Heat/Di¤usion Equation
@w
@t
@2w
=0
k constant
@x2
w(x; 0) = '(x) initial condition
w(0; t) = w(l; t) = 0
boundary conditions
k
Energy estimate:
0 = w(wt
So
2
kwxx ) = ( w2 )t
(kwwx )x + k(wx )2
0=
Zl
=
Zl
w2
( )t dx
2
d
dt
Zl
(
w2
)t
2
(kwwx )x + k(wx )2 dx
0
kwwx jl0
+k
0
or
and therefore E(t) =
0
(wx )2 dx
0
w2
dx =
2
k
0
Rl
Zl
w2
2 dx
Zl
(wx )2 dx
0
0
is a non-increasing function of time or equivalently
0
E(t)
E0
For the initial value problem if '(x) = 0 then E(0) = 0 and E(t) = 0 i.e.
uniqueness.
Stability
Consider
ut = kuxx
vt = kvxx
u(x; 0) = '(x)
v(x; 0) = (x)
Let w(x; t) = u(x; t) v(x; t)
w(x; 0) = u(x; 0) v(x; 0) = '(x)
1
(x)
From the energy estimate we get
Zl
[u(x; t)
Zl
2
v(x; t)] dx
0
['(x)
2
(x; t)] dx
0
So small changes in ' causes small changes in v in the L2 norm.
The maximum principle will give us stability in the maximum norm.
Maximum Principle
Let
Lu =
X
@u X
@2u
@u
+
aij (x; t)
+
bi (x; t)
+ c(x; t)u = f
@t
@xi @xj
@xi
i;j
i
Strong version: Assume Lu 0 , M = max(u). Assume u = M at an interior
point (x0 ; t0 ) and one of the following is true
1. c = 0 and M arbitrary
2. c
0 and M
0
3. M = 0 and c arbitrary
Then u = M everywhere in the domain (i.e. u is constant).
Weak Version: If Lu 0 and c(x; t) = 0 then the maxiumum occurs initially or
on the boundary.
So if juj < M at t = 0; x = 0; x = l then u < M everywhere
2
@u
For a minimum principle consider @t
k @@xu2 = 0
Proof (weak version in one dimension)
@u
@2u
=k 2
@t
@x
2
Intuitively at a maximum the …rst derivatives are zero and @@xu2 < 0 so we have
2
a contradiction. The problem is that we may have a saddle point with @@xu2 = 0.
So instead we supply a perturbation argument.
De…ne:
v(x; t) = u(x; t) + x2
Wish: v(x; t) < M + l
2
>0
with M = max ju(x; t)j so u(x; t) < M + (l2
2
x2 )
Formally:
v(x; t) < M + l2
on t = 0 and x = 0 and x = l
Also
( ) vt
k(u + x2 )xx = ut
kvxx = ut
kuxx
2 k<0
Assume there is a maximum at (x0 ; t0 ) in the interior.
At such a maximum vt = 0 and vxx 0 .
Hence, vt kvxx 0 at (x0 ; t0 ).
This contradicts (*). Hence, v(x; t) < M + l2 .
Note that low order terms destroy this property. Consider
ut = uxx + au a > 0
u(x; 0) = sin x u(0; t) = u(1; t) = 0
has a solution u(x; t) = e(a
2
)t
sin( x) which grows in time if a >
Lemma:
I=
Z1
e
x2
dx =
2
.
p
1
proof:
One proof is by a complex integral and Cauchy’s Theorem.
Another proof is
2
I =
Z1 Z1
1
=
1
Z2 Z1
e
r2
rdrd = 2
1
0
=
2
2
e (x +y ) dxdy
2
Z1
r2 = x2 + y 2
Z1
e
r2
rdr
1
1 d
e
2 dr
r2
dr =
e
1
3
r2 1
j0
=
x = r cos
y sin
Properties of heat equation
Let
@u
@t
2
= k @@xu2
1. u(x
y; t) is a solution (shift)
2. Any derivative of u is a solution
3. Any linear combinations of solutions is a solution
R
4. The integral S(x y; t)g(y)dy is a solution
p
5. u( ax; at) a > 0 is a solution
p
p
= u(x; t).
6. u=u(p)
p = p14k xt . Then u( ax; at) = p14k pax
at
4
( )
@2u
@u
=k 2
@t
@x
u(x; 0) = (x)
proof of #4: If S(x,t) is a solution of (*) then by linearity so is
v(x; t) =
Z1
S(x
y; t)g(y)dy
1
Then
vt =
Z1
St (x
vxx =
Z1
Sxx (x
kvxx =
Z1
[St (x
y; t)g(y)dy
1
y; t)g(y)dy
1
vt
y; t)
kSxx (x
y; t)] g(y)dy = 0
1
proof of (5)
vt
p
v = u( ax; at)
p
vt = aut ( ax; at)
p
vxx = auxx ( ax; at)
kvxx = a(ut kuxx ) = 0
5
De…ne
Qt = kQxx
(
Q(x; 0) = 1 x > 0
Q(x; 0) = 0 x < 0
Assume
then:
Q(x; t) = g(p)
p=
px
4kt
dg @p
1 x 0
p
=
g (p) =
dp @t
2t 4kt
1
dg @p
=p
g 0 (p)
Qx =
dp @x
4kt
1 00
g (p)
Qxx =
4kt
Qt =
So
0 = Qt
kQxx =
1
t
1 0
pg (p)
2
p 0
g (p)
2t
1 00
g (p)
4
Therefore
g 00 (p) + 2pg 0 (p) = 0
g 0 (p) = C1 e
p2
p= p x
Z 4kt
Q(x; t) = g(p) = C1
e
p2
dp + C2
0
We now take the limit as t # 0 . Then
De…ne
8
Z1
>
>
>
>
x > 0 1 = Q(x; 0) = C1 e
>
>
<
So
>
>
>
>
>
>x < 0
:
0 = Q(x; 0) = C1
0
Z1
e
p2
dp + C2 =
p2
dp + C2 =
0
1
C1 = p
C2 =
Therefore
1
2
px
4kt
Q(x; t) =
1
1
+p
2
Z
0
6
p
C1
2
e
p2
dp
+ C2
p
C1
2
+ C2
Error Function
Remember
Z1
e
p2
dp =
Z0
p2
e
Z1
1
dp =
2
1
0
e
p2
dp =
p
2
1
De…ne
2
erf(x) = p
Zx
e
p2
dp
erf c(x) = 1
erf(x)
0
erf(0) = 0
2
p
Zx
1
e
p2
2
dp = p
erf(1) = 1
Z0
e
p2
erf( x) =
2
dp + p
1
Zx
0
Then
px
4kt
1 + erf
Q(x; t) =
2
7
e
p2
erf(x)
dp = 1 + erf(x)
De…ne
S(x; t) =
@Q
@x
Then S is also a solution and
Z1
u(x; t) =
S(x
y; t)'(y)dy
1
is also a solution. Does it satisfy the initial condition u(x; 0) = '(x) ?
u(x; t) =
=
=
Z1
S(x
1
Z1
Z1
y; t)'(y)dy =
Z1
@Q
(x
@x
y; t)'(y)dy
1
@Q
(x
@y
y; t)'(y)dy
since
@Q
=
@x
1
Q(x
y; t)
@'
(y)dy
@y
Q(x
y; t)'(y)j11
1
Assume ' is zero at in…nity then
u(x; t) =
Z1
Q(x
y; t)
@'
(y)dy
@y
1
Using
@Q
@y
(
Q(s; 0) = 1 s > 0
Q(s; 0) = 0 s < 0
We have
u(x; 0) =
Z1
Q(x
=
Zx
@'
(y)dy = 'jx 1 = '(x)
@y
y; 0)
@'
(y)dy
@y
1
1
8
Conclusion
x2
@Q
1
e 4kt
=p
@x
4k t
Z1
(x y)2
1
e 4kt '(y)dy
u(x; t) = p
4k t
S(x; t) =
1
Z1
1
S(x; t)dx = p
1
Z1
1
lim S(x; t) = (x)
t=0
9
e
p2
dp = 1
Importance of Limiting Process
@u
@2u
=
@t
@x2
u(x; 0) = 0
Then a possible solution is
1
u(x; t) = p
e
4 t3
x2
4t
For x …xed and t ! 0 then u(x; t) ! 0.
2
However, for x4t constant and t ! 0 the solution is not bounded !! Hence, this
is not a legitimate solution
10
Main Theorem
Theorem: Assume
@u
@2u
=k 2
@t
@x
u(x; 0) = '(x)
( )
j (x)j < 1
Then
1
u(x; t) = p
4k t
u C1
1<x<1 ;
Z1
(x y)2
4kt
e
'(y)dy
1
0<t<1
each derivative of u(x; t) satis…es ( )
limt#0 u(x; t) = '(x)
proof
u(x; t) =
Z1
S(x
1
S(z; t) = p
Let p =
pz
kt
z=
p
1
e
4k t
y; t)'(y)dy =
S(z; t)'(x
1
z2
4kt
ktp. Then
1
u(x; t) = p
4
Assume j'(x)j
Z1
Zx
p2
4
e
'(x
p
p kt)dp
0
M . Then
ju(x; t)j
M
p
4
Z1
1
11
e
p2
4
dp = M
z)dz
Di¤erentiating we get
@u
=
@x
Z1
@S
(x
@x
1
const
= p
t
Z1
y; t)'(y)dy =
pe
1
Z1
const
p M
t
p2
4
pe
p2
4
Remembering that
z
e
2kt
z2
4kt
'(x
1
const
p M
t
dp
1
@nu
@xn
Z1
p
p kt)dp
'(x
Similarly
Z1
1
p
4 kt
Z1
const
p M
t
p2
4
pn e
dp
1
const
p M
t
S(x; t)dx = 1 we get
1
u(x; t)
'(x) =
Z1
1
1
=p
4
S(x
Z1
y; t) ['(y)
p2
4
e
1
h
'(x
'(x)] dy
p
p kt)
i
'(x) dp
We assume that '(x) is continuous. Hence,
j'(y) '(x)j " when jx yj
. So
p
ju(x; t)
|
bounded by
because
So
Zkt
'(x)j =
j'(y)
p
kt
:::dp +
Z
jpj> p
:::dp
kt
{z } |
{z
}
"
"
2
2
is small
t is small
"
'(x)j
p!1
2
12
z)dz
Backward heat equation
@2u
@u
= k 2
@t
@x
This gives growth in time and is not well posed. Equivalently we are trying to
solve the heat equation backward in time, i.e. given distribution of temperature
at time T …nd original temperature.
Consider
un (x; 0)
=
un (x; t)
=
1
sin(nx) ! 0 as n ! 1
n
2
1
sin(nx)en kt is unbounded
n
Black Scholes
2 2
s
Vss + rsVs rV + Vt = 0
2
s=market value of asset being optioned
t=time
=constant volatility
r=constant interest rate
Given the …nal value we wish to determine how to price it initially. So we
have backward problem ! - but we have additional minus sign =) well posed
13
Exercise
(
'(x; 0) = 1 jxj < l
'(x; 0) = 0 jxj > l
Then
u(x; t) = p
=p
1
4k t
Z1
1
4k t
Zl
(x y)2
4kt
e
=
dy
l
x l
p
4kt
1
=p
4kt
4k t
y
4kt
'(y)dy
1
p
q=px
(x y)2
4kt
e
Z
e
q2
dq
x+l
p
4kt
x l
p
4kt
1
p
4
Z
x+l
p
4kt
e
q2
1
dq = p
4
x+l
p
4kt
= erf
Z
e
x l
p
4kt
x+l
p
4kt
erf
x
p
l
4kt
Note: Solution is di¤erent from zero for all x when t > 0.
14
q2
dq
Lower Order Terms
ut = kuxx
u(x; 0) = '(x)
Let v(x; t) = e t u(x; t) so u(x; t) = e
@u
=e
@t
@2u
=e
@x2
So
e
t @v
@t
e
t
t
u
v(x; t)
t @v
@t
2
@
v
t
@x2
v(x; 0) = '(x)
e
t@
t
2
v
@x2
v=e
v
e
t
or
vt = kvxx
v(x; 0) = '(x)
Hence,
1
v(x; t) = p
4k t
Z1
e
e t
u(x; t) = p
4k t
Z1
e
(x y)2
4kt
'(y)dy
1
1
15
(x y)2
4kt
'(y)dy
v
Convection - Di¤usion
ut + cux = kuxx
u(x; 0) = '(x)
Let v(x; t) = u(x + ct; t) or u(x; t) = v(x ct; t) = v(y; )
| {z }
y
Then using the chain rule
@u
@t
@u
@x
@v @
@v @y
@v
+
=
@ @t
@y @t
@
@v @
@v @y
@v
+
=
@ @x @y @x
@y
=
=
c
@v
@y
So
@v
@
c
@2v
@y 2
@2v
= k 2
@y
@v
@v
+c
@y
@y
@v
@
= k
Therefore
v(x; t)
u(x; t)
=
1
p
4k t
Z1
e
1
4k t
Z1
e
p
=
For example choose
'(x) =
(x y)2
4kt
'(y)dy
1
(x
ct y)2
4kt
'(y)dy
1
(
x<0
jxj > 0
Then
u(x; t) = p
8
<
1
4k t :
Z0
e
(x
ct y)2
4kt
'(y)dy +
1
Z1
0
16
e
(x
ct y)2
4kt
'(y)dy
9
=
;
De…ne p =
u(x; t)
=
=
=
x pct y
4kt
8
>
>
1 <
p
>
>
:
8
>
>
1 <
p
>
>
:
+
2
. Then y =
p
4ktp + x
ct.
9
>
>
=
p2
p2
e dy +
e dy+
>
>
;
x ct
1
p
4kt
0
2
1
3
20
x ct
p
0
4kt
Z1
Z
Z
6B
C p2 7
6B
B
6
C
7
6B
+
A e dy 5 + 4@
4@
x ct
p
4kt
Z
Z0
1
+
2
0
0
x ct
erf( p
)
4kt
17
39
>
>
Z
C p2 7=
C e dy 7
A
5>
>
;
0
x ct
p
4kt
1
Comparison: Wave Equation –Di¤usion Equation
Property
speed
singularity
well posed t > 0
well posed t < 0
maximum principle
Energy
information
Wave
…nite
along characteristics
X
X
X
conserved
transported
18
Di¤usion
in…nite
disappear
X
X
X
decays
decays