases}$ in the weak sense.

Definition

$H^{1} (Ω) = {f : f \in L^{2} (Ω) & \nabla u \in (L^{2} (Ω))^{n} (in the weak sense)}$

Inner product:

⟨ u, v ⟩ = \int_{Ω} u v + \int_{Ω} \nabla u \cdot \nabla v

Definition

$H_{0}^{1} (Ω) = {u \in H^{1} (Ω) : u |_{\partial Ω} = 0} = \overset{―}{C_{0}^{\infty} (Ω)}$

Recall

Different spaces

Definition (Weak Solution)

$u \geq 0$ is called a weak solution of PME

{\begin{cases} u_{t} = Δ u^{m} \\ u |_{δ Ω} = 0 \\ u (x, 0) = u_{0} (x) \in L^{1} (Ω) \end{cases}

if:

$u^{m} \in L^{2} ((0, T); H_{0}^{1} (Ω))$
for any $ϕ \in C^{1} ({\overset{―}{Q}}_{T})$

\underset{Q_{T}}{\int \int} (\nabla u^{m} \nabla ϕ - u ϕ_{t}) d x d t = \int_{Ω} u_{0} (x) ϕ (x, 0) d x

Uniqueness

(Oleinik)
Assume that we have two solutions $u_{1}$ and $u_{2}$ , consider $u_{1} - u_{2}$ , then:

\underset{Q_{T}}{\int \int} \nabla (u_{1}^{m} - u_{2}^{m}) \nabla ϕ - (u_{1} - u_{2}) ϕ_{t} = 0 \forall ϕ

Take your test function to be:

ϕ (x, t) = \int_{t}^{T} (u_{1}^{m} (x, s) - u_{2}^{m} (x, s)) d s, 0 < t < T

Even if $ϕ$ does not have the required smoothness, we may approximate it with smooth functions $ϕ_{ϵ}$ for which the weak formulation of the solution is satisfied for these functions. For notation take $u_{i}^{m} = ω_{i}$ .

We compute:

ϕ_{t} = - (ω_{1} - ω_{2})

\nabla ϕ = \int_{t}^{T} (\nabla u_{1}^{m} - \nabla u_{2}^{m}) d s \in L^{2} (Q_{T})

and moreover $ϕ (t) \in H_{0}^{m} (Ω)$ and $ϕ (x, T) = 0$ .