linear quasi

8/18/2019 Linear Quasi

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First Order Partial Differential Equations, Part - 1:Single Linear and Quasilinear First Order Equations

PHOOLAN PRASAD

DEPARTMENT OF MATHEMATICSINDIAN INSTITUTE OF SCIENCE, BANGALORE

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Denition

First order PDE in two independent variables is a relation

F (x, y ; u ; u x , u y) = 0F a known real function from D3 ⊂ R 5 → R

(1)

Examples: Linear, semilinear, quasilinear, nonlinear equations -

ux + uy = 0u x + uy = ku, c and k are constant

u x + uy = u2

uu x + uy = 0u2

x − u2

y = 0

u 2x + u2y + 1 = 0

ux + 1 − u2y = 0 , dened for |u y | ≤ 1 (2)

A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics 2 / 50

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Symbols for various domains used

In this lecture we denoteby D a domain in R 2 where a solution is dened,by D1 a domain in R 2 where the coefficients of alinear equation are dened andby D2 is a domain in (x,y,u )-space i.e., R 3

nally by D3 a domain in R 5 where the function F

of ve independent variables is dened.




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Meaning of a classical (genuine) solution u(x, y )

u is dened on a domain D ⊂ R 2

u ∈ C 1(D )(x,y,u (x, y ), u x (x, y ), u y(x, y )) ∈ D3 when(x, y ) ∈ DF (x, y ; u(x, y ); ux (x, y ), u y(x, y )) = 0 ∀(x, y ) ∈ D .

We call a classical solution simply a solution .Otherwise, generalized or weak solution .




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ClassicationLinear equation:

a (x, y )ux + b(x, y )uy = c1(x, y )u + c2(x, y ) (3)Non linear equation: All other equations with subclassesSemilinear equation:

a (x, y )ux + b(x, y )uy = c(x,y,u ) (4)Quasilinear equation:

a (x,y,u )ux + b(x,y,u )uy = c(x,y,u ) (5)

Nonlinear equation: F (x, y ; u ; ux , u y) = 0 where F is not linear inux , u y .Properties of solutions of all 4 classes of equations are quitedifferent.


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Example 2

ux = 0

General solution in D = R2

u = f (y), f is an arbitrary C 1 function.

Function f is uniquely determined if u is prescribed on any curve nowhere parallel to x-axis. On a line parallel to x-axis, we can not

prescribe u arbitrarily.A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics 6 / 50



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Example 3

uy = 0 (6)

General solution in D = R 2

u = f (x), f C 1(R ). (7)


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Directional derivativeu x = 0 means rate of change of u in direction (1, 0) parallel to x− axisis zeroi.e. (1, 0).(u

x, u

y) = 0

We say ux = 0 is a directional derivative in the direction (1, 0).Consider a curve with parametric representation x = x(σ), y = y(σ)given by ODE

dx

dσ = a(x, y ),

dy

dσ = b(x, y ) (8)

Tangent direction of the curve at (x, y ):

(a (x, y ), b(x, y )) (9)


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Directional derivative contd..

Rate of change of u(x, y ) with σ as we move along this

curve is dudσ

= uxdxdσ

+ uydydσ

= a(x, y )ux + b(x, y )uy(10)

which is a directional derivative in the direction (a, b)at (x, y )If u satises PDE au x + buy = c(x,y,u ) then

dudσ

= c(x,y,u ) (11)




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Characteristic equation and compatibilityconditionFor the PDE

a(x, y

)u

x + b

(x, y

)u

y = c

(x, y

) (12)Characteristic equationsdxdσ

= a(x, y ), dydσ

= b(x, y ) (13)

and compatibility conditiondudσ

= c(x, y ) (14)

(14) and (15) with a(x, y ) = 0 characteristic equations givedydx =

b(x, y )a (x, y ) (15)

compatibility condition becomesdudx

= c(x, y )a (x, y )

(16)

A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics10 /50

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Example 4

yu x − xu y = 0 (17)Characteristic equations are

dxdσ = y,

dydσ = − x (18)

ordydx = −

xy ⇒ y dy+ x dx = 0 ⇒ x

2

+ y2

= constant(19)

The characteristic curves are circles with centre

at (0,0).A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics 11 /50

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Example 4 contd..

Compatibility conditions along these curves are

dudσ

= 0 ⇒ u = constant . (20)

Hence value of u at (x, y ) = value of u at (− x, − y).u is an even function of x and also of y.


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Example 4 contd..

Will this even function be of the formu = f (x2 + y4)?The informationu = constant on the circles x2 + y2 = constant

⇒ u = f (x2

+ y2)

where f ∈ C 1(R ) is arbitrary.

Every solution is of this form.u is an even function of x and y but of a specialform.


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Linear rst order PDE

a(x, y )u x + b(x, y )uy = c1(x, y )u + c2(x, y ) (21)

Let w(x, y ) be any solution of the nonhomogeneousequation (21). Set u = v + w(x, y )

⇒ v satises the homogeneous equationa (x, y )vx + b(x, y )vy = c1(x, y )v (22)

Let f (x, y ) be a general solution of (22)

⇒ u = f (x, y ) + w(x, y ) (23)


ffi

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Example 5: Equation with constant coefficients

au x + buy = c, a, b, c are constants (24)

For the homogeneous equation, c = 0, characteristicequation (with a = 0 )

dydx = ba ⇒ ay − bx = constant (25)

Along these

dudx

= 0 ⇒ u = constant (26)

Hence u = f (ay − bx) is general solution of thehomogeneous equation.


l h ffi

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Example 5: Equation with constant coefficientscontd..

For the nonhomogeneous equation, the compatibility condition

dudx

= ca

⇒ u = const + ca

x (27)

The constant here is constant along the characteristics

ay − bx = const .Hence general solution

u = f (ay − bx) + ca

x. (28)

Alternatively u = ca x is a particular solution. Hence the result.Solution of a PDE contains arbitrary elements. For a rst orderPDE, it is an arbitrary function.In applications - additional condition ⇒ Cauchy problem.


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The Cauchy Problem for F (x, y ; u ; ux , u y) = 0

Piecewise smooth curve

γ : x = x0(η), y = y0(η), η ∈ I ⊂ Ra given function u0(η) on γ Find a solution u(x, y ) in a neighbourhood of γ

The solution takes the prescribed value u0(η) on γ i.e

u(x0(η), y0(η)) = u0(η) (29)

Existence and uniqueness of solution of a Cauchyproblem requires restrictions on γ , the function F andthe Cauchy data u0(η).


Th C h P bl d

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The Cauchy Problem contd..


E l 6

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Example 6a

Solve yux − xu y = 0 in R 2

with u(x, 0) = x, x ∈R

The solution must be an even function of x and y.

But the Cauchy data is an odd function.

Solution does not exist.

Example 6b

Solve yux − xu y = 0 in a domain Du(x, 0) = x, x ∈R + (30)


E l 6b ti

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Example 6b conti....

Solution is u(x, y ) = ( x 2 + y2)1/ 2 , verify with partialderivatives

ux = x(x 2 + y2)1/ 2, u y = y(x 2 + y2)1/ 2

(31)

Solution is determined in R 2 \ { (0, 0)}.


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Example 7 contd


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Example 7 contd..

If P lies on γ, x = αη, y = βη , the characteristiccurves are

x = αη + 2σ, y = βη + 3σ, η = const , σ varies


Example 7 contd

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Example 7 contd..Compatibility condition

du

dσ = 1 ⇒ u = u0(x , y ) + σ (32)

When P lies on γ

u = u(αη,βη ) + σ = u0(η) + σ (33)

Solution of the Cauchy problemSolve x = αη + 2σ, y = βη + 3σ for σ and η

σ = βx − αy2β − 3α

, η = 2y − 3x2β − 3α

(34)

Substitute in expression for u

u(x, y ) = βx − αy2β − 3α

+ u02y − 3x2β − 3α

(35)


Example 7 contd Existence and Uniqueness

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Example 7 contd.. Existence and Uniqueness

The solution exists as long as 2β − 3α = 0 i.e., thedatum curve is not a characteristic curve

Uniqueness:Compatibility condition carries information on thevariation of u along a characteristic in unique way.This leads to uniqueness.

What happens when 2β − 3α = 0?


Example 8: Characteristic Cauchy problem

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2β − 3α = 0 ⇒ datum curve is a characteristic curveChoose α = 2 , β = 3 ⇒ x = 2η, y = 3η





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Example 8: Characteristic Cauchy problemcontd...

The characteristic Cauchy problem: Solve

2ux + 3u y = 1

with datau (2η, 3η) = u0(η) = u0(

1

2x)

Since

du 0(η)

dη =

d

dηu (2η, 3η) = 2 ux + 3u y = 1 , using PDE (36)

The Cauchy data u0 cannot be prescribed arbitrarily on γ .u 0(η) = η = 12 x , ignoring constant of integration.


Example 8 contd

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Example 8 contd..

u = 12x is a particular solution satisfying theCauchy data and g(3x − 2y) is solution of thehomogeneous equation.Hence

u = 1

2x + g(3x − 2y), g ∈ C 1 and g(0) = 0 (37)

is a solution of the Cauchy problem.Since g is any C 1 function with g(0) = 0 , solution

of the Characteristic Cauchy problem is not unique.We verify an important theorem ” in general,solution of a characteristic Cauchy problemdoes not exist and if exists, it is not unique ”.


Quasilinear equation

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Quasilinear equation

a (x , y , u )u x + b (x , y , u )u y = c(x , y , u ) (38)

Since a and b depend on u, it is not possible to interpret

a (x,y,u ) ∂ ∂x

+ b(x,y,u ) ∂ ∂y

(39)

as a directional derivative in (x, y )- plane.We substitute a known solution u(x, y ) for u in a and b, then atany point (x, y ), it represents directional derivative ∂ ∂σ in thedirection given by

dx

dσ = a(x,y,u (x, y )) ,

dy

dσ = b(x,y,u (x, y )) (40)

Along these characteristic curves, we get compatibility conditiondudσ

= c(x,y,u (x, y )) (41)


Quasilinear equation contd..

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Quasilinear equation contd..

(41) is true for every solution u(x, y ). The Characteristicequations

dxdσ

= a(x,y,u ), dydσ

= b(x,y,u ) (42)

along with the Compatibility condition

dudσ

= c(x,y,u )

forms closed system.

Solution of 3 equations form a 2-parameter family of curves in(x,y,u )-space are called Monge Curves and theirprojections on (x, y )-plane are two parameter family of characteristic curves .


Method of solution of a Cauchy problem

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Method of solution of a Cauchy problem

Solve

a (x,y,u )ux + b(x,y,u )uy = c(x,y,u ) (43)

in a domain D containing

γ : x = x0(η), y = y0(η)

with Cauchy data

u(x0(η), y0(η)) = u0(η) (44)


Method of solution of a Cauchy problem contd..

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y pSolve

dxdσ = a(x,y,u ),

dydσ = b(x,y,u ),

dudσ = c(x,y,u ) (45)

(x,y,u )|σ=0 = ( x0(η), y0(η), u 0(η)) (46)

⇒ x = x(σ, x 0(η), y0(η), u 0(η)) ≡ X (σ, η )y = y(σ, x 0(η), y0(η), u 0(η)) ≡ Y (σ, η )u = u(σ, x 0(η), y0(η), u 0(η)) ≡ U (σ, η ) (47)

Solving the rst two for σ = σ(x, y ), η = η(x, y ) we get the solution

u = U (σ(x, y ), η(x, y )) ≡ u(x, y ) (48)


Method of solution of a Cauchy problem contd..

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y p


Quasilinear equations conti...

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Q qTheorem:

x 0(η), y0(η), u 0(η) ∈ C 1(I ), say I = [0, 1]a (x,y,u ), b(x,y,u ), c(x,y,u ) ∈C 1(D 2), where D2 is a domain in ( x,y,u ) − spaceD 2 contains curve Γ in (x,y,u )-spaceΓ : x = x0(η), y = y0(η), u = u0(η), η ∈ I dy

0dη a(x 0(η), y0(η), u 0(η)) − dx0dη b(x0(η), y0(η), u 0(η)) = 0 , η ∈ I There exists a unique solution of the Cauchy problem in a domain Dcontaining I .

Do not worry about the complex statement in thetheorem. As long as the datum curve γ is nottangential to a characteristic curve and the functionsinvolved are smooth, the solution exist and is unique(see previous slide).


Example 9

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pCauchy problem

ux + uy = u

u(x, 0) = 1 ⇒x0 = η, y0 = 0 , u0 = 1 (49)

Step 1. Characteristic curvesdx

dσ = 1 ⇒ x

= σ

+ η

dydσ

= 1 ⇒ y = σ (50)

Step 2. Therefore σ = y, η = x − y

Step 3. Compatibility conditiondudσ

= u ⇒ u = u0(η)eσ = eσ

Step 4. Solution u = ey

exists on D

=R 2


Example 10

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pCauchy problem

ux + uy = u2u(x, 0) = 1 ⇒

x0 = η, y0 = 0 , u0 = 1 (51)

Step 1. Characteristic equations give

x = σ + η, y = σ

Step 2. Compatibility condition gives

dudσ = u

2

⇒ u = 1

u 0(η) − σ

Step 3. Solution u = 11− yexists on the domain D = y < 1 and u → + ∞ as y → 1− .


Example 11

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Cauchy problemuu x + uy = 0

u(x, 0) = x, 0 ≤ x ≤ 1 (52)

⇒ x = η, y = 0 , u = η, 0 ≤ η ≤ 1 at σ = 0

Step 1. Characteristic equations and compatibilitycondition

dxdσ

= u, dydσ

= 1 , dudσ

= 0 (53)

Step 2. Quasilinear equations, characteristics dependon the solution

u = η


Example 11 contd..

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Step 3.x = η(σ + 1) , y = σ

Step 4. From solution of characteristic equationsσ = y and η = xy+1Step 5. Solution is u = xy+1 , but D =?Step 6. Characteristic curves are straight lines

xy + 1

= η, 0 ≤ η ≤ |

which meet at the point (− 1, 0).Step 7. u is constant on these characteristics (see nextslide).




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Figure: Solution is determined in a wedged shaped regionin (x, y )-plane including the lines y = 0 and y = x + 1 .


Example 12



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Cauchy problemuu x + uy = 0

u(x, 0) = 12 , 0 ≤ x ≤ 1 (54)Step 1. Parametrization of Cauchy data

⇒

x0 =

η, y0 = 0

, u0 =

1

2, 0 ≤

η ≤ 1

.

Step 2. The compatibility condition gives

u = constant = 12

.

Step 3. The characteristic curves arey − 2x = − 2η, 0 ≤ η ≤ 1. (55)

on which solution has the same val ue u = 12.


Example 12 contd..

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Step 4. The solution u = 12 of the Cauchy problem is determined in aninnite strip 2x ≤ y ≤ 2x − 2 in (x, y )-plane.

Important: From examples 11 and 12, we notice that the domain,where solution of a Cauchy problem for a quasilinear equation isdetermined, depends on the Cauchy data.


General solution

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General solution contains an arbitrary function.

Theorem : If φ(x,y,u ) = C 1 and ψ(x,y,u ) = C 2 betwo independent rst integrals of the ODEs

dx

a (x,y,u ) =

dy

b(x,y,u ) =

du

c(x,y,u ) (56)

and φ2u + ψ2u = 0 , the general solution of the PDEau x + buy = c is given by

h (φ(x,y,u ), ψ(x,y,u )) = 0 (57)

where h is an arbitrary function.

A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics 41 /50

Example 13



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uu x + uy = 0 (58)

dxu

= dy1

= du0

(59)

Note 0 appearing in a denominator to be properly interpreted

⇒ u = C 1x − C 1y = C 2

⇒ x − uy = C 2 ⇒ (60)

General solution is given by

φ(u, x − uy ) = 0or u = f (x − uy ) (61)

where h and f arbitrary functions.Note : Solution of this nonlinear equation may be very difficult.Numerical method is generally used.


Example 14C id h diff i l i

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Consider the differential equation

(y + 2ux )ux − (x + 2uy )uy = 1

2(x2 − y2) (62)

The characteristic equations and the compatibility condition are

dxy + 2ux

= dy

− (x + 2uy ) =

du12 (x 2 − y2)

(63)

To get one rst integral we derive from these

xdx + ydy2u(x 2 − y2)

= 2dux2 − y2

(64)

which immediately leads to

ϕ(x,y,u ) ≡ x2 + y2 − 4u 2 = C 1 (65)


Example 14 contd..

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For another independent rst integral we derive asecond combination

ydx + xdyy2 − x2

= 2dux2 − y2

(66)

which leads to

ψ(x,y,u ) ≡ xy + 2u = C 2 (67)

The general integral of the equation (55)is given by

h(x2 + y2 − 4u2, xy + 2u) = 0x2 + y2 − 4u2 = f (xy + 2u) (68)

where h or f are arbitrary function s of their arguments.A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics 44 /50

Example 14 contd..

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Consider a Cauchy problem u = 0 on x − y = 0

⇒ x = η, y = η, u = 0

From (58) and (60) we get 2η2 = C 1 and η2 = C 2 whichgives C 1 = 2C 2. Therefore, the solution of the Cauchyproblem is obtained, when we take h(ϕ, ψ ) = ϕ − 2ψ .This gives, taking only the suitable one,

u = 1

2 (x − y)2 + 1 − 1 (69)

We note that the solution of the Cauchy problem isdetermined uniquely at all points in the (x, y )-plane.


Exercise1 Sh h ll h h i i f h i l

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1. Show that all the characteristic curves of the partialdifferential equation

(2x + u)u x + (2 y + u)uy = u

through the point (1,1) are given by the same straightline x − y = 0

2. Discuss the solution of the differential equation

uu x + uy = 0 , y > 0, −∞ < x < ∞

with Cauchy datau(x, 0) =

α 2 − x2 for |x | ≤ α0 for |x | > α.

(70)


Exercise contd..3 Fi d h l i f h diff i l i



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3. Find the solution of the differential equation

1 − m

r u ux − mMu y = 0

satisfyingu(0, y) =

Myρ − y

where m, r, ρ, M are constants, in a neighourhood of the point x = 0 , y = 0.

4. Find the general integral of the equation

(2x − y)y2ux + 8( y − 2x)x2u y = 2(4 x2 + y2)uand deduce the solution of the Cauchy problem whenthe u(x, 0) = 12x on a portion of the x-axis.


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1

P. Prasad. Nonlinear Hyperbolic Waves in Multi-dimensions .Monographs and Surveys in Pure and Applied Mathematics,Chapman and Hall/CRC, 121 , 2001.

2 P. Prasad. A theory of rst order PDE through propagation of discontinuities. Ramanujan Mathematical Society News Letter,2000, 10 , 89-103; see also the webpage:

3 P. Prasad and R. Ravindran. Partial Differential Equations .Wiley Eastern Ltd, New Delhi, 1985.


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Thank You!

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