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Notes By Ibrahim Al Balushi

Lecture 16

Dirichlet Problem for Poisson eq.u= f in

u= g on (1)

If v= f, then w = u v satisfiesw= 0 in

w= g v on

We needu

C2()

C0(). Our candidate

u(x) =

E(x y)f(y)dy

But f C() is not sufficient; u / C1() s.t u C(). This leads to the notion of Holdercontinuous spaces. In particular, we will show that the integrability and boundedness off can onlyguarantee a function uC1 in the integral expression above. It will be established that a necessarycondition for u C2 which solves u = f is that f must not only be continuous, but also Holdercontinuous. The uniqueness will of u, should it exist, will also depend on the boundary of , asdescribed in the discussion above.

Definition 1. Letf : R, y, 0<

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Newtonian Potential

Definition 2. For any integrablefon domain we define the Newtonian potential offas a functionw defined by

w(x) =

E(x y)f(y)dy, (5)

whereE(x y) defines the fundamental solution for the Laplacian.Suppose f is defined to be the right hand side of the Laplace equation, then w(x) would define

the solution for u= f, ie w := u. By this definition, we may deduce information and estimates onderivatives ofu by studying the behaviour of integral (5). These estimates will form basis for Schauderestimates and the extension of potential theory from the Laplace equation to general linear ellipticPDEs.

Theorem 1. Letfbe bounded and integrable in, and letw be the Newtonian potential off. Then

wC1

() and for anyx,iw(x) =

iE(x y)f(y)dy, i= 1,...,n. (6)

Proof. Choose such thatC(R), 0, = 0on [0, 1] = 1on [2, ) and define(t) =

t

,

for some >0,

w(x) =

E(x y)(|x y|)f(y)dy (7)

is clearly in C1(). now,

|w(x)

w(x)

|=

E(x

y)[(

|x

y

|)

1]f(y)dy (8)

f

|E(x y)||(|x y|) 1|dy (9)

(10)

note that (|x y|) 1 = (1|x y|) 1 = 0 whenever|x y| 2 so we have

=f

|xy|2

|E(x y)|dy

by change of variables|x y|= r we have

=f20

Crn2

rn1drC2f

Henceww uniformly in . ConstantCis a result of higher dimensional polar integration;

d, being change of variables Jacobian determinant. Now let

v(x) =

iE(x y)f(y)dy (11)

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We aim to assert the final statement. Naively, one may immediately attempt to employ Liebnezrule by differentiating under the integral, however E(x

y) is not continuous nor bounded whenever

x y= 0. This is essentially why we make use of(t) where we cut out that region for some >0.This then allows us to invoke Liebnez rule and showing that the result holds at the limits.

|iw(x) vi(x)|= xi

E(x y)(|x y|)f(y)dy

E(x y)xi

f(y)dy

(12)=

E(x y)xi

(|x y|)f(y)dy+

E(x y) (|x y|)xi

xi yi|x y| f(y)dy (13)

E(x y)xi

f(y)dy

(14)

=

Eyxi

[ 1]f(y)dy+

Eyi(|x y|) xi yi|x y| f(y)dy (15)

Cf20

rn1

rn1dr+

2

Eyxi yi|x y| f(y)dy (16)

= iwvi uniformly

A stronger result occurs should fbe Holder continuous on as well. In particular, the Newtonianpotential becomesC2 instead ofC1 by the conditions above.

Theorem 2. Letf be bounded and locally Holder continuous with

[0, 1] in, and letw be the

Newtonian potential off. ThenwC2(), w= f in, and for anyx

ij w(x) =

ij E(x y)(f(y) f(x))dy f(x)

iE(x y)j (y) dSy, i, j = 1,...,n. (17)

where for which the divergence theorem holds andfis extended to vanish outsideProof. The proof of regularity is similar to the previous theorem, however the crucial step is when weuse Holder continuity. Define the RHS of (17) by (x) and set v = iw so for >0 define

v(x) =

iE(x y)(|x y|)f(y)dy (18)

We can write the difference

|(x) j v(x)|=

|xy|2

xj

[1 (|x y|)]

xiE(x y)

(f(y) f(x))dy

(19)Holder continuity implies|f(x) f(y)| [f];x|x y| (20)

[f];x

|xy|2

xj

[1 (|x y|)]

xiE(x y)

|x y|dy (21)

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To assert the solvability of w= fwe will set =BR(x) whereR is sufficiently large in (17) wehave

w=

i

BR(x)

2i E(x y)(f(y) f(x))dy f(x)

BR(x)

iE(x y)i dSy (22)

=

BR(x)/B(x)

E(x y) =0

[f(y) f(x)] (23)

+

B(x)0

as 0

i

2i E(x y)[f(y) f(x)] f(x)

BR(x)

Ey dSy (24)

=

f(x)

BR(x)

Ey (25)

taking i=xi yi|x y| we have, (26)

= f(x)

|Sn1|Rn1

|xy|=R

dSy (27)

=f(x) (28)

Lemma 1. uC1(Rn) ifC() L(). Moreover,uC1 Cf.Proof.

uuiuvi

u(x + hei) u(x) = u(x + hei) u(x) + o(1) (29)=hiu(x + h

ei) + o(1) (30)

=hvi(x + hei) + ho(1) + o(1) (31)

=hvi(x) + ho(1) + o(1) (32)

(33)

vi,(x) =

iE(x y)(|x y|) N

f(y)dy

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vi,vi uniformly.

ivi(x) =B

iN(f(y) f(x))dy+ f(x) B

iN dy F =N eidivF = Nyi

=Nxj

(34)

=

B

[f(y) f(x)]iN+ f(x)

B

N

(35)

for B large enough (36)

=

B

[f(y) f(x)]iN f(x)

B

vj diE (37)

vij (x) =

B

dj diE(x y)[f(y) f(x)]dy

B

f(x)j iE(x y)dyn1

ivi, vij (x) w

=

B

[f(y) f(x)] j [((|x y|) 1)iE(x y)] A

dy

A= xi yi|x y|

|A| Crn1

+ C

r2

suppose|f(y) f(x)| w(|x y|)

|D| 20

Cw(r)rn1dr

rn1 +

20

Cw(r)rn1

rn dr

Cw(r)dr

+

Cw(r)

r dr0as0.

Definition 3. f C() is called Dini continuous ifK compact,w: Ro Ro integrable s.t|f(x) f(y)| w(|x y|) x, yK

and thatR >0, R0 w(r)r1dr

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Iffis Dini in , then

j iu(x) =B

[f(y) f(x)]K(x y)dy f(x) B

j iE(x y)dn1y

B, K= j iE, extendfby 0 outside of . uC2()Theorem 4. Iffis Dini in, thenuC2()C1(Rn)andu= f in. In addition, iff C0,(B)andB, then

uC2 CfC(B).Proof. We have

u(x) =

i

iiu(x) (38)

= BR(x)

( )i

iiE(x y) =0

dy f(x) BR(x)

i

iiE(x y)dn1y (39)

iE(x y) = 1|Sn1|xi yi|x y|n =

i|Sn1||x y|n1

wherei= xiyi|xy|

I2 =

|xy|=R

dn1

|Sn1||x y|n1 =|Sn1|Rn1|Sn1|Rn1 = 1 = u(x) = f(x)

bound for : |j iu(x)| B

[f],x|x y||x y|ndy+ |f(x)|C

[f],x:= supyB|f(x)f(y)||xy| and|f|C(B) = supxB [f],xwe have

fC(B)=fC(B)+ |f|C(B)

Theorem 5. bounded, regular everywhere, f C() Dini in , g C(). Then!uC2() C() s.t

u= f in

u= g on

Proof.

v(x) =

E(x y)f(y)dy = vC2() C1(Rn)

. Then solve w= 0 in

w= g v on

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= u= v+ wC2() C()u= v+ w= f in

u= v+ (g v) = g on

Second Order Elliptic Equations

Lu=

ij

aijij u +

j

bj j u + b0u

aij , bkC(). A= (aij) is Symmetric Positive definite : 1||2 TAn||2 where Rnand1 andn positive constants. Uniformly elliptic.

Given the conditions above, at each x0,QO(n) s.tQA(x0)Q

T = diagonal

= diag(1,...,n), i> 0.

Method of Continuity

Theorem 6. X, YBanach spaces, A0, A1: XY bounded linear operators, At = (1 t)A0+ tA1 :X Y. AssumeC >0 s.t

uX CAtuY uXA0 is surjective = A1 surjective.Proof. Injectivity implies Atu = Atv = uvX 0 = u = v, At injective. Assume Atsurjective. Lets[0, 1]. Asu= f given some arbitrary f Y. We want to use Banach fixed pointtheorem. That is, construct mapping : XX s.t u = u.

Atu= f Asu + Atu= f+ (t s)(A1 A0)u

u= A1t f+ (t s)A1t (A1 A0)u=: u

(u v)X =(t s)A1t (A1 A0)(u v)X (40) |t s|A1t A 1 A0

want

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Schauder Estimates

We will use method of continuity with X=C2,o (), Y =C

0,

().Lu= f in

u= 0 on

idea: Localization, freezing coefficients, estimate the error (L A0)u, partition of unity :Bi. iC, 1]),

i

i = 1in

LA0 = A(x0) from var to constant coeff. We now estimate

uX =iiuX i

iuX CiA(xi)iuY

C

i

LiuY + C

[A(xi) A]iuY + BiuY

C

i

iLuY CLuY

+(Li iL)uY + [A(xi) A]iuY + BiuY

Lecture 18

Motivation from Last time

L=

ijaij ij+

ibij+ b0

Assumeaij , bkC(). > 0 s.t TTA, Rn. Aiming foruC2,C[(1 t) + tL]uC

(continuity of operators)

At = (1 t)I+ tA = TAt(1 t)T+ tTmin(1, )T.sufficient is to show

uC2,CLuCwith C mildly dependence on L. Recall

u(x) = E(x y)f(y)dy, f C(B), Bij u(x) =

[f(y) f(x)]ij E(x y)dy f(x)

B

j iE(x y)dn1y

anduC() C1(Rn), u= f in ,uC2()CfC(B)

ij u(x) iju(x) = f(x)

B

j i[E(x y) E(x y)] + [f(x) f(x)]

B

iiE(x y)+

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consider the midpoint betweenx and x and take ball =|x x| x = x+x2

+

B(x)

[f(y) f(x)]K(x y)dy+ (xx)

ij E(x y) =: K(x y)

+[f(x) f(x)]

B/B(x)

K(x y)dy+

B/B(x)

[f(y) f(x)][K(x y) K(x y)]dy

=BR(z), B= B2R(z)

we find bounds

|I1| |x x|

B

|iE(x y)|dn1y|x x

Rn |B |

where R is the distance between and B whereB

and y

B

C|x x|

R C

|I2| C,

|I3|

B(x)

[f]x,|x y|n

ify was outside ball then|y x| C|y x|

C[f]x,

B3/2(x)

|x y|n

C[f]x,

|I4| C[f]x,

|I5| C+ |

B(x)

jiE(x y) C1n

dn1y| C

|I6| [f]x, |y x||x x||K( y)|dy

[f]x,|x x|

B/B(x)

|y x|| y|n+1 dy

C[f]x,|x x| |y x|n1dy

C[f]x,

rn1rn1dr

C[f]x,1

so|ij u(x) iju(x)| C(|f(x)| + [f]x, + [f]x,)

= |u|C2,()CfC(B)

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up to boundary

Half annuluis on upper half space.R

n+={xRn|xn> 0}

B+r =BrRn+B1 = BR, B2= B2R

=B2{xn= 0}xB+1

ij u(x) =

B+2

[f(y) f(x)]K(x y)dy f(x)

B+2

j iE(x y)dn1y

j has nonzero comp upwards on sigma so

j

=n : ...= 0

same as the interior case.

i=n : by symetry

j i=

ij

i= j = n : u= f = 2nu= fn1i=1

2i u

|2nu|C(B+1 ) |f|C(B+1 ) +n1i=1

|2i u|C(B+1 )

CfC(B+2 ) = |u|C2,(B+1 )CfC(B+2 )Theorem 7. u

C2(B2)

C(B2), f

C(B2), u= f in B2

= uC2,(B1) anduC2,(B1)C(uC(B2)+ fC(B2))Proof.

w= E f = w= f inB2, v= u w = v= 0 in B2.w2,C (B1)CfC(B2)

vC2,(B1)CvC(B2)CuC(B2)+ CwC(B2)whichCwC(B2)CfC(B2)Lemma 2.

XYZ,Banach spaces that are contiuously embedded :

uYC1uX uZC2uYwithXcompactly embbded inY:

{uX|uX1} is compact in Y= > 0, C()0st

uYuX+ C()uZ uX

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Lecture 19

Last time

Theorem 8. uC2(B2) C(B2), f C(B2), u= f in B2; Bi := BiR= uC2,(B1),uC2,(B1)C(uC(B2)+ fC(B2))

whereCdepends onR.

|u|C2,(BR)C(R2uC(B2R)+ RfC(B2B)+ fC(B2R))Analogy: dimensional transformation

x= Rx R = u(x) = u(Rx)

|u|C() = sup

=|u(Rx) u(Ry)|

|x y| = sup|u(x) u(y)|

R|x y| =R+|u|C()

|u|C2,(BR)= R2|u|C2,(B1)Const R2(uC(B2)+ R2fC(B2)+ R2|f|C(B2))=R2(uC(B2R)+ R2fC(B2R)+ R2+|f|C(B2R))

Observation:x= (x)coord transformation

|x y| .|(x) (y)| .|x y| = |u|C() |u|C()XY = X Y X

We provied for upper half annulus thatu = E fuC2,(B+1 ) fC(B+2 )

Theorem 9. uC2(B+2) C(B+2 ), f C(B+2 ), u= f in B+2, u= 0 on

= uC2,(B+1 ) anduC2,(B+1 ) uC(B+2 ) + fC(B+2 )Proof.

w(x) =

B+2

E(xy)f(y)dy

B+2

E(xy)f(y)dy y reflection about (base of upper half annulus)

= w= f in B+1 , w= 0 on v= u w = v= 0 in B+1, v= 0on .

vC2,(B+1 ) vC(B2) uC(B2) conu

C(B+2 )

+wC(B2)

take f(y) = f(y) for yB2 f (y) = f(y)yB+2

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w(x) = 2

B+2E(x y)f(y)dy

B2E(x y)f(y)dy

fC(B2) fC(B+2 ) , fC(B2) fC(B+2 )

|f(x) f(y)||x y|

|f(y) f(x)||x y| for x,yB

+2

|f(x) f(y)||x y| xB

+2 , yB2

Lemma 3. X compactly embedded in Y continuously embedded in Z Banach

> 0,C()> 0 s.tuYuX+ C()uZ uCProof. Let >0 be given.{un} X, unX = 1, stunY > + nuZ nN

unu in Y,unZ< 1n

uY0 = u unZCu unY0 = u= 0 but

uY unY u unY > u unY contradiction.

|uv(x)uv(y)| |u(x)v(x)u(y)v(y)|+|u(y)v(x)u(y)v(y)| |u|C |xy||v(x)|+|u(y)||v|C |xy|

= |uv|C |u|C |v|C0 + |u|C0 |v|C

Global Schauder Estimates

L=

aijij+

bii+ bi, aij , bkC().

A + BLety interior. Ay =

aij (y)ij . r >0 is such that B2r

uC2,(B1) r2uC(B2)+ rAyuC(b2)+ |Ayu|C(B2)AyuC(B2) LuC(B2)+ (Ay A)uC(B2)+ BuC(B2)

BuC biuC bCuC1(Ay A)uC(B2) (aij (y) aij )2uC(B2) aijCuC2(B2)

aij (y) aij ) raijC|Ayu|C(B2) LuC(B2)+ (Ay A)uC+ BuC

Bu

C

bk

C

u

C1,

(Ay A)uC aij(y) aijCuC2+ aij (y) aijCuC2, uC2+ ruC2,(B2)choose r small enough, uC2,0 (B1) =

uC2, LuC+ uC2uC2uC2,+ C()uC0

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Lecture 20

Last time

L=

aijij+

bii+ b0 aij , bkC()Theorem 10. yr >0 small such that

uC2,o (Br(y)) LuCBr(y)+ uC2(Br(y))

for alluC2,o (B+1).Lemma 4.

uC2,(B+1 ) uC(B+2 ) + uC(B+2 )

uC2,

(B+2 ) C(B

+2), u= 0 on. : upper hald annuli with base

Definition 4. we say Ck, isu Uy nbhd of y and : Uy B1 s.t C2,(Uy),1 :B1Uy, 1 C2,(B1) and(Uy ) = B+1 , (Uy ) = B 1+\B1.Theorem 11. C2,, yUnbhd ofy such that

uC2,(U+) LuC(U+)+ uC2(U+)

U+ =U uC2,0 (U), u= 0 onU Proof. By definition,: UyB1; Ur = 1(Br) (r 0 sufficnetly small

uC2,(U+r ) LuC(U+r )+ uC2(U+r )

Theorem 12 (Shauder 1934). Rn bdd domain, C2,,C >0 s.t

uC2,()C(LuC()+ uC()) uC2,() u= 0 on

Proof.y let Uy be nbhd of y s.t local estimates holds{Uy} open cover of . choose a finitesubcover

{Ui

}. we have

{i

}(partion of unitiy ) subordinate to

{Ui

}i : R

n [0, 1] iCo (Ui)

i on

U+i =Ui

uC2,() =

iuC2,()

iuC2,(U+i )+ iuC2(U+i )

(LuC(U+i )+ u2C(U

+i ))

i

(LuC()+ uC2()) LuC()+ uC2()

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iuC2(U+i ) uC2(U+i )

Liu= iLu + (Li iL)u (firstorder)LuC iLuC+ uC1, LuC

= uC2,()C(LuC()+ uC2())uC2,()uC2,+ C()uC

for method of continuity we needuC2, LuC .In general the extra term +uCis unavoid-able because for example

u + ku = 0

u= 0

has non trivial solution.Maximum principal:

b0 Lu0 in uC2() C()= sup

usup

u

Proof. Lu > 0 in . y s.tu(y) = sup u hence Lu0(u0). Define u = u + ex

Corollary 1. b00 Lu0 = umax{0, sup u}Proof. + ={x :u(x)> 0}

Lou= aij iju +

biiu b0u0

in +

= sup+

usup+

umax{0, sup

u}

Corollary 2. b00, Lu= f = |u| sup |u| + Cf.Proof. Choose >0 such that

L0ex1

a11ex1+bex1

ex1

for some >0. Let

v(x) = sup |

u

|+

ed ex1

f

LvL0u fL(u + v)f f0 u + v0 on

L(u + v) f f0 u + v0 on = |u| v = sup |u| + Cf.

Ifb00 :uC2,() LuC()

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Theorem 13 (Shauder). b00 C2, f C(). Then there exists uniqueuC2() C()such that Lu= f in

u= 0 on

moreover,uC2,()anduC() fC()Proof.

X={uC2,()| u = 0 on }yC()

Lecture 21

L=

aij ij +

bii+ b0

where{aij}ij symmetric positive matrix.Theorem 14 (Schauder Regularity Thm). Letaij , bkC(), C2,, f C(), b00.

= !uC2() C() s.tLu= f in

u= 0 on

In addition,u2,; f;Implicitly assume:

Theorem 15 (kellogg). C2,, f C(), uC2() C()satisfiesu= f in

u= 0 on

= uC2,()sketch. Strategy:- first prove for balls then apply Schauder regularity theorem for balls.

Inhomogeneous Dirichlet Boundary Condition

Definition 5. We sayg

C2,() if

v

C2,() s.t

v| = g and g2,,:={v2,;: v|= g} (43)

Corollary 3. fC(), gC2,() = !uC2() C() s.tLu= f in

u= g on and u2,; f;+ g2,;

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Proof. LetvC2,()s.t v| = g, v2,2g2,Lw= f Lv in

w= 0 on and w2, f+ v2,

u= v + wu2, v2,+ w2,

Meaning:F(u) = (Lu,u|), F :C2,()C() C2,()

F1 1

Elliptic Estimates in Holder Spaces

Ck+2,, aij , bkCk,, =uk+2, Luk,+ uk

Proof. ForL = , k= 1

u2, u+ uC u1,+ u1

Elliptic Regularity in Holder spaces

Ck+2,, aij , bk, f

Ck,, g

Ck+2,(), u

C2()

C()

Lu= f in

u= g on = uCk+2().

Proof. ForL = + Id, k= 1 : ifv = iuC2, it should satisfyv= if iu v= ig on

(44)

solving for v , we get vC2,(). Lethu be finite difference approximation ofiu

hu(x) =u(x + hei) u(x)

h

hu= hf hu (45)Subtract (45) from (44)

(v hu) = if hf+ hu iu in v ku= ig hu on

v hu if hf+ ig hu0.

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Meaning: solution is as regular as the data allows. In particular, if everything is smooth, thensolution is smooth. E.g Lu u= 0 in

u= 0 on

andL is C, C = uC.

Interior Regularity

aij, bk, fCk,, uC2. Lu= f in B3r = uCk+2,(Br).Proof.

v= if iu in B2rv= iu on B2,r

= vC3,(Br).

hu= hf hu in B2r= u hu if hf+ iu hu+ iu huC().

Case b0 > 0: In general, L is not invertible with homogeneous B.C, but L Id is invertible forsufficiently large 0 > 0. = (L Id)1 is called the Resolvent ofL. When it exists,

: C()C2,()C()

so : C C is compact.

(L )u= f uu = f u(Id )u= f

Lecture 22

F =I d + k is Fredholm ie. k= dimker(F) 0 such that

XX C(AxY + XY) xX.

= dimkernelA

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Spectral Theorem for Compact Operators

X infinite dimensional Banach space, K: XXcompact.{k} Cs.t k0 (0 = 0) s.t:(Id K)1 bounded if=k for all k .

1dimKer(Id k) 0}u= g on Rn

Heat kernel:E(x, t) = (t)(4t)n/2e|x|/4t

EC(Rn R+). Equation is invariant under transformation:t2tx

t

u(x, t) = (|x|2/t)find radial solution for ODE to find kernel.

Definition 6.

(etg)(x) :=

Rn

E(x y, t)g(y)dny

Heat propagator.

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Theorem 18. gCb(Rn), defineu : Rn R+ R by heat propagator

u(x, t) = (etg)(x)

thenuC(Rn R+) andtu= u inRn R+. Moreover,

etgg locally uniformly as t0 and

u gProof.

tE= E in Rn R+

We use the fact that

E(x, t) = 1, E0E(x, t)

0 as t

0 if x

= 0

let

z =x y

2

t= y = x 2tz

dny= (2

t)ndnz

u(x, t) = (4t)n/2

e|z|2

g(x 2

tz)(2

t)ndnz

n/2g

e|z|2

dz =g

tu= u, E(x, t) = (t)(4t)n/2exp(|x|

2

4t )

(t) =

1 t > 0

0 t < 0

u(x, t) = [etg](x) =

Rn

E(x y, t)g(y)dy

Theorem 19. IfgCb(Rn), thenuC(Rn R+) andtu= u inRn R+etgg as t0,u g

Some notes on the hear propagator operator:

g u : Cb(Rn) Cb(Rn R+) is bounded. One needs|g(x)| C e|x|2 to ensure integrabilityunder the heat proagator defined above. The formula is useless for t < 0. Moreover, the heatpropagator has a smoothing effect; irreversible effect. Disturbances propagate atspeed.

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Duhamels Principal

Ifut = Lu, u|t=0 = g is solved by u(t) = S(t)g, then ut Lu= f, u|t=0 = 0 is solved byu(t) =

t0

S(t s)f(s) ds. (46)

otherwise known as Duhamels formula (DF).

We knowRn

E(x y, t)g(y)dy solves ut = u, u|t=0 = g. Duhamel says

Rn

E(x y, t s)f(y, s) dyds (47)

solves ut

u= f, u

|t=0 = 0, (f(y, s) = 0

s < 0).

Rigorous Treatment

Suppose tu u= f in Rn (0, T).

I= [e(ts)u(s)](x) =

Rn

E(x y, t s)u(y, s)dy ; (0< s < t < T)

assumeuCb(Rn (0, t].I

s =

Rn

E

tu(y, s)dy

e(ts)u(s)

+

Rn

E u

s(y, s)

f+u

dy (48)

E(x y, t s)u(y, s)dy= E u (49)By Greens II vanishing boundary terms by taking very large ball. (50)

Assume the following with 0 < a < b < T (51)

uC2x(Rn (0, T)), (52)2i uCb(Rn [a, b]), (53)tuCb(Rn [a, b]). (54)

(55)

We have

e(tb)u(b) e(ta)u(a) = b

a Rn E(x y, t s)f(y, s)dydsa0, bt. RHS holds provided fCb(Rn [0, t])

e(ta)u(a) =

Rn

E(x y, t a)u(y, a)dy (56)

=

E(x y, t a)u(y, 0)dy+

E(x y, t a)[u(y, a) u(y, 0)]dy (57)

(58)

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[taking balls to attain local uniformly: more detail] Note that[u(y, a) u(y, 0)] is continuous sou(y, a)

u(y, 0) locally uniformly, but since our integration domain is infinite, we take a compact

domain and we have the convergence and outside the ball, we use the decayness of E to ensure theconvergence outside the ball

e(tb)u(b) =

E(x y, t b)u(y, t)dy (59)

=

E(x y, t b)u(y, t)dy+

E(x y, t b)[u(y, b) u(y, t)]dy (60)

Theorem 20. Let u C1(Rn (0, T))C2x(Rn (0, T)). Furthermore, assume u BC(Rn (0, b]),b < T. Also, assumeuBC1(Rn [a, b]) BC2(Rn [a, b]),a, b(0, T). gCb(Rn).

tu u= f Rn (0, T)u= g Rn

{t= 0

}(61)

Then,

u(t) = etg+

t0

e(ts)f(s)ds; (0< t < T) [DF]

In particular solution to (61) is unique in the class considered.

Theorem 21. gCb(Rn), fCb(Rn (0, b]) C1b (Rn [a, b]), (0< a < b < T). Then (DF) solves(61)on(0, T).

Proof.

tu= etg+ f(t) +

t0

e(ts)f(s)ds (62)

=f(t) +

etg+ t

o

e(ts)f(s)ds

(63)

(64)

To justify taking the Laplacian out from under the integral we need the following:

e|x|2

4t e|z|2/4t by|x| |z|t

xE

E|x|

t E

|z|t1/2

,

E f=

t0

E(ts)1/2

f

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Lecture 24

Last time Duhamels Principal

Essentially a formula to solve the inhomogeneous heat equation: Let gCb(Rn)fCb(Rn (0, t]) C1b (Rn [s, t]) 0< s < t < T.

We havec

solves the heat equation tu u= f in Rn (0, T)u= g on Rn {0}

Systems

Letu, f : Rn (0, T)Rm, g : Rn Rm

and consider tui ui= fiui = gi

i= 1,...,m

where explicitly the solution is

ui(t) = etgi+

t0

e(ts)fi(s) ds, i= 1,...,m.

we write for simplicity

u(t) = etg+ t0

e(ts)f(s)ds.

We define theCb(,Rn) byu= max

iui

Nonlinear Reaction-diffusion Equationstu u= f(u)u= g

where f : Rm Rm. (65)

Example

t u= (u v)3tv v= uv

where f([u, v]T) = [(u v)3, uv]T

We have solution

u(t) = etug+

t0

e(ts)f(u(s))dsu = (u) (66)

We attempt to iterate since ui are arguments off so for u1, u2,... where we hope that uk u, withu solving (65).

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Lemma 5. LetgCb(Rn), f C1, uCb(Rn (0, T)) solves (66), thenu also solves (65).

Proof. IfuC1

b (Rn

(s, t))0< s < t < T. Then its obvious because f u satisfies the conditionsof Duhamel.j (e

(ts)f(u(s))) =O((t s)1/2) [previouslectureargumentonLap]= j u satisfies Duhmael. We use X = Cb(Rn (0, T)) for T > 0 to be chosen later [to insure

contraction, we aim to use Banach Fixed point Theorem]. For some large M >0 such that

U=BM X.l= etg.

(u(t)) l(t) t0

e(ts)f(u(s)) ds (67)

t

0

f(u(s)) ds (68)

t uBM, (69)where = sup|v|M|f(v)|. Now, uBM X

(u) lT (u) l+ T g+ T

Choose M 2g. T g= (u)BM for uBM

which asserts the invariance of the ball. Now we need to assert contaction:

(u(t)) (v(t))

t

0

e(ts)[f(u(s)) f(v(s))] ds (70)

t0

f(u(s)) f(v(s)) ds (71)

analogy with the Mean Value Theorem

f(a) f(b) = f()(a b) (a, b)we have

fi(a) fi(b) =fi() (a b)= a + (1 )b, [0, 1], so returning to our problem

(u(t)) (v(t)) t0

u(s) v(s) dstu vwhere

= sup|v|M

maxi,k |ifk(v)|,= (u) (v)Tu v

chooseT < 1 and we have existence on (0, T).

Theorem 22. LetgCb(Rn), f C1. ThenT >0 s.t (65) has a solutionuC1(Rn (0, T)) C2x(R

n (0, T)). It is unique in the class considered in Duhamel. Moreover, unless|u(x, t)| astT as somexRn, u can be continued beyondT.

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A Prioi estimates:

Ifu is a solution on (0, T) thenu(T)

uL2 (t)

u more critical

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Maximum Principal in Rn

Figure needed. E(x, ) = E(ix,) = (4)n/2

e|x|2/4

satisfies

F =f

andtF= F

Definev= u F. Assume:|u(x, t)| M e|x|2

then,

v(x, t)M e|x|2 e

|x|2/4(+)

[4(+ )]n/2 if|x| large.

ifu

0 at{

t= 0}

thentv v= tu 0

then by maximum principal we have

v0 in Rn (0, T) >0.

souF 0 as 0.

Allen-Cahn

tu= u + u

u3

Consider

tu= u + u2

compare withtv= v

2

by taking differencew = u vw0 = uv.

utu= t|u|2

2 =uu + u2 u4 (74)

=12

|u|2 |u|2 + u2 u4 (75)

since(uu) =|u|2 + uu

|u|2 =2(u u) =(2uu) = 2|u|2 + 2uu

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