Parabolic PDEs Notes

PARABOLIC PARTIAL DIFFERENTIAL EQUATIONS Syafiq Johar [email protected]

Contents 1 Heat Equation 1.1 Cauchy Problem for Heat Equation . . . . . . . . . . 1.2 Volume Heat Potential . . . . . . . . . . . . . . . . . 1.3 Heat Potential . . . . . . . . . . . . . . . . . . . . . 1.4 Heat Equation in Half Space . . . . . . . . . . . . . 1.5 Heat Equation in Bounded Domains . . . . . . . . . 1.5.1 Laplace Operator in Bounded Domains . . . 1.5.2 Explicit Solution . . . . . . . . . . . . . . . . 1.5.3 Regularity . . . . . . . . . . . . . . . . . . . . 1.5.4 Uniqueness Results . . . . . . . . . . . . . . . 1.6 Properties of Solutions to Heat Equation . . . . . . . 1.6.1 Local Regularity . . . . . . . . . . . . . . . . 1.6.2 Maximum Principles and Harnack Inequality

. . . . . . . . . . . .

1 2 2 3 4 5 5 6 6 7 7 7 7

2 Measurable Coefficients: Weak Solutions 2.1 Weak Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Existence and Uniqueness of Weak Solutions . . . . . . . . . . . . . . . . . . . . . . . . .

8 8 8

3 Some Properties of Function Spaces for Parabolic PDEs 3.1 Compactness Results for Non-Stationary Problems . . . . . . . . . . . . . . . . . . . . . . 3.2 Parabolic Poincar´e-Sobolev Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 H¨ older Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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4 Measurable Coefficients: Local Regularity 10 4.1 Parabolic PDE with Continuous Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . 11 4.2 Lemmas for Proof of De Georgi-Nash-Moser . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1

Heat Equation

Let Ω ∈ Rn and we consider the space-time cylinder QT = Ω × (0, T ). We denote the parabolic boundary ∂ 0 QT = Ω × {t = 0} ∪ ∂Ω × [0, T ]. We will consider equations of the form ∂t u + Lu = f + div(g) for Lu = −div(a∇u) + b · ∇u + cu. We define anisotropic Lebesgue and Sobolev spaces i.e. Lp,q (QT ) (or

1

Lq (0, T ; Lp (Ω)) or Lq (0, T ; Ω)) and Wp2,1 (QT ) respectively with norms ˆ

! q1

T

||f (·, t)||qp,Ω

||f ||p,q,QT = 0

dt

.

||u||Wp1,2 (QT ) = ||u||p,q,QT + ||∇u||p,q,QT + ||∇2 u||p,q,QT + ||∂t u||p,q,QT . Remark 1.1. 1. The anisotropic Lebesgue space Lp,q (QT ) is the space of all functions from [0, T ] to Lp -functions on Rn which Lp -norms have finite Lq norm wrt t. If p = q, then Lp,p,QT = Lp,QT . 2. In Seregin’s notes, the superscript in Wp1,2 (QT ) denotes the number of derivatives for time and space respectively and the subscript denotes the power of the integrand.

1.1

Cauchy Problem for Heat Equation

Let Ω = Rn and consider the initial value problem ∂t u − ∆u = f

in QT

u(x, 0) = u0 (x)

(1) n

for x ∈ R .

Given functions u0 and f smooth and compactly supported in Rn and QT respectively. We consider u(x, t) = u1 (x, t) + u2 (x, t) where u1 is the heat potential and u2 is the volume heat potential given by ˆ u1 (x, t) = Γ(x − y, t)u0 (y) dy. Rn ˆ tˆ |x|2 1 − 4t for x ∈ Rn , t > 0. u2 (x, t) = Γ(x − y, t − τ )f (y, τ ) dy dτ where Γ(x, t) = n e (4πt) 2 0 Rn The function Γ is known as the heat kernel and its integral over Rn for any t > 0 is equal to 1. Γ also satisfies the heat equation in QT . It can be easily shown that u1 (x, t) solves the homogeneous heat equation with initial condition u1 (x, 0) = u0 (x) whereas u2 (x, t) solves the inhomogeneous heat equation with 0 initial condition in Q+ = Rn × (0, ∞).

1.2

Volume Heat Potential ˆ

Lemma 1.1. Let

|u2 (x, t)|2 dx.

y(t) = Rn

Then, for all t ≥ 0, we have y(t) ≤

√ ˆ tˆ t 0

|f (y, θ)|2 dy dθ =

Rn

√ ˆ t t ||f (·, θ)||22,Rn dθ. 0

By this, it follows that ||u2 (·, t)||22,Rn → 0 as t → 0+ . Lemma 1.2. Let R0 > 0 be so that supp(f ) ∈ BR0 × (0, R02 ). For any k ∈ N∗ , there exists a constant c(n, k) such that for all t ≥ 0 and x ∈ Rn st t + |x|2 ≥ 8R02 , we have |∇k u2 (x, t)| ≤

c(n, k) (t +

|x|2 )

n+2k−2 4

||f ||2,Q+

and |∂t u2 (x, t)| ≤

c(n, f ) (t + |x|2 )

Corollary 1.1. By passing the limit R → ∞, we get the estimate sup ||∇u2 (·, t)||22,Rn + ||∂t u2 ||22,Q+ + ||∇2 u2 ||22,Q+ ≤ 2||f ||22,Q+ . t≥0

This shows that if f ∈ L2 (Q+ ), then u2 ∈ W21,2 (QT ) for any T > 0. 2

n+2 4

.

Now we look at the uniqueness of solution for the inhomogeneous heat equation with 0 initial data. Theorem 1.1 (Weak solutions). Given f ∈ L2 (QT ), there exists a unique v ∈ L2 (QT ) (called the weak solution) st for any smooth test function w ∈ C0∞ (Rn × (−T, T )), ˆ ˆ v(∂t w + ∆w) dx dt = − f w dx dt. QT

QT

Moreover, the following estimates are valid: ˆ ||v(·, t)||22,Rn ≤ t

t

||f (·, s)||22,Rn ds. 0

sup ||∇v(·, t)||2,Rn + ||∂t v||2,Q+ + ||∇2 v||2,Q+ ≤ C||f ||2,Q+ . 0≤t≤T

Theorem 1.2 (Tychonoff). Let v be a measurable function on QT satisfying the assumptions ˆ

|v(x, t)| ≤ M exp(M |x|2 ) ∀(x, t) ∈ QT v(∂t w + ∆w) dx dt = 0 ∀w ∈ C0∞ (Rn × (−T, T )).

QT

Then v ≡ 0.

1.3

Heat Potential

Lemma 1.3. Let u1 be a heat potential. For any t > 0 and p ∈ [1, ∞], we have the following estimate ||u1 (·, t)||p,Rn ≤ ||u0 ||p,Rn . Thus, we can extend the heat potential to all u0 ∈ Lp (Rn ). So, for each t > 0, we define a bounded linear operator S(t) : Lp (Rn ) → Lp (Rn ) by S(t)u0 (·) = u1 (·, t) and S(0) = I. The family of operators {S(t)}t≥0 forms a semigroup (i.e. S(t)S(s) = S(t + s)) which is continuous wrt t (i.e. S(t)u0 → S(s)u0 in Lp (Rn ) as t → s provided 1 ≤ p < ∞). Lemma 1.4. Suppose that u0 is smooth and compactly supported in Rn (so there exists R0 > 0 st ´ supp(u0 ) ⊆ BR0 ), so we have that ∇u1 (x, t) = Rn Γ(x − y, t)∇u0 (y) dy is defined and we get estimates ||∇(u1 (·, t) − u0 (·))||2,Rn → 0

as t → 0

and ||∇u1 (·, t)||2,Rn ≤ ||∇u0 ||2,Rn

for all t ≥ 0.

Lemma 1.5. For k = 1, 2, we have ˆ |∇u1 (x, t)| ≤ c Rn

|∇u0 (y)|2 dy n (t + |x − y|2 ) 2 +k−1

 12 .

Furthermore, for |x| > 2R0 , we have: c||∇u0 ||2,Rn |∇k u1 (x, t)| ≤ √ n ( t + |x|) 2 +k−1

c||∇u0 ||2,Rn and |∂t u1 (x, t)| ≤ √ . n ( t + |x|) 2 +1

Corollary 1.2. From the previous, we get the following estimates sup ||∇u1 (·, t)||2,Rn + ||∂t u1 ||2,Q+ + ||∇2 u1 ||2,Q+ ≤ c||∇u0 ||2,Rn t≥0

sup ||u1 (·, t)||2,Rn ≤ c||u0 (·)||2Rn t≥0

||u1 (·, t)||22,Rn

+ 2||∇u1 ||22,Qt = ||u0 ||22,Rn 3

for any t > 0.

Definition 1.1. Denote ˚ L1p (Ω) for 1 ≤ p < ∞ the completion of C0∞ (Ω) wrt the semi norm ||∇v||p,Ω . ˚ 1 (Ω). For bounded domain with sufficiently smooth boundary, ˚ L1p (Ω) ∼ =W p Lemma 1.6. For each t > 0, u1 (·, t) ∈ ˚ L12 (Ω). Theorem 1.3. 1. Given a function u0 ∈ L2 (Rn ), there exists a unique function v ∈ L2 (QT ) for any T > 0 which satisfies ˆ ˆ v(∂t w + ∆w) dx dt + u0 (x)w(x, 0) dx = 0 (2) Rn

Q+

for any w ∈ C0∞ (Rn+1 ). Moreover, v satisfies the heat equation in the sense of distributions and ||v(·, t)||2,Rn + ||∇v||2,Qt ≤ c||u0 ||2,Rn for all t ≥ 0. Furthermore, v is a continuous function of t ≥ 0 with values in L2 (Rn ). In particular ||v(·, t) − u0 (·)||2,Rn → 0

as t → 0.

2. Given u0 ∈ ˚ L12 (Rn ), there exists a unique function v ∈ L2 (0, T ; ˚ L12 (Rn )) for any T > 0 which satisfies (2). Moreover, v satisfies the heat equation in the sense of distributions and ||∇v(·, t)||2,Rn + ||∇2 v||2,Qt + ||∂t v||2,Qt ≤ c||∇u0 ||2,Rn for all t ≥ 0. Furthermore, ∇v is a continuous function of t ≥ 0 with values in L2 (Rn ). In particular ||∇v(·, t) − ∇u0 (·)||2,Rn → 0

1.4

as t → 0.

Heat Equation in Half Space

Denote the half space Rn+ = {x = (x0 , xn ) : x0 ∈ Rn−1 , xn > 0}. Consider the following initial boundary value problem ∂t u − ∆u = f

in QT = Rn+ × (0, T )

u(x0 , 0, t) = 0

for all 0 ≤ t ≤ T

u(x, 0) = u0 (x)

for all x ∈ Rn .

First, we assume that u0 ∈ C0∞ (Rn+ ) and f ∈ C0∞ (QT ). Extending these to Rn in the odd way, we can get the solution for u. ˆ ˆ tˆ u(x, t) = G(x, y, t)u0 (y) dy + G(x, y, t − s)f (y, s) dy ds Rn +

Rn +

0

where G(x, y, t) = Γ(x − y, t) − Γ(x − y ∗ , t) with y ∗ = (y 0 , −yn ). Using the odd extension, we also define the weak formulation of the problem. For any w ∈ C0∞ (Rn × (−T, T )) with w(x0 , 0, t) = 0, we have ˆ ˆ ˆ u(∂t w + ∆w) dx dt + u0 (x)w(x, 0) dx = − f w dx dt. (3) Rn +

QT

QT

Theorem 1.4. 1. Let f ∈ L2 (QT ) and u0 ≡ 0, there exists a unique function u ∈ L2 (QT ) satisfying the weak formulation (3). Moreover, u ∈ W22,1 (QT ) with the following estimates ˆ t ||u(·, t)||22,Rn+ ≤ t ||f (·, s)||22,Rn+ ds 0

sup ||∇u|| 0≤t≤T

2,Rn +

+ ||∂t u||2,QT + ||∇2 u||2,QT ≤ c||f ||2,QT .

4

2. Given u0 ∈ L2 (Rn ), there exists a unique function u ∈ L2 (QT ) for any T > 0 which satisfies (3). Moreover, u satisfies the heat equation in the sense of distributions and ||u(·, t)||2,Rn + ||∇u||2,Qt ≤ c||u0 ||2,Rn+ for all t ≥ 0 where Qt = Rn+ × (0, t). Furthermore, u is a continuous function of t ≥ 0 with values in L2 (Rn+ ). In particular ||u(·, t) − u0 (·)||2,Rn+ → 0 as t → 0. 3. Given u0 ∈ ˚ L2 (Rn+ ) and f ≡ 0, there exists a unique function v ∈ L2 (0, T : ˚ L12 (Rn+ ) for any T > 0 which satisfies (2). Moreover, v satisfies the heat equation in the sense of distributions and ||∇v(·, t)||2,Rn+ + ||∇2 v||2,Qt + ||∂t v||2,Qt ≤ c||∇u0 ||2,Rn+ for all t ≥ 0. Furthermore, ∇v is a continuous function of t ≥ 0 with values in L2 (Rn+ ). In particular ||∇v(·, t) − ∇u0 (·)||2,Rn+ → 0

1.5

as t → 0.

Heat Equation in Bounded Domains

p 1 0 ˚1 ˚1 0 We denote L−1 p0 (Ω) for p = p−1 an identification of space (Lp ) . In particular, we let H (Ω) = L2 (Ω) and H −1 (Ω) = (˚ L12 )0 with norm given by

||φ||H −1 = sup {|(φ, ψ)| : ||ψ||H 1 = 1}. ψ∈H 1

1.5.1

Laplace Operator in Bounded Domains

Let Ω be a bounded domain with smooth boundary. We look at the Dirichlet boundary value problem ∆u = f ∈ H(Ω) := L2 (Ω) u=0

in Ω on ∂Ω.

The problem has a unique solution u ∈ H 1 (Ω) ∩ W22 (Ω) with the estimate ||∇2 u||2,Ω ≤ C(Ω)||f ||2,Ω . ∆ is an operator with the following properties: 1. It has countable discrete spectrum which diverges to infinity i.e. −∆ϕ = λϕ for ϕ ∈ H(Ω) \ {0} and 0 < λ1 < λ2 < . . . st λm → ∞. 2. dim ker(−∆ − λk 1) is finite for each k ∈ N. 3. The set {ϕk }∞ k=1 of eigenfunctions of the Laplace operator is an orthogonal basis in H(Ω). 1 1 2 4. The set {ϕk }∞ k=1 is an orthogonal system in H (Ω) as well as in dom(∆) = H (Ω) ∩ W2 (Ω) so that λk = ||∇ϕk ||22,Ω = ||∆ϕk ||2,Ω . P∞ P∞ 5. If f ∈ H(Ω), then ||f ||22,Ω = k=1 |ck |2 < ∞ where ck = (f, ϕk ) and k=1 ck ϕk → f in L2 (Ω). P∞ P∞ 6. If f ∈ H 1 (Ω), then ||∇f ||22,Ω = k=1 λk |ck |2 < ∞ and k=1 ck ϕk → f in W21 (Ω). P∞ P∞ 7. If f ∈ dom(∆), then ||∆f ||22,Ω = k=1 λ2k |ck |2 < ∞ and k=1 ck ϕk → f in W22 (Ω).

8. We can get the estimate: ||∆u||H −1 (Ω) ≤ ||∇u||2,Ω = |u|2,1,Ω ≤ ||u||2,1Ω = ||u||H 1 (Ω) . Proposition 1.1. The extension ∆ : H 1 (Ω) → H −1 (Ω) is a bijection. Furthermore, if f ∈ H −1 (Ω), then ∞ X fk2 ||f ||2H −1 (Ω) = where fk = (f, ϕk ). λk k=1

5

1.5.2

Explicit Solution

Consider a bounded domain with smooth boundary Ω. We consider the initial-boundary value problem ∂t u − ∆u = f

in QT = Ω × (0, T ) on ∂Ω × (0, T )

u=0 u(x, 0) = a(x)

(4)

for x ∈ Ω.

Assuming that a ∈ H and f ∈ L2 (0, T ; H −1 ). We construct an explicit solution of the form u(x, t) = P∞ k=1 ck (t)ϕk (x) and ck (0) = ak . Substituting in the equation along with fk , we could derive some estimates via an ODE problem: ||u||2L∞ (0,T ;L2 (Ω)) ≤ 2||a||22 + ||f ||2L2 (0,T ;H −1 ) ˆ TX ∞ ||∇u||2L2 (0,T ;L2 (Ω)) = λk c2k (t) dt ≤ ||a||22 + ||f ||2L2 (0,T ;H −1 ) 0

k=1

||∂t u||2L2 (0,T ;H −1 (Ω))

≤ 2||a||22 + 4||f ||2L2 (0,T ;H −1 )

Letting w ∈ L2 (0, T ; H 1 ), or more precisely, w(x, t) = χ(t)v(x) with v ∈ H 1 and χ(t) ∈ C01 (0, T ), for a.a. t ∈ (0, T ), we get a weak formulation ˆ ˆ (∂t u(x, t) · v + ∇u(x, t) : ∇v(x)) dx = f (x, t) · v(x) dx for all v ∈ H 1 . (5) Ω

Ω

To be precise, the weak formulation above holds at all Lebesgue’s points of the functions t 7→ ∂t u(·, t), t 7→ ∇u(·, t) and t 7→ f (·, t). Theorem 1.5. Assume that u ∈ L2 (0, T ; H 1 ) and ∂t u ∈ L2 (0, T ; H −1 ). Then, u ∈ C([0, T ]; H) and for all t, t1 ∈ [0, T ], we have ˆ tˆ ∂t u · u dx dt = t1

Ω

1 1 ||u(·, t)||22,Ω − ||u(·, t1 )||22,Ω . 2 2

Theorem 1.6. Assume that a ∈ H and f ∈ L2 (0, T ; H −1 ). There exists a unique function u called a weak solution of (4) such that and ∂t u ∈ L2 (0, T ; H −1 ˆ ˆ (∂t u(x, t) · v + ∇u(x, t) : ∇v(x)) dx = f (x, t) · v(x) dx u ∈ L2 (0, T ; H 1 )

Ω

for a.a t ∈ [0, T ]. for all v ∈ H 1

for any v ∈ H 1 .

Ω

u(·, 0) = a(·) fulfilled in the L2 -sense i.e. ||u(·, t) − a(·)||2,Ω → 0 as t → 0+ . Moreover, u ∈ C([0, T ]; H). 1.5.3

Regularity

Recall the weak formulation (5) for the initial-boundary value problem (4). Now assume that a ∈ H 1 and f ∈ L2 (0, T ; H) = L2 (QT ). Then, we have the following estimates: ˆ ||∇u(·, t)||22,Ω + 0

t

||∂t u(·, τ )||22,Ω dτ ≤ ||∇a||22,Ω + ||f ||22,QT

for all t ∈ [0, T ].

||∇2 u||2,QT ≤ c(||f ||2,QT + ||∂t u||2,QT .

6

(6)

Theorem 1.7. Assume that Ω is a bounded domain with smooth boundary and a ∈ H 1 and f ∈ L2 (0, T ; H) = L2 (QT ). Then, u ∈ W22,1 (QT ) with estimates (6). In addition ∇u ∈ C([0, T ]; L2 (Ω)), the equation ∂t u − ∆u = f is satisfied uniquely a.e. in QT and ||∇u(·, t) − ∇a(·)||H → 0 as t → 0+ . Theorem 1.8. Let Ω be a bounded domain with smooth boundary. Consider ∂t u − ∆u = f

inQT = Ω × (0, T )

u|∂ 0 QT = 0. where ∂ 0 QT is the parabolic boundary of QT . Let f ∈ Lp,q (QT ) for some finite p, q > 1. Then, the 2,1 problem has a unique solution such that u ∈ Wp,q (QT ) satisfying the coercive estimate: 2,1 ||u||Wp,q (QT ) ≤ c(Ω, p, q, n)||f ||Lp,q (QT ) .

1.5.4

Uniqueness Results

Lemma 1.7. Let v ∈ L1 (0, T ; L2 (Ω)) and Ω is a bounded domain in Rn with sufficiently smooth ´ boundary. Let QT = Ω × (0, T ). Assume that QT v(∂t w + ∆w) dx dt = 0 for w(x, t) = χ(t)W (x) with arbitrary χ ∈ C0∞ (−T, T ) and W ∈ C 2 (Ω) with W = 0 on ∂Ω. Then, v ≡ 0 in QT . Theorem 1.9. Let v ∈ L1 (0, T ; H 1 ) and Ω is a bounded domain with sufficiently smooth boundary. ´ Assume that QT (v∂t w − ∇v · ∇w) dx dt = 0 for w(x, t) = χ(t)W (x) with arbitrary χ ∈ C0∞ (−T, T ) and W ∈ C0∞ (Ω). Then, v ≡ 0 in QT .

1.6 1.6.1

Properties of Solutions to Heat Equation Local Regularity

Theorem 1.10. Let u ∈ L2 (Q) satisfies the heat equation in Q = B ×(−1, 0) in the sense of distributions ´ (i.e. Q u(∂t w + ∆w) dx dt = 0 for all w ∈ C0∞ (Q)), where B = B1 (0) the unit ball. Then, for any k ∈ N∗ and for any 0 < τ < 1, there exists C(k, n, τ ) > 0 st |∇k u(x, t)| ≤ C

sup

ˆ

(x,t)∈Q(τ )

1.6.2

 21 |u|2 dx dt

where Q(τ ) = Bτ (0) × (−τ 2 , 0).

Q

Maximum Principles and Harnack Inequality

Theorem 1.11 (Weak maximum principle). Let Ω be a bounded domain, T > 0 and QT = Ω × (0, T ). Assume that u ∈ C(QT ) ∩ C 2 (Ω × (0, T )) satisfies ∂t u − ∆u ≤ 0 u≤0

in QT on ∂ 0 QT .

Then, u ≤ 0 in QT . Theorem 1.12 (Strong maximum principle). Assume that u ∈ C(Q) ∩ C 2 (B × (−1, 0)) satisfies ∂t u − ∆u ≤ 0 and u(0, 0) = sup(x,t)∈Q u(x, t). Then, u(x, t) = u(0, 0) for any (x, t) ∈ Q i.e. u is constant in Q. Theorem 1.13 (Harnack’s inequality). Let a non-negative function u ∈ C(Q) ∩ C 2 (B × (−1, 0)) satisfies ∂t u − ∆u = 0. Let z0 = (0, − 21 ). Then, there exists a constant c(n) > 0 st c(n)

sup z∈Q(z0 , 21 )

u(z) ≤

inf

z∈Q(0, 21 )

u(z)

where Q(z, r) = Br (z) × (−r2 + z, z).

7

2

Measurable Coefficients: Weak Solutions

2.1

Weak Solutions

Let Ω be a bounded domain in Rn for n ≥ 2. Consider a linear parabolic PDE in QT = Ω × (0, T ) ∂t u + Lu = f − div(g) u=0

where Lu = −div(a∇u) + b · ∇u + cu

(7)

on ∂Ω × [0, T ]

u(·, 0) = u0 (·) ∈ H = L2 (Ω) where a is a uniformly elliptic (or just elliptic) symmetric matrix. Suppose that a, b and c are bounded measurable functions in QT and f ∈ L2 (QT ) and g ∈ L2 (QT ; Rn ). Definition 2.1 (Weak solution). A function u is a weak solution to (7) if and ∂t u ∈ L2 (0, T ; H −1 ) for a.a. t ∈ (0, T ) ˆ ˆ (∂t u(x, t)w(x) + M(u(x, t), w(x))) dx = (f (x, t)w(x) + g(x, t) · ∇w(x)) dx for all w ∈ H 1 u ∈ C([0, T ]; H) ∩ L2 (0, T ; H 1 )

Ω

Ω

where M(u, v) = (a∇u) · ∇w + b · ∇uw + cuw. Definition 2.2 (Another definition of weak solution). A function u ∈ C([0, T ]; H) ∩ L2 (0, T ; H 1 ) is a weak solution to (7) if for all t ∈ [0, T ] and w ∈ H 1 , we have ˆ

ˆ tˆ (u(x, t) − u0 (x))w(x) dx +

Ω

0

2.2

ˆ tˆ M(u(x, s), w(x)) dx ds =

Ω

(f (x, s)w(x) + g(x, s) · ∇w(x)) dx dt. 0

Ω

Existence and Uniqueness of Weak Solutions

Theorem 2.1. Let a, b, c, f, g satisfy the conditions we stated earlier. Then the problem (7) has a unique weak solution. Lemma 2.1. Assume we are given a sequence uN st ∗

uN − * u in L∞ (0, T ; H)

and ∂t uN * ∂t u inL2 (0, T ; H −1 ).

Then, there exists a subsequence of the sequence uN (denoted as uN ) such that for all w ∈ H, (uN (·, t), w(·)) → (u(·, t), w(·))

3

in C([0, T ]).

Some Properties of Function Spaces for Parabolic PDEs

3.1

Compactness Results for Non-Stationary Problems

Assume that V0 , V and V1 are Banach spaces with the properties: 1. V0 ⊂ V ⊂ V1 and V0 is reflexive. 2. The imbedding V0 ⊂ V is compact. 3. The imbedding V ⊂ V1 is continuous. 4. v ∈ V0 and ||v||V1 = 0 ⇒ ||v||V = 0

8

Lemma 3.1. Given η > 0, there exists C(η) > 0 st for any v ∈ V0 ||v||V ≤ η||v||V0 + C(η)||v||V1 . Proposition 3.1 (Aubin-Lions lemma). Let 1 < p0 , p1 < ∞, V1 be reflexive and define W = {||v||W = ||v||Lp0 (0,T ;V0 ) + ||∂t v||Lp1 (0,T ;V1 ) < ∞}. Then, W is compactly imbedded into Lp0 (0, T ; V ). Proposition 3.2. Let Ω be a bounded domain with sufficiently smooth boundary and let p ≥ 1. The, the space Wp2,1 (QT ) is compactly embedded into Wp1,0 (QT ).

3.2

Parabolic Poincar´ e-Sobolev Inequalities

Proposition 3.3. Let 1 ≤ p < ∞ and u ∈ Wp1,2 (Q(R)) where Q(R) = BR (0) × (−R2 , 0) is the parabolic cylinder. Then, for a constant c independent of R and u, ˆ ˆ ˆ 1 |u−(∇u)Q(R) ·x−(u)Q(R) |p dx dt+ |∇u−(∇u)Q(R) |p dx dt ≤ cR (|∇2 u|p +|∂t u|p ) dx dt. R Q(R) Q(R) Q(R)

3.3

H¨ older Continuity

We first define a bunch of quantities. ˆ ˆ 1 1 |u(z) − (u)z0 ,r | dz. ψ2 (z0 , r) = |∇u(z) − (∇u)z0 ,r | dz. ψ0 (z0 , r) = |Q(r)| Q(z0 ,r) |Q(r)| Q(z0 ,r) ˆ 1 1 ψ1 (z0 , r) = |u(z) − (∇u)z0 ,r · (x − x0 ) − (u)z0 ,r | dz. ψ(z0 , r) = ψ1 (z0 , r) + ψ2 (z0 , r). |Q(r)| Q(z0 ,r) r α

We define H¨ older continuity in the parabolic sense as u ∈ C α, 2 (QT ) if for all z, z 0 ∈ QT , we have p 0 0 0 |u(z) − u(z )| ≤ A|z − z 0 |α |t − t0 | is the parabolic distance. par where |z − z |par = |x − x | + Proposition 3.4. Let u ∈ L1 (QT ) with QT = Ω × (0, T ). Assume that there exists numbers A > 0 and 0 < α < 1 st ψ0 (z0 , r) ≤ Arα whenever Q0 (z0 , r) ⊂ QT . For any Ω0 b Ω and δ > 0, we have: 0

1. For z0 ∈ Ω × [δ, T ], we have limr→0 (u)z0 ,r = u(z0 ). 2. |u(z0 ) − (u)z0 ,r | ≤ c(n, α)Arα . 3. ||u||C α, α2 (Ω0 ×[δ,T ] ≤ c(A + ||u||L1 (QT ) ). Lemma 3.2. Let u ∈ W11,0 (Q). Assume there exist numbers A > 0 and −1 < α < 1 such that ψ(r) ≤ Arα for all 0 < r ≤ 1. We have: 1. If −1 < α < 0, then limr→0 (u)0,r exists and the following estimates hold: |(∇u)0,r | ≤ |(∇u)0,1 | + c(n, α)Arα .

|u0 − (u)0,r | ≤ c(n, α)(Ar1+α + r|(∇u)0,1 |).

2. If α = 0, then limr→0 (u)0,r exists and the following estimates hold: 1 |(∇u)0,r | ≤ |(∇u)0,1 | + c(n, α)A ln . r

1 |u0 − (u)0,r | ≤ c(n, α)((A + |(∇u)0,1 |)r + Ar ln . r

3. If 0 < α < 1, then limr→0 (u)0,r and limr→0 (∇u)0,r = U0 exist and the following estimates hold: |(∇u)0,r | ≤ |(∇u)0,1 | + c(n, α)Arα .

|u0 − (u)0,r | ≤ c(n, α)(Ar1+α + r|(∇u)0,1 |).

|U0 − (∇u)0,r | ≤ c(n, α)Arα . 9

Proposition 3.5. Let u ∈ W11,0 (QT ). Assume there exist numbers A > 0 and −1 < α < 1 such that ψ(r) ≤ Arα whenever Q(z0 , r) ⊂ QT . For any Ω0 b Ω and δ > 0, we have: 1+α 2

1. If −1 < α < 0, then u ∈ C 1+α,

0

(Ω × [δ, T ]) with the estimate

||u||

C 1+α,

1+α 2

0

(Ω ×[δ,T ])

≤ c(A + ||u||W 1,0 (QT ) ). 1

0

β

2. If α = 0, then u ∈ C β, 2 (Ω × [δ, T ]) for any 0 < β < 1 with the estimate ||u||

β

0

C β, 2 (Ω ×[δ,T ])

≤ c(A + ||u||W 1,0 (QT ) ). 1

0

0

α

3. If 0 < α < 1, then u ∈ Lip(Ω × [δ, T ]) and ∇u ∈ C α, 2 (Ω × [δ, T ]) with the estimate ||u||Lip(Ω0 ×[δ,T ]) + ||∇u||C α, α2 (Ω0 ×[δ,T ]) ≤ c(A + ||u||W 1,0 (QT ) ). 1

4

Measurable Coefficients: Local Regularity

Consider the equation for u ∈ L2,∞ (Q) ∩ W21,0 (Q) (8) ´ with symmetric and elliptic a, in the weak sense i.e. Q (u∂t w−(a∇u)·∇w) dx dt = 0 for any w ∈ C0∞ (Q). ∂t u − div(a∇u) = 0

α

Theorem 4.1 (De Giorgi-Nash-Moser). Any function satisfying (8) belongs to C α, 2 (Q(r)) for any 0 < r < 1 and some 0 < α < 1. This is very difficult to prove. We require some proposition and lemmas. Proposition 4.1. If u satisfies (8), then, ∂t u ∈ L2 (−1, 0; H −1 (B)). Proposition 4.2. Assume that u satisfies (8). Let ψ ∈ C0∞ (B × (−1, 1)). Then, for any t ∈ S, we have ˆ ˆ t ˆ ˆ t ˆ ˆ t ˆ 2 2 2 2 2 u ψ dx + 2 ψ (a∇u) · ∇u dx ds = u ∂t ψ dx ds − 2 (a∇u) · ∇ψ 2 dx ds. (9) B

−1

−1

B

−1

B

B

Remark 4.1. If u ∈ C([−1, 0]; L2 (B)), then the identity (9) holds for all t ∈ [−1, 0]. Another point to notice is if from the fact that u ∈ L2 (−1, 0; H −1 (B)). So, for any ϕ ∈ H 1 (B), ˆ ˆ tˆ ˆ tˆ (u(x, t)ϕ(x) − u(x, s)ϕ(x)) dx = ∂τ u(x, τ )ϕ(x) dx = − (a(x, τ )∇u(x, τ )) · ∇ϕ(x) dx dτ. B

s

B

s

B

We now want to prove Caccioppoli-type inequality. Fix a non-negative cutoff function ϕ ∈ C0∞ (B(2)) and χ0 (t) st ϕ(x) = 1 for x ∈ B and χ0 (x) = 0 for t ≤ −4, χ0 (x) = 1 for t ≥ −1, and linearly interpolated in −4 < t < 1. For a point z0 = (x0 , t0 ) and R > 0 st Q(z0 , 2R) ⊂ Q, we let     x − x0 t − t0 . ϕ (x) = ϕ . χt0 ,2R (t) = χ0 x0 ,2R R2 R Also we introduce a mean value of u as: ˆ u ˆx0 ,2R (t) = B(x0 ,2R)

ˆ u(x, t)ϕ2x0 ,2R (x) dx

B(x0 ,2R)

!−1 ϕ2x0 ,2R (x) dx

.

Lemma 4.1 (Caccioppoli-type inequality). Under our standing assumptions, the following holds ˆ ˆ c 2 |∇u| dz ≤ 2 |u(x, t) − u ˆx0 ,2R (t)|2 dz. R Q(z0 ,2R) Q(z0 ,R) 10

Lemma 4.2. Under our standing assumptions, if Q(z0 , 2R) ⊂ Q, the following holds ˆ ˆ 2 2 |u − (u)z0 ,R | dz ≤ c(n, ν, ϕ)R |∇u|2 dz. Q(z0 ,R)

Q(z0 ,2R)

Lemma 4.3. Assume that Q(z0 , 4R) ⊂ Q. Then ˆ ˆ 2 |u − u ˆz0 ,2R | dz ≤ c(n, ϕ) Q(z0 ,2R)

4.1

|u − (u)z0 ,4R |2 dz.

Q(z0 ,4R)

Parabolic PDE with Continuous Coefficients

Suppose that u satisfies (8) in Q and a ∈ C(Q) is uniformly elliptic. Fix 0 < r1 < 1. Given δ > 0, we can find 0 < µ < (1 − r1 )/2 such that |a(z) − a(z0 )| < δ whenever z ∈ Q(z0 , µ) and z0 ∈ Q(r1 ). Given z0 ∈ Q(r1 ) and 0 < r < µ let v ∈ C([t0 − r2 , t0 ]; H(B(x0 , r)) ∩ W21,0 (Q(z0 , r)) with ∂t u ∈ L2 (t0 − r2 , t0 ; H −1 (B(x0 , r))) as a solution to ∂t v − div(a(z0 )∇v) = ∂t u − div(a(z0 )∇u) := f

in Q(z0 , r) on ∂ 0 Q(z0 , r).

v=0

By using energy estimates and continuity assumptions, we get ˆ ˆ 2 2 |∇v| dz ≤ cδ |∇u|2 dz. Q(z0 ,r)

Q(z0 ,r)

By setting w = v − u where w is the distributional solution to the parabolic equation with constant coefficients ∂t w − div(a(z0 )∇w) = 0, the following estimate is valid using Caccioppoli: ˆ

 ρ n+2 ˆ

|∇w|2 dz for any 0 < ρ ≤ r ≤ µ r Q(z0 ,r)   ˆ ˆ ρ n+2 |∇u|2 dz ≤ c(n, ν) + δ2 |∇u|2 dz r Q(z0 ,ρ) Q(z0 ,r)  n+2−γ ˆ ˆ r |∇u|2 dz ≤ c1 |∇u|2 dz for z0 ∈ Q(r1 ) and any 0 < r ≤ µ. µ Q(z0 ,r) Q |∇w|2 dz ≤ c(n, ν)

Q(z0 ,ρ)

⇒ ⇒

The last inequality is obtained by the technical lemma: Lemma 4.4. Assume that Φ : [0, R0 ] → [0, ∞) is an increasing function with    r n+2 Φ(r) ≤ c + ε Φ(R) for any 0 < r ≤ R ≤ R0 with c, ε > 0 constants. R For any 0 < γ < n + 2, there exists ε0 (n, γ, c) st if ε ≤ ε0 , then  n+2−γ r Φ(R0 ) for all 0 < r < R0 . Φ(r) ≤ c1 (n, γ, c) R0 Then, by using Lemma 4.2, we get ˆ ˆ |u − (u)z0 ,r/2 |2 dz ≤ c2 rn+4−γ |∇u|2 dz. Q(z0 ,r/2)

Q

This proves the following theorem α

Theorem 4.2. Assume all of the above, then for any 0 < α < 1 and 0 < r < 1, we have u ∈ C α, 2 (Q(r)). β

Remark 4.2. Assuming that a is H¨ older continuous with exponent 0 < β < 1 i.e. a ∈ C β, 2 (Q), we can β1 improve the theorem to show that ∇u ∈ C β1 , 2 (Q(r)) for any 0 < β1 < β and 0 < r < 1. 11

4.2

Lemmas for Proof of De Georgi-Nash-Moser

We require several lemmas for the proof. Lemma 4.5 (Moser iteration). For any 0 < R < 1, we have ˆ |u(z)| ≤ C(R, ν, n)

sup z=(x,t)∈Q(R)

|u(z)2 dz

 21 .

Q

Lemma 4.6 (Harnack’s inequality). Let a non-negative function u ∈ C(Q(R)) ∩ C 2 (B(R) × (−R2 , 0]) satisfies ∂t u − div(a∇u) = 0 in Q(R) with ellipticity condition. Let z0R = (0, −R2 /2). Then, c(n, ν)

u(z) ≤

sup

inf

u(z).

z∈Q(R/2)

z∈Q(z0R ,R/2)

Lemma 4.7 (Estimate of H¨ older norms). Let u ∈ C(Q) ∩ C 2 (B × (−1, 0]) be a solution of ∂t u − div(a∇u) = 0 in Q(R). Then, for any 0 < r < 1, there are two constants A(n, ν, r) > 0 and 0 < α(n, ν) < 1 such that |u(z) − u(z0 )| ≤ A|z − z0 |α par sup |u(z)| z∈Q

12

for any z, z0 ∈ Q(r).