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Part I

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

Part I

20401

Tony Shardlow

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Part I

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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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Part I

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

Outline


NotationExamples

2 Three famous PDEs

3 Basics

4 Well posedness

5 Linearity

6 Classifying PDEs

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Part I

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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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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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Part I

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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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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


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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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Part I

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs


where

u(t, x) = x2t2 + A(t) + B(x)


ut(t, x) = 2x2t + A(t)

utx = 4xt


utxtx = 4

utxtxt = 0.

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs


where

u(t, x) = x2t2 + A(t) + B(x)


ut(t, x) = 2x2t + A(t)

utx = 4xt


utxtx = 4

utxtxt = 0.

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Part I

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs


where

u(t, x) = x2t2 + A(t) + B(x)


ut(t, x) = 2x2t + A(t)

utx = 4xt


utxtx = 4

utxtxt = 0.

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs


where

u(t, x) = x2t2 + A(t) + B(x)


ut(t, x) = 2x2t + A(t)

utx = 4xt


utxtx = 4

utxtxt = 0.

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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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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


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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


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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


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Part I

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

Outline


2 Three famous PDEsLaplaces equationHeat equationWave equation

3 Basics

4 Well posedness

5 Linearity

6 Classifying PDEs

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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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Part I

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs






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Part I

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs






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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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Part I

20401

Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

Laplaces Equation

In two dimensions,

2ux2

+ 2uy2

= 0



2u

x2+

2u

y2+

2u

z2= 0


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L l E i

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Part I

20401

Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

Laplaces Equation

In two dimensions,

2ux2

+ 2uy2

= 0



2u

x2+

2u

y2+

2u

z2= 0


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L l E i

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Part I

20401

Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

Laplaces Equation

In two dimensions,

2ux2

+ 2uy2

= 0



2u

x2+

2u

y2+

2u

z2= 0


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Show u = 1/r is a soln of

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Part I

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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


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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




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Part I

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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




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Part I

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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



Soln of Laplaces eqn in 2d

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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


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Part I

20401

Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs


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



dimensions. 30/132


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Part I

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs


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



dimensions. 31/132

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The Heat Equation

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Part I

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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

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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

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Part I

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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

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Part I

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

The Wave Equation


2u

t2 c2

2u

x2= 0 utt c

2uxx = 0


2ut2

c2( 2u

x2+

2u

y2) = 0 utt c

2(uxx + uyy) = 0

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The Wave Equation

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Part I

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

e a e quat o


2u

t2 c2

2u

x2= 0 utt c

2uxx = 0


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

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Part I

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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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Part I

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs



ut =A

y

y


2A(x ct)


(x ct)


utt c2uxx = 0,


ut + cux = 0,

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs



ut =A

y

y


2A(x ct)


(x ct)


utt c2uxx = 0,


ut + cux = 0,

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Outline

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Part I

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs


2 Three famous PDEs

3 BasicsPDE and orderExamplesProblem 2a

4 Well posedness

5 Linearity

6 Classifying PDEs

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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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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

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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


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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


order 3

HOMEWORK


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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


order 3

HOMEWORK


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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


order 3

HOMEWORK


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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


order 3

HOMEWORK


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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


order 3

HOMEWORK


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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


order 3

HOMEWORK


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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


order 3

HOMEWORK


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Problem 2a: Find a PDE foru(t, x) = A(x+ ct) +B(x ct),

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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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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

( , ) ( ) ( )


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


(x+ct)+B(xct)


utt = c2uxx.

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

( ) ( ) ( )


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


(x+ct)+B(xct)


utt = c2uxx.

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs


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


(x+ct)+B(xct)


utt = c2uxx.

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Problem 3a: determine A,Busing u(0, x) = f0(X).

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Part I

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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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Part I

20401

Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs


u(0, x) = A(x) + B(x)


A(x) = B(x) = f0(x)/2


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Part I

20401

Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs


u(0, x) = A(x) + B(x)


A(x) = B(x) = f0(x)/2


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Part I

20401

Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs


u(0, x) = A(x) + B(x)


A(x) = B(x) = f0(x)/2


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Outline

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Part I

20401

Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs


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




dt= 2t

General solution is

u(t) = t2 + C ,



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




dt= 2t

General solution is

u(t) = t2 + C ,



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




dt= 2t

General solution is

u(t) = t2 + C ,



du

dt= 2t, u(0) = u0.


+ u0 64/132

Well posedness

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Part I

20401

Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs




dt= 2t

General solution is

u(t) = t2 + C ,



du

dt= 2t, u(0) = u0.


+ 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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Part I

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs


as the domain. e.g.,ut = uxx, (t, x) [0, T] [0, 1].




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Part I

20401

Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs


as the domain. e.g.,ut = uxx, (t, x) [0, T] [0, 1].




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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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Part I

20401

Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs



u(t, x) = f(x), t = 0

or/and

boundary conditions



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Part I

20401

Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs



u(t, x) = f(x), t = 0

or/and

boundary conditions



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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


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


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


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Even with initial and boundary conditions some PDEs are ill

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Part I

20401

Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs




ut+uxx = 0,




HOMEWORK


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Part I

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs




ut+uxx = 0,




HOMEWORK


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P I



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Part I

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs




ut+uxx = 0,




HOMEWORK


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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

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

Backward heat equation is illposed

W h h th t

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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


We have show that

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Part I

20401

Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

We have show that



u(0, x) = cos(x/)


u(t, x) = et/2

cos(x/)


2 as 0.

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Part I


We have show that

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Part I

20401

Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

We have show that



u(0, x) = cos(x/)


u(t, x) = et/2

cos(x/)


2 as 0.

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Part I

Outline


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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

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

ut = uxx



L(u) = f




L(u) = ut uxx

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Part I

Linearity

The heat equation

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

ut = uxx



L(u) = f




L(u) = ut uxx

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Part I

fi ( )

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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,

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

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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20401

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


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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


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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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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

(uxx +

uyy) =

f.


utt c2uxx = f.




ut +1


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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

(uxx

+ uyy

) = f.


utt c2uxx = f.




ut +1


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Part I



( )

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20401

Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

(uxx

+ uyy

) = f.


utt c2uxx = f.




ut +1


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Part I

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

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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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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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Superposition with particularsoln


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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

( ) p y


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

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Superposition with particularsoln


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Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs

( ) p y


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


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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


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


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


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


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


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

8/3/2019 PDEs - Slides (1)

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Partialderivatives

Three famousPDEs

Basics

Wellposedness

Linearity

ClassifyingPDEs


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

8/3/2019 PDEs - Slides (1)

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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


8/3/2019 PDEs - Slides (1)

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Partialderivatives

Three famousPDEs

Basics

Well

posedness

Linearity

ClassifyingPDEs

2



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Part I

20401

Examples

ut +1


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Partialderivatives

Three famousPDEs

Basics

Well

posedness

Linearity

ClassifyingPDEs

2



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Part I

20401

Examples

ut +1


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Partialderivatives

Three famousPDEs

Basics

Well

posedness

Linearity

ClassifyingPDEs

2



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Part I

20401


We discuss again the three classical second order PDEs

1 heat equation:

8/3/2019 PDEs - Slides (1)

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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


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


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Part I

20401

Type of PDE


d b d f i d d f

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Partialderivatives

Three famousPDEs

Basics

Well

posedness

Linearity

ClassifyingPDEs




discriminant b2





HOMEWORK


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Part I

20401

Type of PDE


d b d f i d d f

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Partialderivatives

Three famousPDEs

Basics

Well

posedness

Linearity

ClassifyingPDEs




discriminant b2





HOMEWORK


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Part I

20401

Heat equation ut uxx = f isparabolic


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Partialderivatives

Three famousPDEs

Basics

Well

posedness

Linearity

ClassifyingPDEs


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


8/3/2019 PDEs - Slides (1)

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Partialderivatives

Three famousPDEs

Basics

Well

posedness

Linearity

ClassifyingPDEs


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


8/3/2019 PDEs - Slides (1)

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Partialderivatives

Three famousPDEs

Basics

Well

posedness

Linearity

ClassifyingPDEs


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


ffi

8/3/2019 PDEs - Slides (1)

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Partialderivatives

Three famousPDEs

Basics

Well

posedness

Linearity

ClassifyingPDEs


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


i h ffi i

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Partialderivatives

Three famousPDEs

Basics

Well

posedness

Linearity

ClassifyingPDEs


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


i h ffi i

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Partialderivatives

Three famousPDEs

Basics

Well

posedness

Linearity

ClassifyingPDEs


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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