Lecture 3 Heat equations
invariance, explicit solutions
self-similar way
“Green’s”function for
8 the “whole”space time and “half”space time
maximum principle
>
>
>
>
< higher order derivative estimates
mean value formulas regularity-analytic in x, smooth in t
>
>
Harnack
>
>
:
Liouville
8
< mean value
weak formulation weak formulation
:
distribution
uniqueness with exponential growth, Tikhonov non-uniqueness, nonexistence
with super quadratic exponential growth, nonanalytic in t; backward uniqueness
Invariance of ut 4u = 0
u (x + x0 ; t + t0 )
u eA x; t
u ( x; 2 t)
u + v; au
Dx Dtl u still temperature/caloric
d
Du Ax = d"
u e"A x; t "=0 ; A =
AT
Dx u (x; t) x + 2Dt u (x; t) = dd u ( x;
R
u (x y; t s) ' (y; s) dyds
2
Kelvin t n=2 e jxj =4t u (x=t; 1=t)
2
t)
=1
Examples.
All harmonic functions are caloric
x1 x2 x3 ; jxj2 n ; Re = Im e3x1 +4x2 +i5x3
1
jxj2 ; t2 + t jxj2 =n + (x41 +
+ x4n ) = (12n) ; more later
t + 2n
2
2
Re = Im ei x j j t ; e x+j j t
RMK. e x+j jt ; Re = Im ei( x+j jt) ; u ( x + j j t) are wave functions; ex1 +t are both
caloric and wave functions.
More examples.
Caloric polynomials
(x; t) =
1
t
now
(x; t) Dx Dtl
0
December 1, 2016
1
e
n=2
jxj2
4t
x 1
;
t t
is a caloric polynomial of degree j j + 2l:
eg.
D111
=
=
=
1
tn=2
e
jxj2
4t
1
jxj2 x1 x2 x3
x1 x2 x3
! ( t)n=2 e 4t
3
8
( 2t) ( x ; 1 )
t t
x1 x2 x3
(x;t)
! ( 1)n=2
:
8
D123
=
(
jxj2
4t
e
tn=2
x1
2t
2
x1
2t
+
x1 x1
"
x1
2t
+
x1
x1
2t
+
x1
x1
1
x21
x1
+
+
2t2
2t
2t
4t2
3
3
2
6tx1 x31 ( xt ; t1 )
n=2 jxj 6tx1 + x1 (x;t)
n=2 6tx1 + x1
4t
!
(
t)
e
!
(
1)
:
8t3
8
8
Radial ones
ut
4u =
(0;0)
(x; t) =
0
(x)
Fourier transform way in space and time
Z
1
u (x; t) e
u^ ( ; s) =
n+1
(2 ) 2 Rn+1
satis…es
is^
u + j j2 u^ =
Then
1
(2 )
n+1
2
1
u (x; t) =
(2 )n+1
i( x+st)
(2 )
Rn+1
(t)
1
and u^ =
Z
0
n+1
2
1
i(x
2e
is + j j
dxdt
1
:
is + j j2
+ts)
d ds:
We see the radiality u (x; t) = u (jxj ; t) : Question, how to invert?
Fourier transform way in space only
Z
1
u^ ( ; t) =
u (x; t) e i x dx
n
2
n
(2 ) R
satis…es
u^t + j j2 u^ =
1
(2 )n=2
Then
e
j j2 t
u^
=
t
j j2 t
e
(2 )n=2
0
(t) =
2
8
< 0
:
0
(t) :
1
(2 )n=2 0
0
for t > 0
(0) for t = 0
for t < 0
x1
2t
2
#)
and
(
1
(2 )n=2
for t > 0
:
0
for t < 0
(
0:1 (2
2
RMK. If the jump function at t = 0 is chosen as ej j t u^ =
0:9
2
j t
ej
u^ =
1
)n=2
1
(2 )n=2
for t > 0
;
for t < 0
2
then u^ = 0:9e j j t is not invertible for t < 0:
We proceed with the conversion.
For t < 0; u = 0:
For t > 0
Z
Z
1
1
1
1
j j2 +i px
j j2 t+ix
t d
u=
e
d =
e
n n=2
n=2
n=2
(2 ) t
(2 )
Rn
Rn (2 )
Z
Z
2
ix
jxj2
jxj2 Y
1
1
1
1
p
2
t
4t
4t
e
=
e
e
e k
d =
(2 )n tn=2
(2 )n tn=2
1
Rn
R
{z
|k
ixk 2
p
2 t
d
n=2
=
1
jxj2
4t
e
(4 t)n=2
:
}
RMK. In switching each of the line integrals to the real axis, we take holomorphic
2
function h (z) = e z and domain as an “in…nite”rectangle with bottom y = c and
top y = 0 in the following divergence formula (FTC):
Complex version
Z
Z
h (z) dz =
hz (z) dz ^ dz = 0:
@
Real version
Z
Z
(dy; dx)= dA
(udx vdy) + i (udy + vdx)
=
(dx;dy)=T dA
@
@
Z
(vx + uy ) + i (ux vy ) dxdy = 0:
=
(v; u)
+ i (u; v)
RMK. Splitting way for Gaussian integral
I=
Z
jxj2
e
dx =
Rn
=
Z
1
Z
n=2
e
x21 x22
dx1 dx2
R2
n=2
e
r2
2 rdr
=
e
r2
1 n=2
=
n=2
:
0
0
Radial way
I=
Z
0
1
e
r2
j@B1 j r
n 1
j@B1 j
dr =
2
=r2
3
Z
0
1
e
n
2
1
d =
j@B1n j
(n=2) :
2
dA
Consequently
8
2
>
>
<
2 n=2
2 n=2
n
j@B1 j =
=
( n2 1)!
(n=2) >
(n
>
: n 2 n
( 2 1)( 2
In chart
n
1 2
j@B1n j 2 2
3
4
4
5
4
3
2
2
n=1
n even
1)=2
2)
6
2
n odd
( 32 )
7
8
15
3
:
8
1
2
3
9
4
&
16
105
4
:
Self-similar way
self similar solution u (x; t) =
1
ut =
1
t
1
t
4u =
=
v 00
r
p
+1
t
@rr +
1 00
v
t
+
t
Set =
v 00 ( ) +
v+
r
p
+
r 0
1
v
t 2t3=2
n 1
@r v
r
n 1p
1 0
v
r
t
n 1 0
r
p
t
v
r
p
+
t
=
1
t
r
p
v
1
t
r
p
v0
2 t
v
+1
p
4u = \ (x; t)00 or 0
to ut
t
r= t
=
1
t
1p
r 0
v
2 t
n p1 0
v
r= t
v 00 +
+1
r
p
r
p
+ v
t
=0
t
; then
v ( ) + 12 v 0 ( ) + v ( ) = 0
1 n 0
0
( n 1v0) +
v + n 1v = 0
2
|
{z
}
0
=( 12 n v ) when =n=2
* Let = n=2
n 1 0
t
n 1 0
2
1 n
v
2
= c or v 0 + 2 v = c= n 1 or ve 4
2
2
2 R
Then v ( ) = c20 e 4 + c e 4 e 4 = n 1 d and
Z 2
jxj2
jxj2
1 6 0
4t
4t
u (x; t) = tn=2 4c e
+ ce
e 4 = n 1d
v +
0
2
= ce 4 =
p
=r= t
n 1
3
7
5
|
{z
}
After some testing for the fundamental solution, we …nd c = 0:
Green’s function for the whole space-time and “half”space-time
Whole space-time
(
jx yj2
1
1
4(t s)
1<s<t :
n=2
n=2 e
(4 )
(t s)
(x; t; y; s) =
0
s>t
u (x; t) =
=
f=
1
(4 )n=2
1
(4 )n=2
Z 1Z
0
Rn
Z
s
t
1
1
Z
1
Rn
e
n=2
4
(t
jyj2
4s
e
s)n=2
f (x
y; t
jx yj2
4(t s)
f (y; s) dyds
s) dyds
8
< f 2 C0
f 2 C =2
for
:
f 2 C12
solving
When
C12
with compact support
not enough
just …ne, little involved calculus
more than enough
to get u 2 C12 ;
4u = f (x; t) :
ut
function f having compact support in Rn+1 ; we see the convolution with
the heat kernel (4 t)
n=2
e
jxj2
4t
leads
Dt f =Dx f =Dxx f : C 0 ! C 0 :
Exercise (f (y; s) = 0 (y))
Z 1
Z 1
2
jxj2
jxj2
s= 4t
1
1 1
n=2
s jxj
4t
e
s e
dt =
ds
n=2 jxjn
4s2
(4 t)n=2
0
0
Z 1
Z 1
n
n
1
1
1
1
1
2
s
1
2
2
s
s
e
ds
=
e s ds
=
n
n
2
n
2
n=2
1
4 n=2 jxj
4
jxj
0
0
2
(n=2)
1
1
1
! for n 3:
=
n 2 =
n=2
2
(n 2) jxj
(n 2) j@B1 j jxjn 2
Question. n = 1; 2?
Rt
As for the equation ut 4u = f (x; t) ; when Dt hits the upper limit in 1 ; one
just substitute s with t in the integrand, we reach f (x; t) or
Z
jx yj2
1
lim
e 4(t s) f (y; s) dy = f (x; t) ;
s!t
s)n=2
Rn (t
when Dt
4 hits the integrand, as the kernel is caloric, one gets 0;
(Dt
4x )
(x; t; y; s) = 0:
RMK. We indicate the Schauder estimates C1;2; =1 for mere C a=2 function f with
compact support in the heat potential. The idea is to split o¤ integrable factors in
space or time from the bounded function t e 1=t : The order is D12 ; D11 ; and then
by equation Dt u = 4u:
Z t Z
jx yj2
(x1 y1 ) (x2 y2 )
1
D12 u (x; t) = cn
e 4(t s) f (y; s) dyds
4 (t s)2
(t s)n=2
1 Rn
"
#
Z t Z
jx yj2
(x1 y1 ) (x2 y2 )
1
e 4(t s) f (y; s)
f (x; t)
dyds
= cn
2
n=2
n
4 (t s)
(t s)
by cancellation
1 R
Z t Z
h
i
jx yj2
jx yj2
1
=2
4(t
s)
jx yj + (t s)
dyds
e
s)2 (t s)n=2
1 B1 4 (t
Z 1Z
jyj2
jyj2 1
4s
=
e
jyj + s =2 dyds change variables to move singularity to (0; 0)
2 n=2
0
B1 4s s
! n+1
! n2 +1 4
+4
Z 1Z
2
jyj2
jyj2
1
1
jyj2
1
1
jyj2
4s +
4s dyds
e
e
n
=2 s1
=4
s
s
jyjn =2 s1 =4
0
B1 jyj
|
{z
}|
{z
}|
{z
} |
{z
}
integrable
integrable
bounded
5
bounded
D11 u = cn
= cn
Z
=
Z
0
Z
Z
t
1
1
t
1
t
1
Rn
Z
Rn
B1
"
"
"
(x1 y1 )2
4 (t s)2
#
1
2 (t
(x1 y1 )2
4 (t s)2
1
2 (t
s) (t
#
s) (t
#
jx yj2
1
+
s)2 2 (t s) (t
B1 4 (t
0
1
Z
Z
Z
B
B
@
jyj2
4s2
|{z}
+
similar to D12
1C
C 1 e
2s A sn=2
1
1
jx yj2
4(t s)
1
jx yj2
4(t s)
e
s)n=2
s)n=2
e
s)n=2
jyj2
4s
e
jx yj2
4(t s)
jyj + s
h
=2
jx
f (y; s) dyds
"
f (y; s)
f (x; t)
by cancellation
yj + (t
s)
=2
i
#
dyds
dyds:
We …nish the estimate involving 1s term, whose singularity is still of order n 2 +
in space-time, then integrable.
Z 1Z
jyj2
1 1
4s
e
jyj + s =2 dyds
n=2
0
B1 s s
! n2 + 4
! n2 4
Z 1Z
jyj2
jyj2
1
1
jyj2
1
jyj2
1
4s +
4s dyds
e
e
=
1
=4
s
s1 =4 jyjn =2
s
jyjn =2
0
B1 s
|
{z
}|
{z
}|
{z
} |
{z
}
integrable
integrable
bounded
bounded
Space–half time
Physical observation: temperature/density evolves forward as time goes forward.
Mathematical observation: initial “velocity”ut already determined by initial “position”4u: So only initial temperature can be prescribed:
(IVP)
ut 4u = 0 for t > 0
:
u (x; 0) = ' (x)
It is similar to, but much better (behaved) than harmonic functions over half space
4u = 0 for xn > 0
u (x0 ; 0) = ' (x0 ) 2 L1 (Rn )
say C k (Rn ) \ L1 (Rn )
:
Bonus
n=2
Z
( ; t) ' = (4 t)
exp
jx yj2 =4t ' (y) dy
Rn
Z
p
= ( ) n=2
exp
jyj2 ' x 2 ty dy
u (x; t) =
Rn
solves the (IVP). Namely
t!0+
i) u (x; t) ! ' (x) ;
ii) u (x; t) analytic in x&t for t > 0; as long as ' (x) 2 L1 (Rn ) ;
6
dyds
8
< ' 2 C 0 ) u 2 C00 (Rn [0; 1)
1
(Rn [0; 1) :
' 2 C 1 ) u 2 C1=2
iii)
:
' 2 C 2 ) u 2 C12 (Rn [0; 1)
Proof.
i) "
way, Lebesgue dominate convergence way.
R
ii) Rn (4 t) n=2 exp
jx yj2 =4t ' (y) dy is a sum of x-t analytic function for
t > 0:
iii) We only need to show the continuity near t = 0: We skip the C 0 & C 1 case.
Now ' 2 C 2 and Dx2 u takes the form
Z
p
t!0+
n=2
2
Dx u (x; t) = ( )
exp
jyj2 Dx2 ' x 2 ty dy ! Dx2 ' (x) :
Rn
The limit can be done via "- way assuming kD2 'kL1 (Rn ) < 1; or Lebesgue dominated convergence way assuming kD2 'kL1 (Rn ) < 1:
As for the continuity of ut ; one way is to use the continuity of Dx2 u coupled with
t!0+
the equation: ut (x; t) = 4u (x; t) ! 4' (x) : Another way is direct and fun. We
also include here.
2
3
Z
odd integral0
p
!
y
ut (x; t) = ( ) n=2
exp
jyj2 p 4Dx ' x 2 ty
Dx ' (x) + Dx ' (x) 5 dy
t
Rn
Z
i
p
y h
n=2
Dx ' (x) dy
=( )
exp
jyj2 p
Dx ' x 2 ty
t
Rn
p
2 tykD2 'kL1
Z
exp
jyj2 2 jyj2 dy D2 ' L1 :
( ) n=2
Rn
So either "- way or Lebesgue dominated convergence theorem implies
Z
t!0+
n=2
ut (x; t) ! ( )
exp
jyj2 2 y; Dx2 ' (x) y dy
n
R
Z
n=2
+ 'nn (x) yn2 dy
= 2( )
exp
jyj2 '11 (x) y12 +
Rn
"
#
Z
jyj2
jyj2
n=2
2
= 2( )
exp
jyj
'11 (x)
+
+ 'nn (x)
dy
n
n
Rn
Z
2 4 ' (x)
=
exp
jyj2 jyj2 dy
n=2
n
n
Z 1R
Z 1
4' (x)
r2 = 4' (x)
2
2 n 1
=
2 exp r r r
j@B1 j dr =
j@B1 j
exp (
n n=2 0
n n=2
0
4' (x) 2 n=2
n
4' (x) 2 n=2
n n
=
+
1
=
= 4' (x)!
n=2
n=2
n
(n=2)
2
n
(n=2)
2 2
7
)
n=2
d
4u = 0 for xn > 0
u (x0 ; 0) = ' (x0 )
RMK. In contrast, recall for harmonic function in the upper half space
' 2 C 1 Rn
1
Rn
' 2 C 1;
1
n
; uxn 2 C 1 R+
; uxn 2 C
n
:
R+
Nonhomogeneous problem
ut 4u = f (x; t) in Rn
u (x; 0) = ' (x) ; say C 0
(0; 1) ; say C12 compact support
:
has A solution
u (x; t) =
1
(4 )n=2
Z tZ
0
where we assumed f
i) u 2 C12 for t > 0;
t!0+
e
jx yj2
4(t s)
s)n=2
f (y; s) dyds +
1
Z
e
jx yj2
4t
(4 t)n=2 Rn tn=2
Rt
0 with t < 0 for the integral 1 ; satisfying
(t
Rn
' (y) dy;
ii) u (x; t) ! ' (x) :
We only prove ii) with
1
(4 )n=2
1
n=2
Z tZ
Rn
0
e
Rn
Z
1
t kf kL1
n=2
s)n=2
(t
Z tZ
0
jx yj2
4(t s)
e
jyj2
f x
e
jyj2
f (y; s) dyds
p
2 t
sy; s dyds
dy
Rn
t!0+
= t kf kL1 ! 0:
Mean value equality
Recall the derivation of the “solid" mean value formula for harmonic functions
Z
Z
u4v v4u=
uv
vu
BR
Set v =
1
(n
|
2) j@B1 j
{z
}
@BR
1
jxjn
2
=
1
1
cn jxjn
2
cn
u (0) =
Z
@BR
where
1
Rn
2
uv dA =
Z
:
8
=
1
=l
cn Rn 2
u jD j dA
As R goes from 1 to 0; the level of runs from 1=cn to
R 1=c
weight w (l) = c ( l) satisfying 1 = 1 n w (l) dl; and then
u (0) =
=
Z
1=cn
w (l)
1
Z
B1
Z
1; we seek a power
u jD j dAdl
=l
u w (l) jD j2 dx;
where we used change of variable or co-area formula: dx = dvol = dA
dl
jD j
…gure gradient length=dl= jD j
Now we choose w so that the weight w (l) jD j2 = 1= jB1 j ; then
w (l) =
At R = 1 and l =
cn
1
Rn
jB1 j (n 2)
2
1
:
1=cn ; we have
c (1=cn ) =
1
cn
jB1 j (n 2)
2
:
The integral for the weight implies
1
c
1+
Thus
(1 + ) =
cn
jB1 j(n 2)2
1
cn
n
n
= 1:
2) ; then
= n= (n
w (l) =
1+
2
(cn )
R1
RMK. Old weight way, 1 = 0 nrn 1 dr
And pleasantly w (l) jD j2 = 1= jB1 j!
n=(n 2)
( l)
r=( cn l)
2(n 1)
n 2
1=(n 2)
=
n
n
:
cn
2 n
n
2
R
1=cn
1
( l)
2(n 1)
n 2
Mean value equality for caloric functions (sphere version)
Z
1
jxj2 (0; 0; x; t)
q
u (0; 0) =
u
(x;
t)
dA:
2
2 2
(4 R2 )n=2 =(4 R2 ) n=2
2
4t jxj + 2nt + jxj
…gure: heat sphere
(0; 0; x; t) = [4 ( t)]
n=2
jxj2
exp
= (4 )
4t
One “solid”version
u (0; 0) =
1
(4 R2 )n=2
Z
(4 R2 )
9
jxj2
u (x; t) 2 dA;
n=2
4t
n=2
dl:
where
(x0 ; t0 ; x; t) =
1
jx0 xj2
4(t0 t)
e
t)]n=2
[4 (t0
…gure: backward heat kernel graphs
Derivation of the hollow version.
Z
UT
)
Z
Du Dv + u 4 v =
Z
div (uDv) =
UT
(div; Dt ) (uDv; 0) =
UT
Z
Z
(uDv; 0) ( x ; t ) dA
@UT
Z
Z
Dv Du + v 4 u =
div (vDu) =
(div; Dt ) (uvDu; 0) =
(vDu; 0) ( x ; t ) dA
UT
UT
@UT
Z
Z
Z
Z
uDt v + vDt u =
Dt (uv) =
(div; Dt ) (0; uv) =
(0; uv) ( x ; t ) dA
+)
UT
UT
)
Z
UT
Z
UT
u Dt v + 4v + v Dt u 4u
| {z }
| {z }
00
jxj
1
4t
n=2 e
[4 ( t)]
E=
(
@UT
=
1
;
[4 ]n=2
1
n=2
[4 ( t)]
UT = E \ ft
Z
uv
x
@UT
0
\ (0;0)
2
Take v =
UT
then
e
1
jxj2
4t
[4 ]
n=2
)
, jxj2
v u
|{z}
x
+ uv
t
dA:
0
2nt ln ( t)
sg :
…gure UT
We have from the Green’s identity
Z
0=
uv x +
@E\ft sg
v
x
=D
x
=D
2
=r
2
x 2
2t
+
2
Z
uvdA
| {z }
! u (0; 0)
as s ! 0 ; say for u 2 C 0
Is
D
jD j2
=
j(D ; Dt )j
j(D ; Dt )j
x 2
2t
n 1
2 t
jxj2
4t2
2
10
=
jxj2
q
:
2
4t2 jxj2 + 2nt + jxj2
Therefore
u (0; 0) =
u (0; 0) =
=
1
(4 )n=2
Z
=(4 )
1
(4 R2 )n=2
Z
=(4 R2 )
Z
jxj2 (0; 0; x; t)
u (x; t) q
dA
n=2
2
2 2
2
4t jxj + 2nt + jxj
=(4 R2 )
jxj2 (0; 0; x; t)
u (x; t) q
dA
n=2
2
2 2
2
4t jxj + 2nt + jxj
jD j2
u (x; t)
dA
n=2
j(D ; Dt )j
=l
Having this sphere version, let us get a “solid”mean value formula, by integrating
level by level of the heat
R 1 kernel.
Set w (l) s.t. 1 = 1 n=2 w (l) dl;
(4 )
Z 1
Z
jD j2
u (0; 0) =
u
w (l)
dA dl
n=2
=l j(D ; Dt )j
( 41 )
Z
dl
2
uw
(l)
jD
j
dxdt
recall
dA
= dvol
=
n=2
jr j
( 41 )
Z
x 2
2
u
w
(l)
=
dxdt
n=2
| {z } 2t
( 41 )
constant
Let w (l) =
1
4
n=2 1
;
l2
R1
1 n=2 1 1
n=2 = 1: Thus
n=2 w (l) dl =
4
l ( 41 )
( 41 )
Z
1
jxj2
u (0; 0) =
u
dxdt
n=2
4t2
(4 )n=2
( 41 )
then
or
u (x0 ; t0 ) =
1
(4 R2 )n=2
Z
(x0 ;t0 ;x;t)
jxj2
u
dxdt
n=2
4t2
( 4 1R2 )
…gure heat ball
rescale the unit heat ball by R2 in time and R in space
In particular, for u
1=
1
(4 R2 )n=2
1
Z
[4 ( t)]
n=2
e
jxj2
4t
(4
1
n=2
R2 )
jxj2
dxdt =
4t2
Z
jxj2
dxdt:
2
1 4t
| {z }
say 4 R2 =1
RMK. Other choices of weights. For example w (l) = c=l2
Z
jxj2
u (0; 0) = c
u
dxdt;
4t2
1
11
;0<
<1
the kernel is still singular, though the kernel/weight has integrable singularity of order
n
2 in space-time.
Applications of mean value formulas
App1. Strong max principle:
Let u 2 C12 solution to ut 4u = 0 in UT : THEN
max u only attains at the parabolic boundary of UT ;
otherwise, if max u = u (x0 ; t0 ) ; where (x0 ; t0 ) is an interior or non-parabolic
boundary of UT ; then we have u (x; t)
u (x0 ; t0 ) for all (x; t) in the closure of the
connected set of UT \ ft t0 g by chain of downward heat balls.
Def: Parabolic boundary points cannot center any heat ball inside the domain UT :
Examples of UT
…gure parabolic bdry
Proof of the strong max principle.
Suppose u (x0 ; t0 ) = maxUT u and (x0 ; t0 ) is not a parabolic boundary point, that
is, (x0 ; t0 ) centers a heat ball in UT : By the mean value formula in this ball
Z
1
u (x0 ; t0 ) =
(4 R2 )n=2
kernel 0
1
R
Thus u (x; t)
(x0 ;t0 ;x;t) (4 R2 )
(4 R2 )n=2
kernel=1
=
Z
jxj2
u (x; t)
dxdt
n=2
4t2
(x0 ;t0 ;x;t) (4 R2 )
n=2
u (x0 ; t0 )
jxj2
dxdt
4t2
u (x0 ; t0 ) :
u (x0 ; t0 ) in
(
(x; t)j
1
(t0
t)n=2
e
jx0 xj2
4(t0 t)
1
Rn
)
:
RMK. The closure includes the points at horizontal level ft = t0 g ; this is because
such a point (y; t0 ) is the limit of (y; s) as s " t0 ; and the segment (x0 ; t0 ) -(y; s) can
be covered a chain of heat balls.
…gure downward segment
Then u (y; t) = lim u (y; s) = lim u (x0 ; t0 ) :
Uniqueness of caloric function on bounded domains
12
Let u; v be two C12 (UT ) \ C00 UT solutions to wt 4w = 0 in UT ; and u = v on
the parabolic boundary of UT : THEN u v.
RMK. UT including U1 domains like ft > convex (x)g ; say
jxj4 :
…gure t
1
[0; T ]?
Question. What happens to Rn [0; T ] or R+
App2. Regularity
Faking space dimension Rn ! Rn+m will lead us to a C12 (even better ones for
larger m) kernel in the mean value formula:
u (x0 ; t0 ) =
Z
=
ut (x; t)
(4x + 4y ) u (x; t) = 0
(x0 ;t0 ;x;t) (4 R2 )
Z
n=2
u (x; t) K (x0
u (x; t) K (x0
x; t0
x; t0
t) dxdt
t) dxdt
Rn R1
where K is the kernel of Kuptsov (c.f. Neil A. Waston 2002). Then starting from
L1 function, satisfying the parabolic mean value formula, we immediately have C12
solution to the heat equation (no need existence). We can also get interior estimates.
RMK. One way to verify those C12 functions (out of L1 ; enjoying solid, then hollow
mean value formulas) satisfy the heat equation, comes from the derivation of the heat
sphere mean value derivation
Z
Z
uv x dA + u (x0 ; t0 ) = 0:
v (Dt u 4u) dxdt =
c
=c
Next, by using a di¤erent argument via Green’s identity over a cylinder UT = BR
[0; T ] ; we show the C12 solutions are C 1 in x; t and C ! in x: Recall the fundamental
solution is not C ! in t:
Green’s identity
Z
Z
u Dt v + 4v + v Dt u 4u =
uv x vu x + uv t dA:
| {z }
| {z }
UT
@UT
\ (0;0)00
v=
jx yj2
s)
(x; t; y; s) = e 4(t
0
= [4 (t
s)]n=2
13
u (x; t) =
Z
u (y; s)
@U [0;t]
+
Z
+
y
(x; t; y; s)
{z
}
dA
C ! in x not in t; C 1 in t
u y (y; s)
@U [0;t]
Z
|
|
(x; t; y; s)
{z
}
dA
C ! in x not in t; C 1 in t
u (y; 0)
|
U
(x; t; y; 0) dy:
{z
}
C ! in x;t for t
So we conclude u is C ! (x) and C 1 (t) in UT0
0
UT :
Interior estimates
max Dxk Dtl u
C1
C (k; l; K) max Dx2 u + jDt uj
C2
or
max Dxk Dtl u
CR
C (k; l; K) max juj
C3
C (k; l; K)
max juj
C3R
Rk+2l
via scaling v (x; t) = u (Rx; R2 t) ; Dxk Dtl v (x; t) = Rk+2l Dxk Dtl uj(Rx;R2 t) : Here CR =
BR (x0 ) (t0 R2 ; t0 ) :
Liouville Theorem: Global (eternal) solution, say C12 to ut
( 1; +1) satisfying
ju (x; t)j
A jxjk + jtjl
4u = 0 in Rn
for large jxj + jtj
must be a caloric polynomial of degree less than k + 2l:
Proof.
C (k 0 ; l0 ; K)
R!1
Dxk Dtl u (0; 0)
A Rk + R2l ! 0
Rk0 +2l0
for k 0 + 2l0 > k + 2l: Note (0; 0) could be anywhere, so u (x; t) is a caloric polynomial
of degree k + 2l:
App3. Harnack inequality
u 0 solution to ut 4u = 0; then
max u
Cr
C (n) min
u:
+
Cr
…gure Harnack:
14
One proof is via “fake” dimension mean value formula, it is little “involved” in calculating the positive weight. However, everything is at calculus level.
Uniqueness revisited
If a caloric function is analytic in terms of x and t; like
Z
jx yj2
1
4t
u (x; t) =
' (y) dy for t > 0;
e
(4 t)n=2 Rn
then u (x; 0:9) determines all the temperature for t > 0 : t > 0:9 or 0 < t < 0:9; also
like
u (x; t) = ex+t or e t sin x;
u (x; 0) determines all temperature for 1 < t < 1:
In general, caloric functions are not analytic in t; like
(
x2
1
4t
for x 6= 0 and t > 0
n=2 e
(4 t)
;
u (x; t) =
0
for x 6= 0 and t 0
then u (x; 0) cannot determine u (x; t) for t > 0:
Recall we have from the maximum principle, that two caloric functions agree on
the parabolic boundary of a bounded space-time domain, they agree everywhere.
…gure cylinder domain and
t > jxj2
intersecting t
T
Energy proof of the uniqueness: w 2 C12 UT solution to wt 4w = 0; vanishing
on the parabolic boundary, then w 0:
C1 way. We calculate/estimate the energy
Z Z
Z TZ
Z T Z
Z
2
!
0
0
jDwj dxdt =
div (wDw) w 4 wdxdt =
w w dA
w 4 wdx dt
=
Z
UT
T Z
0
0
wwt dxdt =
t
Z
t
T
Z
T
t(x)
1 2
w t dtdx =
2
Z
0
T
@
t
1 2
w (x; T ) dx
2
t
0:
So jDwj = 0 and then w 0:
To prepare for a proof of backward uniqueness, we present another
R
C0 way. We calculate/estimate the L2 norm of w: Set E (t) = t w (x; t)2 dx; say
t = :
Z
Z
d
_
E (t) = E (t) =
2wwt dx =
2w 4 wdx
dt
Z
Z
Z
2
0
=2
w
~ w dA 2 jDwj dx = 2 jDwj2 dx 0:
Then
R
@
w2 (x; t) dx = E (t)
E (0) = 0: So w (x; t)
0:
8
< wt 4w = 0 in
2
u=v
on @
Backward uniqueness: Let u and v be two C1 solutions to
:
u=v
on
then u v on
[0; T ] :
15
[0; T ]
[0; T ] ;
T
Proof. By linearity, we only need to show from wt 4w = 0 in
w = 0 on
T and @
[0; T ] ; we have w 0 or w (x; 0) = 0:Set
Z
w (x; t)2 dx:
E (t) =
[0; T ] and
The proof is based on the observation that ln E (t) is convex in terms of t:
Step1. As in the C 0 way in the above, we have
Z
_
E (t) = 2 jDwj2 dx 0:
R
R 2
1=2 R
w=0 on @ R
We will also need jDwj2 dx
=
w4wdx
w dx
(4w)2 dx
Take one more derivative
Z
Z
Z
!
0
E• (t) = 4 Dw Dwt dx = 4
w wt dA + 4 4w wt dx
@
Z
= 4 (4w)2 dx:
Step2. Suppose E (t) > 0 for 0
t < T0
1=2
T and E (T 0 ) = 0: Then
E_
d
ln E =
dt
E
and
•
d2
EE
E_ 2
ln
E
=
0;
dt2
E2
R
R
R
2
•
jDwj2 dx
4 w2 dx (4w)2 dx = EE:
since E_ 2 = 4
Now the convex function ln E (t) cannot go to 1 as t goes to T 0 ; for it should
stay above the tangent line at (0; ln E (0)) : This contradiction shows E (t) 0:
Uniqueness for Cauchy problem with constraints in Rn R+ :
Max Principle: Let u be C12 (Rn (0; T )) \ C (Rn [0; T ]) solution to
ut 4u = 0
:
u (x; 0) = g (x)
Suppose ju (x; t)j
2
Aeajxj in Rn
ju (x; t)j
[0; T ] : Then
sup g (x) in Rn
[0; T ] :
Rn
Proof. We only need to prove u (x; t) supRn g (x) , M for subcaloric solution
2
ut 4u 0 with sub quadratic-exponential growth u (x; t) Aeajxj :
Step1.
…gure t-direction thin domain:
16
:
For any
> 0; set
v=M+
4v = 0
vt
jxj2
1
4 8a
t
1
1
8a
n=2
t
4u
ut
v
u on @BR
v
u on Rn
0;
e (
)
1
16a
for R large
f0g :
RMK. Invariance
way to construct this barrier:
p
jxj2
jxj2
1
* p1t e 4t 99K p1 t e 4t still sol.
* the time shift sol v is quadratic-exponential growth at t = 0; it goes to 1 as
1
for every x:
t % 8a
It follows from the maximum principle on the bounded domain,
u (x; t)
M+
1
8a
t
1
;
16a
So for any …xed (x0 ; t0 ) with t0
u (x0 ; t0 )
jxj2
1
4 8a
t
1
n=2
e (
) in B
R
1
:
16a
we have
jx0 j2
1
4 8a
t0
1
M+
0;
1
8a
n=2
t0
e (
!0
) !
M:
2
1
Step2. The above argument works equally well on 16a
; 16a
;
[0; T ] :
Corollary. The Cauchy problem with growth constraint
ut 4u = f
u (x; 0) = g (x)
in Rn
2
; 3
16a 16a
;
; still
(0; T )
2
with ju (x; t)j Aeajxj in Rn [0; T ] ; has at most one solution.
Proof. The di¤erence of any two solutions satis…es the condition in the max
2
principle with g = 0 and di¤erence is less 2Aeajxj ; so the di¤erence is 0:
ut 4u = 0
2
u (x; 0) = eajxj
eg. The caloric function, or a solution to
Integral way to construct the barrier:
u (x; t) =
=
=
1
n=2
(4 t)
Z
1
n=2
Z
jx yj2
4t
2
eajyj dy
Rn
e
Rn
1
(1
e
p
2
jyj2 +aj2 ty xj
e1
4at)n=2
17
4at
jxj2
dy
is
1
[0; 4a
) with constraint ju (x; t)j
* a > 0 unique in Rn
2
1
e1
(1 4at)n=2
4at
jxj2
and
grows faster than eajxj for t > 0;
2
* a < 0 uniqueness in Rn [0; 1) with constraint ju (x; t)j
e100jxj and grows
2
faster than eajxj for t > 0:
The message: the growth/decay rate is not preserved precisely.
ut 4u = 0 in Rn
u (x; 0) = 0
Nonuniqueness of Cauchy problem
[0; 1)
Tikhonov’s counterexample.
Idea of construction:
* along t = 0; position u (x; 0) alone determines all the derivatives (if analytic).
* along x = 0; position u (0; t) and velocity ux (0; t) determines all the derivatives
in x; (if analytic in x):
Now we solve a “real”Cauchy problem along the t-axis
u (0; t) = g (t)
ux (0; t) = 0
ux (0; t) = 0; uxxx (0; t) = uxt (0; t) = 0;
; Dx2k+1 u (0; t) = Dtk ux (0; t) = 0
uxx (0; t) = ut (0; t) = g 0 (t) ; uxxxx (0; t) = utt (0; t) = g 00 (t) ;
; Dx2k u (0; t) =
k
k
Dt u (0; t) = Dt g (t) :
Assuming u is C ! in terms of x; then
u (x; t) = g (t) +
1
X
g (k) (t)
k=1
(2k)!
x2k :
Technical realization:
graph for g (t) =
e
0
1
t
t>0
t 0
How to control the derivatives
1st Direct try.
g (t) = e t
1
g 0 = e t ht
i
00
t
1 2
2
g =e
( t
)
( + 1) t
h
1 3
g 000 = e t
( t
) +
+ ( + 1) ( + 2) t
h
(k)
t
1 k
g =e
( t
) +
( + 1) ( + 2)
e
t
k! ( t
1 k
)
18
need
3
i
> 1:
( +k
1) t
k
i
g (k)
(2k)!
1:
e
t
ju (x; t)j
k
k!
(2k)!
k!
1
( t
) ; and (2k)!
= 2k 1 3 5 1 (2k 1)
P
e t
1 k
) x2k = e t e
g (t) + 1
k=1 k! 22k ( t
1
22k k!
1
t
4
x2
=e
1
t
+
x2
1
4 t +1
t!0+
!
2nd complex try
Observe e t is analytic when t > 0
…gure complex plan z = t + is
g (t) =
g (k) (t) =
R
g(z)
dz
Rz t g(z)
k!
dz
2 i
(z t)k+1
1
2 i
Now by continuity of z at 1; Re z
1
then Re (tz)
t for jtz tj
t
2
1
or Re z
t
for
jz
tj
t
2
and
e z
= eRe z
e
So
g
(k)
k!
2
(t)
Z
(z
1
t
2
for jz
; where
1j
for jz
tj
t)k+1
dz
k! e 2 t
2
2 ( t)k+1
=
(1=2) < 1;
t:
1
g (z)
jz tj= t
1
2
t=
k! e
1
t
2
( t)k
and
ju (x; t)j
g (t) +
1
X
k=1
e
e
t
1
t
2
+e
2
+ xt
1
1 k! e 2 t
(2k)! ( t)k
1
t
2
1
X
1
k!
k=1
x2k
x2
t
k
< 1 for all (x; t) with t > 0;
! 0 as t ! 0 + for each …xed x;
provided
> 1:
Then the Tikhonov’s series converges, we’ve constructed a “super”quadratic-exponential
caloric function such that ut uxx = 0 and u (x; 0) = 0; u (x; t) is not identically 0
for t > 0:
RMK. Choosing
1
1
t
(1 t)
e
t>0 ;
g (t) =
0 t 0 or t 1
…gure complex plan z = t + is for this g
19
we have another Tikhonov solution/caloric function vanishes before 0 or after 1:
…gure Tikhonov double sided vanishing
RMK. Let v = u2 ; where u is the above Tikhonov caloric function, then vt 4v =
2uut 2u 4 u 2 jDuj2 = 2 jDuj2 0: This v is non-negative sub-caloric function
vanishing at t = 0 and t = 1; yet doesn’t vanish identically between (0; 1) :
Nonanalytic, yet smooth solution in Rn [0; 1)
eg.
ut uxx = 0
4
u (x; 0) = e x
has the bounded solution/quadratic-exponential growth solution
Z
jx yj2
4
1
4t
u (x; t) =
e jyj dy
e
1=2
(4 t)
R1
Z
p 4
1
2
e y e jx 2 tyj dy
= 1=2
R1
which is C 1 (R1 [0; 1)) ; but NOT C ! at t = 0:
In fact, if u (0; t) is analytic in terms of t near t = 0; then
u (0; t) =
1
X
ak tk ;
k=0
where
1
m
1 k
1 2k
1 2k X ( x4 )
ak = Dt u (0; 0) = Dx u (0; 0) = Dx
k!
k!
k!
m!
m=0
2k=4m
=
So
1 (4m)!
> m!:
(2m)! m!
m!1
a2m t2m > m!t2m ! 1 for any …xed t > 0:
Then the series diverges, u (x; t) cannot be analytic in t at (0; 0) :
Nonexistence of nonnegative solution to Cauchy problem
ut uxx = 0
4
u (x; 0) = ex
20
x=0
First note the representation
1
1=2
(4 t)
Z
jx yj2
4t
e
R1
4
ey dy = 1:
Overheated, the nonnegative solution blows up once time starts.
Now the proof.
2
…gure for ex and gk
4
gk (x) 2 C01 (Bk+1 ) and gk (x) = ex on Bk : The bounded C12 solution to
ut uxx = 0
u (x; 0) = gk
is
uk (x; t) =
1
1=2
…gure for e
Z
e
y2
gk x
p
2 ty dy:
R1
y2
p
2 ty
and gk R
For each …xed k and say, 0.9, there exists Rk = R (k; 0:9) large so that
0
0:9 for 0
uk ( R; t)
This is because
2
1:
k+1
p :
t
Then as u is nonnegative, uk
0:9 + u on the parabolic boundary of the cylinder
BRk [0; 1] : The maximum principle implies uk 0:9+u in BRk [0; 1] : In particular
uk ( Rk ; t)
uk (0; 1)
1
t
1=2
e
R p
k 1
2 t
ek
4
0:9 + u (0; 1) for all k:
But uk (0; 1) goes to +1; as k goes to +1: A contradiction!
In fact, u (0; l) is forced to be 1 for all small l > 0:
Question: Existence of sign-changing solutions? Answer: YES, F. B. Jr. Jones
1977.
21