8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Part I
20401
Tony Shardlow
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Outline
1 Partial derivatives
2 Three famous PDEs
3 Basics
4 Well posedness
5 Linearity
6 Classifying PDEs
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Outline
1 Partial derivatives
NotationExamples
2 Three famous PDEs
3 Basics
4 Well posedness
5 Linearity
6 Classifying PDEs
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Notation for partial derivatives
A function u(t, x) of variables t, x has partial derivatives
u
t,
u
x
also denotedu
t = utu
x = ux.
For example, u(t, x) = x3t2 then
ut = 2tx3, ux = 3x
2t2 .
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Notation for partial derivatives
A function u(t, x) of variables t, x has partial derivatives
u
t,
u
x
also denotedu
t = utu
x = ux.
For example, u(t, x) = x3t2 then
ut = 2tx3, ux = 3x
2t2 .
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Higher order derivatives
We also need higher order partial derivatives. As ut = 2tx3 and
ux = 3x2t2,
utt = 2x3, utx = 6tx
2 , uxx = 6 x t2.
HOMEWORK
You can now try Problem 1
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Higher order derivatives
We also need higher order partial derivatives. As ut = 2tx3 and
ux = 3x2t2,
utt = 2x3, utx = 6tx
2 , uxx = 6 x t2.
HOMEWORK
You can now try Problem 1
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Show that utxtxt = 0,
where
u(t, x) = x2t2 + A(t) + B(x)
and A(t) be B(x) denote some differentiable functions.
ut(t, x) = 2x2t + A(t)
utx = 4xt
Two specific differentiations kill the two functions.
utxtx = 4
utxtxt = 0.
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Show that utxtxt = 0,
where
u(t, x) = x2t2 + A(t) + B(x)
and A(t) be B(x) denote some differentiable functions.
ut(t, x) = 2x2t + A(t)
utx = 4xt
Two specific differentiations kill the two functions.
utxtx = 4
utxtxt = 0.
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Show that utxtxt = 0,
where
u(t, x) = x2t2 + A(t) + B(x)
and A(t) be B(x) denote some differentiable functions.
ut(t, x) = 2x2t + A(t)
utx = 4xt
Two specific differentiations kill the two functions.
utxtx = 4
utxtxt = 0.
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Show that utxtxt = 0,
where
u(t, x) = x2t2 + A(t) + B(x)
and A(t) be B(x) denote some differentiable functions.
ut(t, x) = 2x2t + A(t)
utx = 4xt
Two specific differentiations kill the two functions.
utxtx = 4
utxtxt = 0.
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Show that utxtxt = 0,
where
u(t, x) = x2t2 + A(t) + B(x)
and A(t) be B(x) denote some differentiable functions.
ut(t, x) = 2x2t + A(t)
utx = 4xt
Two specific differentiations kill the two functions.
utxtx = 4
utxtxt = 0.
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Show that ut =14uxx
Let
u(t, x) = t1/2ex2/t
ut = 1
2t3/2ex
2t1 + x2t5/2ex2t1
ux = 2xt3/2ex
2t1
uxx = 2t3/2ex
2t1 + 4x2t5/2ex2t1
Thus
ut1
4uxx = 0
This is called the heat equation and is one of the mostimportant PDEs.
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Show that ut =14uxx
Let
u(t, x) = t1/2ex2/t
ut = 1
2t3/2ex
2t1 + x2t5/2ex2t1
ux = 2xt3/2ex
2t1
uxx = 2t3/2ex
2t1 + 4x2t5/2ex2t1
Thus
ut1
4uxx = 0
This is called the heat equation and is one of the mostimportant PDEs.
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Show that ut =14uxx
Let
u(t, x) = t1/2ex2/t
ut = 1
2t3/2ex
2t1 + x2t5/2ex2t1
ux = 2xt3/2ex
2t1
uxx = 2t3/2ex
2t1 + 4x2t5/2ex2t1
Thus
ut1
4uxx = 0
This is called the heat equation and is one of the mostimportant PDEs.
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Show that ut =14uxx
Let
u(t, x) = t1/2ex2/t
ut = 1
2t3/2ex
2t1 + x2t5/2ex2t1
ux = 2xt3/2ex
2t1
uxx = 2t3/2ex
2t1 + 4x2t5/2ex2t1
Thus
ut1
4uxx = 0
This is called the heat equation and is one of the mostimportant PDEs.
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Outline
1 Partial derivatives
2 Three famous PDEsLaplaces equationHeat equationWave equation
3 Basics
4 Well posedness
5 Linearity
6 Classifying PDEs
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
The three famous PDEs
We introduce the three classical second order PDEs
Heat equation model of heat diffusion in space and time.
Laplaces equation model of steady heat distribution (notime dependence)
Wave equation model of wave motion in space and time.
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
The three famous PDEs
We introduce the three classical second order PDEs
Heat equation model of heat diffusion in space and time.
Laplaces equation model of steady heat distribution (notime dependence)
Wave equation model of wave motion in space and time.
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
The three famous PDEs
We introduce the three classical second order PDEs
Heat equation model of heat diffusion in space and time.
Laplaces equation model of steady heat distribution (notime dependence)
Wave equation model of wave motion in space and time.
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Laplaces Equation
In two dimensions,
2ux2
+ 2uy2
= 0
Also writen uxx + uyy = 0
In three dimensions,
2u
x2+
2u
y2+
2u
z2= 0
Also written uxx + uyy + uzz = 0
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8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Laplaces Equation
In two dimensions,
2ux2
+ 2uy2
= 0
Also writen uxx + uyy = 0
In three dimensions,
2u
x2+
2u
y2+
2u
z2= 0
Also written uxx + uyy + uzz = 0
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L l E i
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Laplaces Equation
In two dimensions,
2ux2
+ 2uy2
= 0
Also writen uxx + uyy = 0
In three dimensions,
2u
x2+
2u
y2+
2u
z2= 0
Also written uxx + uyy + uzz = 0
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L l E i
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Laplaces Equation
In two dimensions,
2ux2
+ 2uy2
= 0
Also writen uxx + uyy = 0
In three dimensions,
2u
x2+
2u
y2+
2u
z2= 0
Also written uxx + uyy + uzz = 0
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Show u = 1/r is a soln of
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
/Laplaces eqn in R3
where r = (x2 + y2 + z2)1/2.
Thenux = x(x
2 + y2 + z2)3/2
and
uxx = (x2 + y2 + z2)3/2 + 3x2(x2 + y2 + z2)5/2
uyy = (x2 + y2 + z2)3/2 + 3y2(x2 + y2 + z2)5/2
uzz = (x2 + y2 + z2)3/2 + 3z2(x2 + y2 + z2)5/2
Thusuxx + uyy + uzz = 0
and u satisfies Laplaces equation in three dimensions. 25/132
Show u = 1/r is a soln of
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
/Laplaces eqn in R3
where r = (x2 + y2 + z2)1/2.
Thenux = x(x
2 + y2 + z2)3/2
and
uxx = (x2 + y2 + z2)3/2 + 3x2(x2 + y2 + z2)5/2
uyy = (x2 + y2 + z2)3/2 + 3y2(x2 + y2 + z2)5/2
uzz = (x2 + y2 + z2)3/2 + 3z2(x2 + y2 + z2)5/2
Thusuxx + uyy + uzz = 0
and u satisfies Laplaces equation in three dimensions. 26/132
Show u = 1/r is a soln of
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
/Laplaces eqn in R3
where r = (x2 + y2 + z2)1/2.
Thenux = x(x
2 + y2 + z2)3/2
and
uxx = (x2 + y2 + z2)3/2 + 3x2(x2 + y2 + z2)5/2
uyy = (x2 + y2 + z2)3/2 + 3y2(x2 + y2 + z2)5/2
uzz = (x2 + y2 + z2)3/2 + 3z2(x2 + y2 + z2)5/2
Thusuxx + uyy + uzz = 0
and u satisfies Laplaces equation in three dimensions. 27/132
Show u = 1/r is a soln of
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
/Laplaces eqn in R3
where r = (x2 + y2 + z2)1/2.
Thenux = x(x
2 + y2 + z2)3/2
and
uxx = (x2 + y2 + z2)3/2 + 3x2(x2 + y2 + z2)5/2
uyy = (x2 + y2 + z2)3/2 + 3y2(x2 + y2 + z2)5/2
uzz = (x2 + y2 + z2)3/2 + 3z2(x2 + y2 + z2)5/2
Thusuxx + uyy + uzz = 0
and u satisfies Laplaces equation in three dimensions. 28/132
Soln of Laplaces eqn in 2d
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Soln of Laplace s eqn in 2d
Let r = (x2 + y2)1/2 and
u(x, y) =1
2ln(x2 + y2) = ln(r).
ux = x(x2 + y2)1
uxx = (x2 + y2)1 2x2(x2 + y2)2
uyy = (x2 + y2)1 2y2(x2 + y2)2
Thus we have thatuxx + uyy = 0
hence u(x, y) = ln r satisfies Laplaces equation in two
dimensions. 29/132
Soln of Laplaces eqn in 2d
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Soln of Laplace s eqn in 2d
Let r = (x2 + y2)1/2 and
u(x, y) =1
2ln(x2 + y2) = ln(r).
ux = x(x2 + y2)1
uxx = (x2 + y2)1 2x2(x2 + y2)2
uyy = (x2 + y2)1 2y2(x2 + y2)2
Thus we have thatuxx + uyy = 0
hence u(x, y) = ln r satisfies Laplaces equation in two
dimensions. 30/132
Soln of Laplaces eqn in 2d
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Soln of Laplace s eqn in 2d
Let r = (x2 + y2)1/2 and
u(x, y) =1
2ln(x2 + y2) = ln(r).
ux = x(x2 + y2)1
uxx = (x2 + y2)1 2x2(x2 + y2)2
uyy = (x2 + y2)1 2y2(x2 + y2)2
Thus we have thatuxx + uyy = 0
hence u(x, y) = ln r satisfies Laplaces equation in two
dimensions. 31/132
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The Heat Equation
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
The Heat Equation
For a parameter > 0,
u
t
2u
x2= 0 ut uxx = 0
In one dimension, we saw example solution
u(t, x) =1
t1/2 e
x
2
/4t
In two dimensions,
u
t
(2u
x2
+2u
y2
) = 0 ut (uxx + uyy) = 0
with example soln
u(t, x) =1
t3/2e(x
2+y2)/4t
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The Heat Equation
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
The Heat Equation
For a parameter > 0,
u
t
2u
x2= 0 ut uxx = 0
In one dimension, we saw example solution
u(t, x) =1
t1/2 e
x
2
/4t
In two dimensions,
u
t
(2u
x2
+2u
y2
) = 0 ut (uxx + uyy) = 0
with example soln
u(t, x) =1
t3/2e(x
2+y2)/4t
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The Wave Equation
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
The Wave Equation
In one spatial dimension,
2u
t2 c2
2u
x2= 0 utt c
2uxx = 0
In two spatial dimensions,
2ut2
c2(2ux2
+2uy2
) = 0 utt c2(uxx + uyy) = 0
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The Wave Equation
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
The Wave Equation
In one spatial dimension,
2u
t2 c2
2u
x2= 0 utt c
2uxx = 0
In two spatial dimensions,
2ut2
c2( 2u
x2+
2u
y2) = 0 utt c
2(uxx + uyy) = 0
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The Wave Equation
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
e a e quat o
In one spatial dimension,
2u
t2 c2
2u
x2= 0 utt c
2uxx = 0
In two spatial dimensions,
2ut2
c2( 2u
x2+
2u
y2) = 0 utt c
2(uxx + uyy) = 0
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For c> 0, u(t, x) = A(x ct)i l f
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
is a soln of wave eqn
Then u(t, x) = A(y) with y = x ct so
ut =A
y
y
t= cA(x ct), utt = c
2A(x ct)
ux = A(x ct), uxx = A
(x ct)
Thus u satisfies the
utt c2uxx = 0,
and also uni-directional wave equation
ut + cux = 0,
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For c> 0, u(t, x) = A(x ct)i l f
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
is a soln of wave eqn
Then u(t, x) = A(y) with y = x ct so
ut =A
y
y
t= cA(x ct), utt = c
2A(x ct)
ux = A(x ct), uxx = A
(x ct)
Thus u satisfies the
utt c2uxx = 0,
and also uni-directional wave equation
ut + cux = 0,
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For c> 0, u(t, x) = A(x ct)i l f
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
is a soln of wave eqn
Then u(t, x) = A(y) with y = x ct so
ut =A
y
y
t= cA(x ct), utt = c
2A(x ct)
ux = A(x ct), uxx = A
(x ct)
Thus u satisfies the
utt c2uxx = 0,
and also uni-directional wave equation
ut + cux = 0,
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Outline
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
1 Partial derivatives
2 Three famous PDEs
3 BasicsPDE and orderExamplesProblem 2a
4 Well posedness
5 Linearity
6 Classifying PDEs
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PDE and order
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Definition (PDE)
A PDE is a relationship of the form
F(u, t, x, y, . . . , ut, ux, uy, . . . , utt, utx, uty, . . .) = 0
where u is the solution and is a function of the independentvariables t, x, y, z, ...
Definition (order)
The order of the PDE is the highest degree of differentiation
that appears in the equation.
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PDE and order
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Definition (PDE)
A PDE is a relationship of the form
F(u, t, x, y, . . . , ut, ux, uy, . . . , utt, utx, uty, . . .) = 0
where u is the solution and is a function of the independentvariables t, x, y, z, ...
Definition (order)
The order of the PDE is the highest degree of differentiation
that appears in the equation.
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Find the order of
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
1 uxut12 u
2x = e
u
order 1
2 uxy + uxxyy = 0
order 4
3 uxuyutxy + uxx uyy = 0
order 3
HOMEWORK
You can now try Problem 24
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Find the order of
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
1 uxut12 u
2x = e
u
order 1
2 uxy + uxxyy = 0
order 4
3 uxuyutxy + uxx uyy = 0
order 3
HOMEWORK
You can now try Problem 24
45/132
Find the order of
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
1 uxut12 u
2x = e
u
order 1
2 uxy + uxxyy = 0
order 4
3 uxuyutxy + uxx uyy = 0
order 3
HOMEWORK
You can now try Problem 24
46/132
Find the order of
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
1 uxut12 u
2x = e
u
order 1
2 uxy + uxxyy = 0
order 4
3 uxuyutxy + uxx uyy = 0
order 3
HOMEWORK
You can now try Problem 24
47/132
Find the order of
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
1 uxut12 u
2x = e
u
order 1
2 uxy + uxxyy = 0
order 4
3 uxuyutxy + uxx uyy = 0
order 3
HOMEWORK
You can now try Problem 24
48/132
Find the order of
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
1 uxut12 u
2x = e
u
order 1
2 uxy + uxxyy = 0
order 4
3 uxuyutxy + uxx uyy = 0
order 3
HOMEWORK
You can now try Problem 24
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Find the order of
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
1 uxut12 u
2x = e
u
order 1
2 uxy + uxxyy = 0
order 4
3 uxuyutxy + uxx uyy = 0
order 3
HOMEWORK
You can now try Problem 24
50/132
Find the order of
8/3/2019 PDEs - Slides (1)
51/132
Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
1 uxut12 u
2x = e
u
order 1
2 uxy + uxxyy = 0
order 4
3 uxuyutxy + uxx uyy = 0
order 3
HOMEWORK
You can now try Problem 24
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Problem 2a: Find a PDE foru(t, x) = A(x+ ct) +B(x ct),
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
( , ) ( ) ( )
where c is a constant and A(y), B(y) are given functions.
u(t, x) = A(x + ct) + B(x ct)
Then
ut = cA
(x+ ct) cB
(x ct), ux = A
(x+ ct) + B
(x ct)
and
utt = c2A(x+ct)+c2B(xct), uxx = A
(x+ct)+B(xct)
Hence we have the second order PDE
utt = c2uxx.
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Problem 2a: Find a PDE foru(t, x) = A(x+ ct) +B(x ct),
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
( , ) ( ) ( )
where c is a constant and A(y), B(y) are given functions.
u(t, x) = A(x + ct) + B(x ct)
Then
ut = cA
(x+ ct) cB
(x ct), ux = A
(x+ ct) + B
(x ct)
and
utt = c2A(x+ct)+c2B(xct), uxx = A
(x+ct)+B(xct)
Hence we have the second order PDE
utt = c2uxx.
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Problem 2a: Find a PDE foru(t, x) = A(x+ ct) +B(x ct),
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
( ) ( ) ( )
where c is a constant and A(y), B(y) are given functions.
u(t, x) = A(x + ct) + B(x ct)
Then
ut = cA
(x+ ct) cB
(x ct), ux = A
(x+ ct) + B
(x ct)
and
utt = c2A(x+ct)+c2B(xct), uxx = A
(x+ct)+B(xct)
Hence we have the second order PDE
utt = c2uxx.
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Problem 2a: Find a PDE foru(t, x) = A(x+ ct) +B(x ct),
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
where c is a constant and A(y), B(y) are given functions.
u(t, x) = A(x + ct) + B(x ct)
Then
ut = cA
(x+ ct) cB
(x ct), ux = A
(x+ ct) + B
(x ct)
and
utt = c2A(x+ct)+c2B(xct), uxx = A
(x+ct)+B(xct)
Hence we have the second order PDE
utt = c2uxx.
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Problem 3a: determine A,Busing u(0, x) = f0(X).
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
We have u(t, x) = A(x + ct) + B(x ct). For t = 0,
u(0, x) = A(x) + B(x)
Then U(0, x) = f0(x) implies that A(x) + B(x) = f0(x).There are solutions, for example
A(x) = B(x) = f0(x)/2
but the soln is NOT unique.
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Problem 3a: determine A,Busing u(0, x) = f0(X).
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
We have u(t, x) = A(x + ct) + B(x ct). For t = 0,
u(0, x) = A(x) + B(x)
Then U(0, x) = f0(x) implies that A(x) + B(x) = f0(x).There are solutions, for example
A(x) = B(x) = f0(x)/2
but the soln is NOT unique.
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Problem 3a: determine A,Busing u(0, x) = f0(X).
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
We have u(t, x) = A(x + ct) + B(x ct). For t = 0,
u(0, x) = A(x) + B(x)
Then U(0, x) = f0(x) implies that A(x) + B(x) = f0(x).There are solutions, for example
A(x) = B(x) = f0(x)/2
but the soln is NOT unique.
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Problem 3a: determine A,Busing u(0, x) = f0(X).
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
We have u(t, x) = A(x + ct) + B(x ct). For t = 0,
u(0, x) = A(x) + B(x)
Then U(0, x) = f0(x) implies that A(x) + B(x) = f0(x).There are solutions, for example
A(x) = B(x) = f0(x)/2
but the soln is NOT unique.
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Outline
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
1 Partial derivatives
2 Three famous PDEs
3 Basics
4 Well posednessAn ODEInitial conditions for a PDEBoundary conditions for a PDE
Definition
5 Linearity
6 Classifying PDEs60/132
Well posedness
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
When does a PDE have a solution? When is that solution
unique? When is it a good model?ODE
Find u(t) such thatdu
dt= 2t
General solution is
u(t) = t2 + C ,
where C is the constant of integration . For unique
solution,Initial value problem for ODE
du
dt= 2t, u(0) = u0.
The u0 is can be found from C , so that u(t) = t2
+ u0. 61/132
Well posedness
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
When does a PDE have a solution? When is that solution
unique? When is it a good model?ODE
Find u(t) such thatdu
dt= 2t
General solution is
u(t) = t2 + C ,
where C is the constant of integration . For unique
solution,Initial value problem for ODE
du
dt= 2t, u(0) = u0.
The u0 is can be found from C , so that u(t) = t2
+ u0. 62/132
Well posedness
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
When does a PDE have a solution? When is that solution
unique? When is it a good model?ODE
Find u(t) such thatdu
dt= 2t
General solution is
u(t) = t2 + C ,
where C is the constant of integration . For unique
solution,Initial value problem for ODE
du
dt= 2t, u(0) = u0.
The u0 is can be found from C so that u(t) = t2
+ u0 63/132
Well posedness
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
When does a PDE have a solution? When is that solution
unique? When is it a good model?ODE
Find u(t) such thatdu
dt= 2t
General solution is
u(t) = t2 + C ,
where C is the constant of integration . For unique
solution,Initial value problem for ODE
du
dt= 2t, u(0) = u0.
The u0 is can be found from C so that u(t) = t2
+ u0 64/132
Well posedness
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
When does a PDE have a solution? When is that solution
unique? When is it a good model?ODE
Find u(t) such thatdu
dt= 2t
General solution is
u(t) = t2 + C ,
where C is the constant of integration . For unique
solution,Initial value problem for ODE
du
dt= 2t, u(0) = u0.
The u0 is can be found from C so that u(t) = t2
+ u0 65/132
Initial conditions for a PDE
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Now consider the PDE: find u(t, x) such that
ut = 2t
Integrating gives the solution u = t2 + A(x) where A(x) is thefunction of integration.
For a unique solution, we specify initial condition
Initial value problem
ut = 2tu(0, x) = f(x)
()
where f(x) is a given function.The solution is u(t, x) = t2 + f(x).
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Boundary conditions for a PDE
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Often we consider u(t, x) for x restricted to some set , known
as the domain. e.g.,
ut = uxx, (t, x) [0, T] [0, 1].
When has a boundary, we need boundary conditions foruniqueness.
u(t, 0) = g0(t) u(t, 1) = g1(t) for all t > 0
where g0(t) and g1(t) are given functions.
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Boundary conditions for a PDE
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Often we consider u(t, x) for x restricted to some set , known
as the domain. e.g.,ut = uxx, (t, x) [0, T] [0, 1].
When has a boundary, we need boundary conditions foruniqueness.
u(t, 0) = g0(t) u(t, 1) = g1(t) for all t > 0
where g0(t) and g1(t) are given functions.
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Boundary conditions for a PDE
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Often we consider u(t, x) for x restricted to some set , known
as the domain. e.g.,ut = uxx, (t, x) [0, T] [0, 1].
When has a boundary, we need boundary conditions foruniqueness.
u(t, 0) = g0(t) u(t, 1) = g1(t) for all t > 0
where g0(t) and g1(t) are given functions.
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A PDE is an equation in thederivatives ofu
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
and to be well posed need some
initial conditionsIf u depends on time, an initial conditions is
u(t, x) = f(x), t = 0
or/and
boundary conditions
When u depends on x , condition on u at the boundary, e.g.the Dirichlet condition is
u(t, x) = g(x), x on boundary of
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A PDE is an equation in thederivatives ofu
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
and to be well posed need some
initial conditionsIf u depends on time, an initial conditions is
u(t, x) = f(x), t = 0
or/and
boundary conditions
When u depends on x , condition on u at the boundary, e.g.the Dirichlet condition is
u(t, x) = g(x), x on boundary of
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A PDE is an equation in thederivatives ofu
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
and to be well posed need some
initial conditionsIf u depends on time, an initial conditions is
u(t, x) = f(x), t = 0
or/and
boundary conditions
When u depends on x , condition on u at the boundary, e.g.the Dirichlet condition is
u(t, x) = g(x), x on boundary of
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Example: wave equation
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Find u(t, x) such that
utt c2uxx = 0 with (x, t) (0, 1) [0, T].
Two initial conditions are needed because of second derivativewith respect to t:
The following is well posed.
utt c2uxx = 0 in (0, 1) [0, T]
u(0, x) = f1(x) ; ut(0, x) = f2(x) for all x (0, 1)u(t, 0) = g0(t) ; u(t, 1) = g1(t) for all t > 0
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Example: wave equation
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Find u(t, x) such that
utt c2uxx = 0 with (x, t) (0, 1) [0, T].
Two initial conditions are needed because of second derivativewith respect to t:
The following is well posed.
utt c2uxx = 0 in (0, 1) [0, T]
u(0, x) = f1(x) ; ut(0, x) = f2(x) for all x (0, 1)u(t, 0) = g0(t) ; u(t, 1) = g1(t) for all t > 0
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Laplaces equation on R2
( )
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Find u(x, y) such that
uxx + uyy = 0, (x, y) (0, 1) (0, 1) .
For a unique solution, we need a boundary condition on
= {(0, 1) 0} {(0, 1) 1} {0 (0, 1)} {1 (0, 1)}.
1 For a Dirichlet boundary condition we specify
u( x) = gD( x) for all x .
2 For a Neumann boundary condition we specify the(outward) normal derivative of the function value:
un ( x) = gN( x) for all x .
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Wave equation on R2
Fi ll id h h i i 2 fi d ( ) h
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Finally consider the heat equation in R2: find u(t, x, y) suchthat
utt (uxx + uyy) = 0, (t, x) [0, T].
For a unique solution, we need an initial condition and aDirichlet or a Neumann boundary condition on .
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Definition
H i h d fi i i h h ll b i i f
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Here is the definition that we have all been waiting for ...
Definition (well posed)
A PDE problem is well posed if
existence there exists a solution
uniqueness the solution is uniquestability the solution depends continuously on the data
(initial and boundary conditions).
Definition (ill posed)If a problem is not well posed, we say it is ill posed.
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Definition
H i th d fi iti th t h ll b iti f
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Here is the definition that we have all been waiting for ...
Definition (well posed)
A PDE problem is well posed if
existence there exists a solution
uniqueness the solution is uniquestability the solution depends continuously on the data
(initial and boundary conditions).
Definition (ill posed)If a problem is not well posed, we say it is ill posed.
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Example of ill posed problems
E ith i iti l d b d diti PDE ill
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Even with initial and boundary conditions, some PDEs are illposed.
backward heat equation
Find u(t, x) such that
ut+uxx = 0,
for all x R, together with the initial data u(0, x) = 0.
This problem has the unique solution, u(t, x) = 0.
However it is ill posed, because small changes in the initial datagive large changes in the solution
HOMEWORK
You can now try Problem 56
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Example of ill posed problems
Even with initial and boundary conditions some PDEs are ill
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Even with initial and boundary conditions, some PDEs are illposed.
backward heat equation
Find u(t, x) such that
ut+uxx = 0,
for all x R, together with the initial data u(0, x) = 0.
This problem has the unique solution, u(t, x) = 0.
However it is ill posed, because small changes in the initial datagive large changes in the solution
HOMEWORK
You can now try Problem 56
80/132
Example of ill posed problems
Even with initial and boundary conditions some PDEs are ill
8/3/2019 PDEs - Slides (1)
81/132
Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Even with initial and boundary conditions, some PDEs are illposed.
backward heat equation
Find u(t, x) such that
ut+uxx = 0,
for all x R, together with the initial data u(0, x) = 0.
This problem has the unique solution, u(t, x) = 0.
However it is ill posed, because small changes in the initial datagive large changes in the solution
HOMEWORK
You can now try Problem 56
81/132
P I
Example of ill posed problems
Even with initial and boundary conditions some PDEs are ill
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Even with initial and boundary conditions, some PDEs are illposed.
backward heat equation
Find u(t, x) such that
ut+uxx = 0,
for all x R, together with the initial data u(0, x) = 0.
This problem has the unique solution, u(t, x) = 0.
However it is ill posed, because small changes in the initial datagive large changes in the solution
HOMEWORK
You can now try Problem 56
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P t I
u(t, x) = et/2
cos(x/) is asoln of backward heat eqn
h i Th
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
where is parameter. Then
ut = (1/2)et/2 cos(x/) = u/2
and
ux =(1/)et/2 sin(x/)
uxx =(1/2)et/
2cos(x/) = u/2.
Hence, ut = uxx.
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Part I
u(t, x) = et/2
cos(x/) is asoln of backward heat eqn
h i t Th
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Part I
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Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
where is parameter. Then
ut = (1/2)et/2 cos(x/) = u/2
and
ux =(1/)et/2 sin(x/)
uxx =(1/2)et/
2cos(x/) = u/2.
Hence, ut = uxx.
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Part I
Backward heat equation is illposed
W h h th t
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
We have show that
u(t, x) = et/2 cos(x/)
is a soln of the backward heat equation ut = uxx. Note thatif is tiny, then the intial data
u(0, x) = cos(x/)
is close to the zero initial data. But t/2 is large so that
u(t, x) = et/2
cos(x/)
is large for any t > 0.We have used the fact that if t > 0 then et/
2 as 0.
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Part I
Backward heat equation is illposed
We have show that
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
We have show that
u(t, x) = et/2 cos(x/)
is a soln of the backward heat equation ut = uxx. Note thatif is tiny, then the intial data
u(0, x) = cos(x/)
is close to the zero initial data. But t/2 is large so that
u(t, x) = et/2
cos(x/)
is large for any t > 0.We have used the fact that if t > 0 then et/
2 as 0.
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Part I
Backward heat equation is illposed
We have show that
8/3/2019 PDEs - Slides (1)
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Part I
20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
We have show that
u(t, x) = et/2 cos(x/)
is a soln of the backward heat equation ut = uxx. Note thatif is tiny, then the intial data
u(0, x) = cos(x/)
is close to the zero initial data. But t/2 is large so that
u(t, x) = et/2
cos(x/)
is large for any t > 0.We have used the fact that if t > 0 then et/
2 as 0.
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Part I
Outline
1 Partial derivatives
8/3/2019 PDEs - Slides (1)
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
2 Three famous PDEs
3 Basics
4 Well posedness
5 LinearityLinear BVPHeat equationSuperposition
6 Classifying PDEs
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Part I
Linearity
The heat equation
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
qut = uxx
is an example of a linear PDE and these have many specialproperties.
To test linearity, we express the PDE and any boundaryconditions as
L(u) = f
where L is a differential operator, u(t, x) is the solution withx Rd (typically the spatial dimension d = 1, 2, or 3) and
f(t, x), is the right hand side.
For heat equation ut = uxx,
L(u) = ut uxx
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Part I
Linearity
The heat equation
8/3/2019 PDEs - Slides (1)
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
ut = uxx
is an example of a linear PDE and these have many specialproperties.
To test linearity, we express the PDE and any boundaryconditions as
L(u) = f
where L is a differential operator, u(t, x) is the solution withx Rd (typically the spatial dimension d = 1, 2, or 3) and
f(t, x), is the right hand side.
For heat equation ut = uxx,
L(u) = ut uxx
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Part I
Linearity
The heat equation
8/3/2019 PDEs - Slides (1)
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
ut = uxx
is an example of a linear PDE and these have many specialproperties.
To test linearity, we express the PDE and any boundaryconditions as
L(u) = f
where L is a differential operator, u(t, x) is the solution withx Rd (typically the spatial dimension d = 1, 2, or 3) and
f(t, x), is the right hand side.
For heat equation ut = uxx,
L(u) = ut uxx
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Part I
fi ( )
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
Definition (linear operator)
The operator L is linear if for any two functions u and v andany R,
1 L(u+ v) = L(u) + L(v);
2 L(u) = L(u).
We show the heat equation operator
L(u) = ut uxx
is linear.
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Part I
Show heat equation operator islinear
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
L(u+ v) = (u+ v)t (u+ v)xx
= (ut + vt) (uxx + vxx)
= (ut uxx) + (vt vxx)
= L(u) + L(v)
L(u) = (u)t (u)xx
= ut uxx
= (ut uxx)
= L(u)
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Part I
Show heat equation operator islinear
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
L(u+ v) = (u+ v)t (u+ v)xx
= (ut + vt) (uxx + vxx)
= (ut uxx) + (vt vxx)
= L(u) + L(v)
L(u) = (u)t (u)xx
= ut uxx
= (ut uxx)
= L(u)
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Part I
Linear BVP
We assume the boundary conditions are linear. For example,
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
1 u(x, 0) = f(x) initial condition
2 u(x, t) = g(x) for x on boundary (Dirichlet condition)
3 ux(x, t) = g1(x) for x on boundary (Neumann condition)
Linear Boundary Value Problem (BVP)
is a PDEL(u) = f
subject to linear boundary conditions, where L is linear.
We often speak of nonlinear PDEs, where L or the boundaryconditions are not linear.
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Part I
Linear BVP
We assume the boundary conditions are linear. For example,
8/3/2019 PDEs - Slides (1)
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
1 u(x, 0) = f(x) initial condition
2 u(x, t) = g(x) for x on boundary (Dirichlet condition)
3 ux(x, t) = g1(x) for x on boundary (Neumann condition)
Linear Boundary Value Problem (BVP)
is a PDEL(u) = f
subject to linear boundary conditions, where L is linear.
We often speak of nonlinear PDEs, where L or the boundaryconditions are not linear.
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Part I
Example the heat equation
ut uxx = 0 in (0, 1) [0, T ]
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Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
ut uxx 0 in (0, 1) [0, T]
u(0, x) = f(x) for all x (0, 1)u(t, 0) = g0(t) ; u(t, 1) = g1(t) for all t > 0
()
HOMEWORK
You can now try Problem 7
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Part I
Example the heat equation
ut uxx = 0 in (0, 1) [0, T]
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
t xx ( , ) [ , ]
u(0, x) = f(x) for all x (0, 1)u(t, 0) = g0(t) ; u(t, 1) = g1(t) for all t > 0
()
HOMEWORK
You can now try Problem 7
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Part I
Other Linear PDE problems
* The Poisson equation
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
(uxx +
uyy) =
f.
* The wave equation, with wave speed c
utt c2uxx = f.
* The steady-state convection-diffusion equation with viscosity > 0 and horizontal wind w
(uxx + uyy) + wux = f.
* The Black-Scholes equation with stock price x, interest rater and volatility
ut +1
22x2uxx + rxux ru = 0.
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Part I
Other Linear PDE problems
* The Poisson equation
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
(uxx +
uyy) =
f.
* The wave equation, with wave speed c
utt c2uxx = f.
* The steady-state convection-diffusion equation with viscosity > 0 and horizontal wind w
(uxx + uyy) + wux = f.
* The Black-Scholes equation with stock price x, interest rater and volatility
ut +1
22x2uxx + rxux ru = 0.
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Part I
Other Linear PDE problems
* The Poisson equation
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
(uxx
+ uyy
) = f.
* The wave equation, with wave speed c
utt c2uxx = f.
* The steady-state convection-diffusion equation with viscosity > 0 and horizontal wind w
(uxx + uyy) + wux = f.
* The Black-Scholes equation with stock price x, interest rater and volatility
ut +1
22x2uxx + rxux ru = 0.
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Part I
Other Linear PDE problems
* The Poisson equation
( )
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
(uxx
+ uyy
) = f.
* The wave equation, with wave speed c
utt c2uxx = f.
* The steady-state convection-diffusion equation with viscosity > 0 and horizontal wind w
(uxx + uyy) + wux = f.
* The Black-Scholes equation with stock price x, interest rater and volatility
ut +1
22x2uxx + rxux ru = 0.
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Part I
Nonlinear PDE problems
* The inviscid Burgers equation
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
ut + uux = 0.
* The Korteweg-de Vries (KdV) equation
ut + 6uux + uxxx = 0.
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Part I
20401
Nonlinear PDE problems
* The inviscid Burgers equation
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Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
ut + uux = 0.
* The Korteweg-de Vries (KdV) equation
ut + 6uux + uxxx = 0.
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Part I
20401
Example
For the PDE,ut + uux = 0
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
ut + uux 0
let L(u) = ut + uux and it is non-linear as
L(u) = (u)t + (u)(u)x
= ut + 2uux
= (ut + uux)
= L(u) ( = 1)
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Part I
20401
Superposition
If the PDE and the associated boundary conditions are of theform L(u) = 0 and L is a linear operator then the boundary
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Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
( ) p y
value problem is said to be homogeneous.
Theorem (superposition)
If u1 and u2 are any two solutions of a homogeneous boundary
value problem, then any linear combination v = u1 + u2with , R is also a solution.
Proof.
L(v) = L(u1 + u2) = L(u1) 0
+L(u2) 0
= 0.
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Part I
20401
Superposition with particularsoln
If the PDE and the associated boundary conditions are of theform L(u) = 0 and L is a linear operator then the boundary
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
( ) p y
value problem is said to be homogeneous.
Theorem
If up is a particular soln of the linear BVPLu = f and v is asoln of the homogeneous problem Lv = 0,then w = up + v is a soln of Lu = f .
Proof.
L(w) = L(up + v) = L(up) f
+L(v)0
= f.
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Part I
20401
Superposition with particularsoln
If the PDE and the associated boundary conditions are of theform L(u) = 0 and L is a linear operator then the boundary
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20401
Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
( ) p y
value problem is said to be homogeneous.
Theorem
If up is a particular soln of the linear BVPLu = f and v is asoln of the homogeneous problem Lv = 0,then w = up + v is a soln of Lu = f .
Proof.
L(w) = L(up + v) = L(up) f
+L(v)0
= f.
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Part I
20401
Outline
1 Partial derivatives
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Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
2 Three famous PDEs
3 Basics
4 Well posedness
5 Linearity
6 Classifying PDEsSecond order PDEsLinear constant coefficient second order PDEs
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Part I
20401
General second orderPDEnonlinear
Here is the generic nonlinear second order PDE in twoindependent variables
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Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
autt + butx + cuxx + dut + eux + gu = f
with coefficients :
a(
u, x, t, ux
, ut, u
xx, u
xt, u
tt)b(u, x, t, ux, ut, uxx, uxt, utt)c(u, x, t, ux, ut, uxx, uxt, utt)
depend on 2nd order derivsd(u, x, t, ux, ut)e(u, x, t, ux, ut)
g(u, x, t)
f(x, t)
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Part I
20401
Second order PDE quasi-linear
Here is the generic quasi-linear second order PDE in twoindependent variables
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Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
autt + butx + cuxx + dut + eux + gu = f
with coefficients :
a(u, x, t, ux
, ut)
b(u, x, t, ux, ut)c(u, x, t, ux, ut)
independent of 2nd order derivsd(u, x, t, ux, ut)e(u, x, t, ux, ut)
g(u, x, t)
depend on u and 1st order derivs
f(x, t)
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Part I
20401
Second order PDE semi-linear
Here is the generic semi-linear second order PDE in twoindependent variables
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Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
autt + butx + cuxx + dut + eux + gu = f
with coefficients :
a(x, t)b(x, t)c(x, t)
independent of ud(u, x, t, ux, ut)e(u, x, t, ux, ut)
g(u, x, t)
depends on u and 1st order derivs
f(x, t)
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Part I
20401
Example
ut + uux uxx = f
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Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
autt + butx + cuxx + dut + eux + gu = f
with coefficients
a = 0
b = 0c =
d = 1e = u
g = 0f = f
Semi-linear
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Part I
20401
Example
ut + uux uxx = f
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Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
autt + butx + cuxx + dut + eux + gu = f
with coefficients
a = 0
b = 0c =
d = 1e = u
g = 0f = f
Semi-linear
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Part I
20401
Second order PDE linear
Here is the generic linear second order PDE in twoindependent variables
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Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
autt + butx + cuxx + dut + eux + gu = f
with variable coefficients :
a(x, t)b(x, t)c(x, t)d(x, t)e(x, t)
g(x, t)f(x, t)
independent of u
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Part I
20401
Second order PDE linearconstant coefficient
Here is the generic linear second order PDE in twoindependent variables
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Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
autt + butx + cuxx + dut + eux + gu = f
with constant coefficients :
a
b
c
d
e
gf
constant
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Part I
20401
Examples
ut +1
u2x uxx = f(x, t) semi-linear
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Partialderivatives
Three famousPDEs
Basics
Wellposedness
Linearity
ClassifyingPDEs
2
utt + uxuxx = f(x, t) quasi-linear
u2tt + uxuxx + u2 = f(x, t) non-linear
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Part I
20401
Examples
ut +1
u2x uxx = f(x, t) semi-linear
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Partialderivatives
Three famousPDEs
Basics
Well
posedness
Linearity
ClassifyingPDEs
2
utt + uxuxx = f(x, t) quasi-linear
u2tt + uxuxx + u2 = f(x, t) non-linear
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Part I
20401
Examples
ut +1
u2x uxx = f(x, t) semi-linear
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Partialderivatives
Three famousPDEs
Basics
Well
posedness
Linearity
ClassifyingPDEs
2
utt + uxuxx = f(x, t) quasi-linear
u2tt + uxuxx + u2 = f(x, t) non-linear
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Part I
20401
Examples
ut +1
u2x uxx = f(x, t) semi-linear
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Partialderivatives
Three famousPDEs
Basics
Well
posedness
Linearity
ClassifyingPDEs
2
utt + uxuxx = f(x, t) quasi-linear
u2tt + uxuxx + u2 = f(x, t) non-linear
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Part I
20401
The three famous PDEs
We discuss again the three classical second order PDEs
1 heat equation:
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Partialderivatives
Three famousPDEs
Basics
Well
posedness
Linearity
ClassifyingPDEs
ut = uxx
2 Laplaces equation:
uxx + uyy = 0
3 wave equation:utt + c
2uxx = 0
Any linear constant coefficient second order PDE is related toone of these.
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Part I
20401
Type of PDE
autt + butx + cuxx + dut + eux + gu = f
b d f f
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Partialderivatives
Three famousPDEs
Basics
Well
posedness
Linearity
ClassifyingPDEs
and a, b, c, d, e, g, f are independent of u.
Definition (PDE type)
There are three generic types of PDE associated the
discriminant b2
4ac. These are associated with conicsections:
hyperbolic b2 4ac > 0;
parabolic b2 4ac = 0;
elliptic b2 4ac < 0.
HOMEWORK
You can now try Problem 8
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Part I
20401
Type of PDE
autt + butx + cuxx + dut + eux + gu = f
d b d f i d d f
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Partialderivatives
Three famousPDEs
Basics
Well
posedness
Linearity
ClassifyingPDEs
and a, b, c, d, e, g, f are independent of u.
Definition (PDE type)
There are three generic types of PDE associated the
discriminant b2
4ac. These are associated with conicsections:
hyperbolic b2 4ac > 0;
parabolic b2 4ac = 0;
elliptic b2 4ac < 0.
HOMEWORK
You can now try Problem 8
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Part I
20401
Type of PDE
autt + butx + cuxx + dut + eux + gu = f
d b d f i d d f
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Partialderivatives
Three famousPDEs
Basics
Well
posedness
Linearity
ClassifyingPDEs
and a, b, c, d, e, g, f are independent of u.
Definition (PDE type)
There are three generic types of PDE associated the
discriminant b2
4ac. These are associated with conicsections:
hyperbolic b2 4ac > 0;
parabolic b2 4ac = 0;
elliptic b2 4ac < 0.
HOMEWORK
You can now try Problem 8
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Part I
20401
Heat equation ut uxx = f isparabolic
autt + butx + cuxx + dut + eux + gu = f
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Partialderivatives
Three famousPDEs
Basics
Well
posedness
Linearity
ClassifyingPDEs
with constant coefficients :
1 a = 0
2 b = 0
3
c = 4 d = 1; e = 0; g = 0
b2 4ac = 0 parabolic
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Part I
20401
Heat equation ut uxx = f isparabolic
autt + butx + cuxx + dut + eux + gu = f
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Partialderivatives
Three famousPDEs
Basics
Well
posedness
Linearity
ClassifyingPDEs
with constant coefficients :
1 a = 0
2 b = 0
3 c=
4 d = 1; e = 0; g = 0
b2 4ac = 0 parabolic
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Part I
20401
Laplaces eqn (utt + uxx) = fis elliptic
autt + butx + cuxx + dut + eux + gu = f
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Partialderivatives
Three famousPDEs
Basics
Well
posedness
Linearity
ClassifyingPDEs
with constant coefficients :
1 a = 1
2 b = 0
3 c= 1
4 d = 0; e = 0; g = 0
b2 4ac = 4 < 0 elliptic
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Part I
20401
P i l
Laplaces eqn (utt + uxx) = fis elliptic
autt + butx + cuxx + dut + eux + gu = f
ffi
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Partialderivatives
Three famousPDEs
Basics
Well
posedness
Linearity
ClassifyingPDEs
with constant coefficients :
1 a = 1
2 b = 0
3 c= 1
4 d = 0; e = 0; g = 0
b2 4ac = 4 < 0 elliptic
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Part I
20401
P ti l
Wave eqn utt c2uxx = f is
hyperbolic
autt + butx + cuxx + dut + eux + gu = f
i h ffi i
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Partialderivatives
Three famousPDEs
Basics
Well
posedness
Linearity
ClassifyingPDEs
with constant coefficients :
1 a = 1
2 b = 0
3 c =
c2
4 d = 0; e = 0; g = 0
b2 4ac = 4c2 > 0 hyperbolic
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Part I
20401
Partial
Wave eqn utt c2uxx = f is
hyperbolic
autt + butx + cuxx + dut + eux + gu = f
i h ffi i
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Partialderivatives
Three famousPDEs
Basics
Well
posedness
Linearity
ClassifyingPDEs
with constant coefficients :
1 a = 1
2 b = 0
3 c = c2
4 d = 0; e = 0; g = 0
b2 4ac = 4c2 > 0 hyperbolic
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