part 5 laplace equationjmb/lectures/pdelecture5.pdf · steady state stress analysis problem, which...
TRANSCRIPT
Laplace’s Equation
• Separation of variables – two examples• Laplace’s Equation in Polar Coordinates
– Derivation of the explicit form– An example from electrostatics
• A surprising application of Laplace’s eqn– Image analysis– This bit is NOT examined
0yx 2
2
2
2
=∂∂
+∂∂ φφ
02 =∇ φ
Laplace’s Equation
In the vector calculus course, this appears as where
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
∂∂∂∂
=∇
y
x
Note that the equation has no dependence on time, just on the spatial variables x,y. This means that Laplace’s Equation describes steady statesituations such as:
• steady state temperature distributions
• steady state stress distributions
• steady state potential distributions (it is also called the potential equation
• steady state flows, for example in a cylinder, around a corner, …
Steady state stress analysis problem, which satisfies Laplace’sequation; that is, a stretched elastic membrane on a rectangular former that has prescribed out-of-plane displacements along the boundaries
y
x
w = 0
w = 0w = 0
axsinww 0π
=
a
b
w(x,y) is the displacement in z-direction
x
yz
0yw
xw
2
2
2
2
=∂∂
+∂∂
Stress analysis example: Dirichlet conditions
Boundary conditionsTo solve:
axxa
wbxw
byyawaxxwbyyw
≤≤=
≤≤=≤≤=≤≤=
0sin),(
00),(00)0,(00),0(
0 for ,
for , for , for ,
π
Solution by separation of variables
kYY
XX
YY
XX
YXYXyYxXyxw
=′′
−=′′
=′′
+′′
=′′+′′=
0
0)()(),(
from which
and so
as usual …
where k is a constant that is either equal to, >, or < 0.
Case k=0)()(),()( DCyyYBAxxX +=+=
)(),(:00),(0)(0
000),0(
DCyAxyxwByxwyYDC
DCByw
+==≡≡==
===⇒=
withContinue so , then , if
or
ACxyyxwDwA
ADxxw
==≡=
=⇒=
),(0)00
00)0,(
withContinue or (so Either
0),(0000),( ≡⇒==⇒=⇒= yxwCAACayyaw or
That is, the case k=0 is not possible
Case k>0
)sincos)(sinhcosh(),(,2
yDyCxBxAyxwk
ααααα
++==
that so that Suppose
)sincos(sinh),(00),(00)sincos(0),0(
00sinh,10cosh
yDyCxByxwAyxwDCyDyCAyw
ααα
αα
+=⇒=≡⇒===+⇒=
==
withContinue
that Recall
yxBDyxwCyxwB
xBCxw
αα
α
sinsinh),(00),(0
0sinh0)0,(
=⇒=≡⇒=
=⇒=
withContinue
0),(000sinsinh0),(
≡⇒===⇒=
yxwDByaBDyaw
or either so αα
Again, we find that the case k>0 is not possible
Final case k<0
)sinhcosh)(sincos(),(
2
yDyCxBxAyxwk
ααααα
++=−=
that Suppose
)sinhcosh(sin),(000
0)sinhcosh(0),0(
yDyCxByxwAwDC
yDyCAyw
ααα
αα
+=⇒=≡⇒===+⇒=
withcontinue usual, as
yxBDyxwCwB
xBCxw
αα
α
sinhsin),(000
0sin0)0,(
=⇒=≡⇒==⇒=
withcontinue
ya
nxa
nBDyxwa
na
wDByaBDyaw
nπππαα
αα
sinhsin),(0sin
0000sinhsin0),(
=⇒=⇒=
≡⇒===⇒=
or
Solution
ayn
axnKyxw
nn
ππ sinhsin),(1∑∞
=
=Applying the first three boundary conditions, we have
absinh
wK 01 π=We can see from this that n must take only one value, namely 1, so that
which gives:a
bna
xnKaxw
nn
πππ sinhsinsin1
0 ∑∞
=
=
and the final solution to the stress distribution is
ay
ax
ab
wyxw πππ sinhsin
sinh),( 0=
axsinw)b,x(w 0π
=The final boundary condition is:
Check out: http://mathworld.wolfram.com/HyperbolicSine.html
abn
axnKxfw
nn
ππ sinhsin)(1
0 ∑∞
=
=Then
and as usual we use orthogonality formulae/HLT to find the Kn
y
x
w = 0
w = 0w = 0
)x(fww 0=
a
b
More general boundary condition
Types of boundary condition
1. The value is specified at each point on the boundary: “Dirichlet conditions”
2. The derivative normal to the boundary is specified at each point of the boundary: “Neumann conditions”
3. A mixture of type 1 and 2 conditions is specified
),( yxφ
),( yxn∂∂φ
Johann Dirichlet (1805-1859)http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Dirichlet.html
Carl Gottfried Neumann (1832 -1925)http://www-history.mcs.st-
andrews.ac.uk/history/Mathematicians/Neumann_Carl.html
A mixed condition problemy
x
w = 0
w = 0
xa
ww2
sin0π
=
a
b
0=∂∂
xw
0yw
xw
2
2
2
2
=∂∂
+∂∂
To solve:
Boundary conditions
axxa
wbxw
byxw
axxwbyyw
ax
≤≤=
≤≤=∂∂
≤≤=≤≤=
=
02
sin),(
00
00)0,(00),0(
0 for ,
for ,
for , for ,
π
A steady state heat transfer problem
There is no flow of heat across this boundary; but it does not necessarily have a constant temperature along the edge
Solution by separation of variables
kYY
XX
YY
XX
YXYXyYxXyxw
=′′
−=′′
=′′
+′′
=′′+′′=
0
0)()(),(
from which
and so
as usual …
where k is a constant that is either equal to, >, or < 0.
Case k=0)()(),()( DCyyYBAxxX +=+=
)(),(:00),(0)(0
000),0(
DCyAxyxwByxwyYDC
DCByw
+==≡≡==
===⇒=
withContinue so , then , if
or
ACxyyxwDwA
ADxxw
==≡=
=⇒=
),(0)00
00)0,(
withContinue or (so Either
0),(0000 ≡⇒==⇒=⇒=∂∂
=
yxwCAACyxw
ax
or
That is, the case k=0 is not possible
Case k>0
)sincos)(sinhcosh(),(,2
yDyCxBxAyxwk
ααααα
++==
that so that Suppose
)sincos(sinh),(00),(00)sincos(0),0(
00sinh,10cosh
yDyCxByxwAyxwDCyDyCAyw
ααα
αα
+=⇒=≡⇒===+⇒=
==
withContinue
that Recall
yxBDyxwCyxwB
xBCxw
αα
α
sinsinh),(00),(0
0sinh0)0,(
=⇒=≡⇒=
=⇒=
withContinue
0),(00
0sincosh0
≡⇒==
=⇒=∂∂
=
yxwDB
yaBDxw
ax
or either so
ααα
Again, we find that the case k>0 is not possible
Final case k<0
)sinhcosh)(sincos(),(
2
yDyCxBxAyxwk
ααααα
++=−=
that Suppose
)sinhcosh(sin),(000
0)sinhcosh(0),0(
yDyCxByxwAwDC
yDyCAyw
ααα
αα
+=⇒=≡⇒===+⇒=
withcontinue usual, as
yxBDyxwCwB
xBCxw
αα
α
sinhsin),(000
0sin0)0,(
=⇒=≡⇒==⇒=
withcontinue
ya
nxa
nBDyxwa
na
wDB
yaBDxw
n
ax
2)12(sinh
2)12(sin),(
2)12(0cos
000
0sinhcos0
πππαα
ααα
−−=⇒
−=⇒=
≡⇒==
=⇒=∂∂
=
or
Solution y
anx
anKyxw
nn 2
)12(sinh2
)12(sin),(1
ππ −−=∑
∞
=
Applying the first three boundary conditions, we have
ba
wK
2sinh
01 π=We can see from this that n must take only one value, namely 1, so that
which gives: ba
nxa
nKaxw
nn 2
)12(sinh2
)12(sin2
sin1
0πππ −−
=∑∞
=
and the final solution to the stress distribution is
ay
ax
ab
wyxw πππ sinhsin
sinh),( 0=
axwbxw
2sin),( 0
π=The final boundary condition is:
Check out: http://mathworld.wolfram.com/HyperbolicSine.html
PDEs in other coordinates…
• In the vector algebra course, we find that it is often easier to express problems in coordinates other than (x,y), for example in polar coordinates (r,Θ)
• Recall that in practice, for example for finite element techniques, it is usual to use curvilinear coordinates … but we won’t go that far
We illustrate the solution of Laplace’s Equation using polar coordinates*
*Kreysig, Section 11.11, page 636
A problem in electrostatics
Thin strip of insulating material
0V
rθ
Radius ahalfupper on the ),,( UzrV =θ
0112
2
2
2
22
22 =
∂∂
+∂∂
+∂∂
+∂∂
=∇zVV
rrV
rrVV
θ
I could simply TELL you that Laplace’sEquation in cylindrical polars is:
This is a cross section of a charged cylindrical rod.
… brief time out while I DERIVE this
2D Laplace’s Equation in Polar Coordinates
yθ
r
x
θcosrx =θsinry =
22 yxr +=
⎟⎠⎞
⎜⎝⎛= −
xytan 1θ
02
2
2
22 =
∂∂
+∂∂
=∇yu
xuu ),,( θrxx = ),r(yy θ=where
0),(),(),(
2 =∇
=
θ
θ
ruruyxu
So, Laplace’s Equation is
We next derive the explicit polar form of Laplace’s Equation in 2D
xu
xr
ru
xu
∂∂
∂∂
+∂∂
∂∂
=∂∂ θ
θ
xu
xxu
xr
ru
xxr
ru
xu
∂∂
⎟⎠⎞
⎜⎝⎛∂∂
∂∂
+∂∂
∂∂
+∂∂
⎟⎠⎞
⎜⎝⎛∂∂
∂∂
+∂∂
∂∂
=∂∂ θ
θθ
θ 2
2
2
2
2
2
Use the product rule to differentiate again
and the chain rule again to get these derivatives
xru
xr
ru
rru
x ∂∂
⎟⎠⎞
⎜⎝⎛∂∂
∂∂
+∂∂
⎟⎠⎞
⎜⎝⎛∂∂
∂∂
=⎟⎠⎞
⎜⎝⎛∂∂
∂∂ θ
θ xru
xr
ru
∂∂
∂∂∂
+∂∂
∂∂
=θ
θ
2
2
2
xu
xru
ru
x ∂∂
⎟⎠⎞
⎜⎝⎛∂∂
∂∂
+∂∂
⎟⎠⎞
⎜⎝⎛∂∂
∂∂
=⎟⎠⎞
⎜⎝⎛∂∂
∂∂ θ
θθθθ xu
xr
ru
∂∂
∂∂
+∂∂
∂∂∂
=θ
θθ 2
22
(*)
Recall the chain rule:
θcosrx = θsinry = 22 yxr += ⎟⎠⎞
⎜⎝⎛= −
xytan 1θ
The required partial derivatives
rx
xrx
xrryxr =
∂∂
⇒=∂∂
⇒+= 22222
ry
yr=
∂∂
Similarly,
42
2
42
2
22
3
2
2
2
3
2
2
2
2,2
,
,
rxy
yrxy
x
rx
yry
x
rx
yr
ry
xr
=∂∂
=∂∂
=∂∂
−=∂∂
=∂∂
=∂∂
θθ
θθin like manner ….
4
2
2
2
43
2
3
2
2
2
2
2
2
2 22ryu
rxyu
rxy
ru
ry
ru
rx
ru
xu
θθθ ∂∂
+∂∂
+−
∂∂∂
+∂∂
+∂∂
=∂∂
Similarly, 4
2
2
2
43
2
3
2
2
2
2
2
2
2 22rxu
rxyu
rxy
ru
rx
ru
ry
ru
yu
θθθ ∂∂
+∂∂
−∂∂
∂+
∂∂
+∂∂
=∂∂
So Laplace’s Equation in polars is 011
2
2
22
2
2
2
2
2
=∂∂
+∂∂
+∂∂
=∂∂
+∂∂
θu
rru
rru
yu
xu
Back to Laplace’s Equation in polar coordinates
Plugging in the formula for the partials on the previous page to the formulae on the one before that we get:
011
0
2
2
22
2
2
2
2
2
=∂∂
+∂∂
+∂∂
=∂∂
+∂∂
θu
rru
rru
yu
xu
to equivalent is
Example of Laplace in Cylindrical Polar Coordinates (r, Θ, z)
Consider a cylindrical capacitor
yθ
r
z
x
zP(r,θ,z)
0112
2
2
2
22
22 =
∂∂
+∂∂
+∂∂
+∂∂
=∇zVV
rrV
rrVV
θ
Laplace’s Equation in cylindrical polars is:
Thin strip of insulating material
0V
rθ
Radius ahalfupper on the ),,( UzrV =θ
In the polar system, note that the solution must repeat itself every θ = 2π
πθπθθπθθθ2:0),a(V
0:U),a(V≤≤∀=≤≤∀=
Boundary conditions
V should remain finite at r=0
There is no variation in V in the z-direction, so 0=∂∂
zV
0Vr1
rV
r1
rV
2
2
22
2
=∂∂
+∂∂
+∂∂
θUsing separation of variables
( ) ( )θΘrRV =
0Rr1R
r1R 2 =′′+′+′′ ΘΘΘ
2rRrRR ′+′′
=′′
−ΘΘ
As before, this means
constant a ,krR
rRR=
′+′′=
ΘΘ′′
− 2
This means we can treat it as a 2D problem
The case k=0
( )
( )DrCbarV
DrCrRrRR
ba
++=
+=⇒=′
+′′
+=Θ⇒=Θ′′
ln)(),(
ln)(0
)(0
θθ
θθ
so and
The solution has to be periodic in 2π: a=0
The solution has to remain finite as r→0: c=0
constant a ,gbdrV ==),( θ
The case k<0
( )
( )mm
mm
mm
rDrCrRRrm
rRR
mBmAmmk
+=⇒=+′
+′′
+=Θ⇒=Θ+Θ′′
−=
−)(0
sinhcosh)(0
2
2
2
2
that Suppose
θθθ
The solution has to be periodic in θ, with period 2π. This implies that 0),(0 ≡⇒== θrVBA mm
The case k>0
( )( )n
nn
nnn
nn
nn
nn
rDrCnBnArV
rDrCrRRrn
rRR
nBnAnnk
−
−
++=
+=⇒=−′
+′′
+=Θ⇒=Θ+Θ′′
=
)sincos(),(
)(0
)sincos()(0
2
2
2
2
θθθ
θθθ
that Suppose
Evidently, this is periodic with period 2π
To remain finite as r→0 0=nD
The solution( )∑ ++=
nnn
n nBnArgrV θθθ sincos),(
( )⎩⎨⎧
≤<≤≤
==++∑ πθππθ
θθθ20
0),(sincos
UaVnBnAag
nnn
n
Notice that we have not yet applied the voltage boundary condition!! Now is the time to do so
∫ =π
πθθ2
0
),( UdaVIntegrating V from 0 to 2π:
Left hand side: ( )
2
2sincos2
0
Ug
gdnBnAagn
nnn
=
=++∫ ∑
so and
πθθθπ
Solving for Am and Bm
( )∑ ++=n
nnn nBnAaUaV θθθ sincos
2),(
mAaA
aAdmU
dmnBnAadmUdmaV
mm
m
mm
nnn
n
all for so and ,
,00
0cos
cos)sincos(cos2
cos),(
0
2
0
2
0
2
0
==
+=
++=
∫
∫ ∫ ∫∑
π
πθθ
θθθθθθθθθ
π
π π π
)12(2
cos2
sin
sin)sincos(sin2
sin),(
2
00
2
0
2
0
2
0
−==
+⎥⎦⎤
⎢⎣⎡=
++=
∫
∫ ∫ ∫∑
nmaBmU
aBmmUdmU
dmnBnAadmUdmaV
mm
mm
nnn
n
odd for ,
π
πθθθ
θθθθθθθθθ
ππ
π π π
We apply the orthogonality relationships:
So far, the solution is
∑ −−
+= −−
n
nn nr
anUUrV θπ
θ )12sin()12(
22
),( )12()12(
( ) θπ
θ )1n2sin(a)1n2(
rU22U,rV
1n)1n2(
)1n2(
−−
+= ∑∞
=−
−
Check for r = a, θ = π/2:
( )2
)1n2sin(a)1n2(
aU22U,rV
1n)1n2(
)1n2( ππ
θ −−
+= ∑∞
=−
−
⎥⎦⎤
⎢⎣⎡ ++++= ......
25sin
51
23sin
31
2sinU2
2U πππ
π
⎥⎦⎤
⎢⎣⎡ +−+−+= ......
71
51
311U2
2U
π
U4
U22U
=⎥⎦⎤
⎢⎣⎡+=π
π
0V
rθ
U
An application in image analysis
• We saw that the Gaussian is a solution to the heat/diffusion equation
• We have studied Laplace’s equation• The next few slides hint at the application
of what we have done so far in image analysis
• This is aimed at engaging your interest in PDEs … it is not examined
Laplace’s Equation in image analysis
Image fragment Edge map
How do we compute the edges?
inte
nsity
Signal position, x
d in
tens
ity/ d
x
Constant gradient + max at the step. Note amplified noise
d2in
tens
ity/ d
x2
Zero crossing at step. But doubly amplified noise
• Remove the noise by smoothing • Find places where the second derivative of the image is zero
Gaussian smoothing
Blurring with a Gaussian filter is one way to tame noise
Zero crossings of a second derivative, isotropic operator, after Gaussian smoothing
02 =∇ smoothIAn application of Laplace’s Equation!
Limits of isotropic Gaussian blurring
A noisy image Gaussian blurring
Gaussian is isotropic – takes no account of orientation of image features – so it gives crap edge features
0)(2 =∗∇ IGσ
As the blurring is increased, by increasing the standard deviation of theGaussian, the structure of the image is quickly lost.
Can we do better? Can we make blurring respect edges?
Anisotropic diffusion
22
||
/||11);(
or k,constant somefor ,);(
));((
2
2
kItxg
etxg
ItxgI
kI
Tt
σ
σ
∇+=
=
∇∇=∂∇
−
This is a non-linear version of Laplace’s Equation, in which the blurring is small across an edge feature (low gradient) and large along an edge.
Anisotropic blurring of the noisy image
Top right: Gaussian
Bottom left and right: different anisotropic blurrings
Example of
anisotropic diffusion –brain MRI images, which are very noisy.
Top: Gaussian blur
Middle and bottom: anistropic blur
Anisotropic blurring of the house image retaining important structures at different degrees of non-linear blurring
Two final examples:
Left – original image
Right – anistropic blurring