first order partial differential equations, part - 1...

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

PHOOLAN PRASAD

DEPARTMENT OF MATHEMATICSINDIAN INSTITUTE OF SCIENCE, BANGALORE

Definition

First order PDE in two independent variables is a relation

F (x, y;u;ux, uy) = 0

F a known real function from D3 ⊂ R5 → R(1)

Examples: Linear, semilinear, quasilinear, nonlinear equations -

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

ux + uy = u2

uux + uy = 0u2x − u2y = 0

u2x + u2y + 1 = 0

ux +√

1− u2y = 0, defined for |uy| ≤ 1 (2)

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

Symbols for various domains used

In this lecture we denote

by D a domain in R2 where a solution is defined,

by D1 a domain in R2 where the coefficients of alinear equation are defined and

by D2 is a domain in (x, y, u)-space i.e., R3

finally by D3 a domain in R5 where the function Fof five independent variables is defined.


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

u is defined on a domain D ⊂ R2

u ∈ C1(D)

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

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


Classification

Linear 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, uy) = 0 where F is not linear inux, uy.Properties of solutions of all 4 classes of equations are quitedifferent.


Example 2

ux = 0

General solution in D = R2

u = f(y), f is an arbitrary C1 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 notprescribe u arbitrarily.


Example 3

uy = 0 (6)

General solution in D = R2

u = f (x), fεC1(R). (7)


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


Directional derivative contd..

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

du

dσ= ux

dx

dσ+ uy

dy

dσ= a(x, y)ux + b(x, y)uy

(10)

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

du

dσ= c(x, y, u) (11)


Characteristic equation and compatibilityconditionFor the PDE

a(x, y)ux + b(x, y)uy = c(x, y) (12)

Characteristic equations

dx

dσ= a(x, y),

dy

dσ= b(x, y) (13)

and compatibility condition

du

dσ= c(x, y) (14)

(14) and (15) with a(x, y) 6= 0 characteristic equations give

dy

dx=b(x, y)

a(x, y)(15)

compatibility condition becomes

du

dx=c(x, y)

a(x, y)(16)

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

Example 4

yux − xuy = 0 (17)

Characteristic equations are

dx

dσ= y,

dy

dσ= −x (18)

or

dy

dx= −x

y⇒ y dy+x dx = 0⇒ x2+y2 = constant.

(19)The characteristic curves are circles with centreat (0,0).


Example 4 contd..

Compatibility conditions along these curves are

du

dσ= 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.


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 ∈ C1(R) is arbitrary.

Every solution is of this form.

u is an even function of x and y but of a specialform.


Linear first order PDE

a(x, y)ux + 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 satisfies the homogeneous equation

a(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)


Example 5: Equation with constant coefficients

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

For the homogeneous equation, c = 0, characteristicequation (with a 6= 0)

dy

dx=b

a⇒ ay − bx = constant (25)

Along these

du

dx= 0⇒ u = constant (26)

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


Example 5: Equation with constant coefficientscontd..

For the nonhomogeneous equation, the compatibility condition

du

dx=c

a⇒ u = const +

c

ax (27)

The constant here is constant along the characteristicsay − bx = const.Hence general solution

u = f(ay − bx) +c

ax. (28)

Alternatively u = cax is a particular solution. Hence the result.

Solution of a PDE contains arbitrary elements. For a first orderPDE, it is an arbitrary function.

In applications - additional condition ⇒ Cauchy problem.


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

Piecewise smooth curveγ : 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 γi.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).


The Cauchy Problem contd..


Example 6a

Solve yux − xuy = 0 in R2

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 6bSolve yux − xuy = 0 in a domain D

u(x, 0) = x, x ∈ R+ (30)


Example 6b conti....

Solution is u(x, y) = (x2 + y2)1/2, verify with partialderivatives

ux =x

(x2 + y2)1/2, uy =

y

(x2 + y2)1/2(31)

Solution is determined in R2 \ {(0, 0)}.


Example 7Cauchy problem: Solve

2ux + 3uy = 1

with data u|γ = u(αη, βη) = u0(η) on

γ : βx− αy = 0; α, β = constant

Characteristic curve through an arbitrary point P ′(x′, y′) in(x, y)- plane.

dx

dσ= 2,

dy

dσ= 3

⇒ x = x′ + 2σ, y = y′ + 3σ

is a straight line3x− 2y = 3x′ − 2y′


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..

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 problem

Solve x = αη + 2σ, y = βη + 3σ for σ and η

σ =βx− αy2β − 3α

, η =2y − 3x

2β − 3α(34)

Substitute in expression for u

u(x, y) =βx− αy2β − 3α

+ u0

(2y − 3x

2β − 3α

)(35)


Example 7 contd.. Existence and Uniqueness

The solution exists as long as 2β − 3α 6= 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

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


Example 8: Characteristic Cauchy problemcontd...

The characteristic Cauchy problem: Solve

2ux + 3uy = 1

with data

u(2η, 3η) = u0(η) = u0(1

2x)

Since

du0(η)

dη=

d

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

The Cauchy data u0 cannot be prescribed arbitrarily on γ.u0(η) = η = 1

2x, ignoring constant of integration.


Example 8 contd..

u = 12x is a particular solution satisfying the

Cauchy data and g(3x− 2y) is solution of thehomogeneous equation.

Hence

u =1

2x+ g(3x− 2y), g ∈ C1 and g(0) = 0 (37)

is a solution of the Cauchy problem.

Since g is any C1 function with g(0) = 0, solutionof 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

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

du

dσ= c(x, y, u(x, y)) (41)


Quasilinear equation contd..

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

dx

dσ= a(x, y, u),

dy

dσ= b(x, y, u) (42)

along with the Compatibility condition

du

dσ= 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 ofcharacteristic curves.


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..

Solve

dx

dσ= a(x, y, u),

dy

dσ= b(x, y, u),

du

dσ= c(x, y, u) (45)

(x, y, u)|σ=0 = (x0(η), y0(η), u0(η)) (46)

⇒ x = x(σ, x0(η), y0(η), u0(η)) ≡ X(σ, η)y = y(σ, x0(η), y0(η), u0(η)) ≡ Y (σ, η)u = u(σ, x0(η), y0(η), u0(η)) ≡ U(σ, η) (47)

Solving the first 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..


Quasilinear equations conti...

Theorem:

x0(η), y0(η), u0(η) ∈ C1(I), say I = [0, 1]

a(x, y, u), b(x, y, u), c(x, y, u) ∈C1(D2), where D2 is a domain in (x, y, u)− space

D2 contains curve Γ in (x, y, u)-spaceΓ : x = x0(η), y = y0(η), u = u0(η), η ∈ Idy0dη a(x0(η), y0(η), u0(η))− dx0

dη b(x0(η), y0(η), u0(η)) 6= 0, η ∈ IThere 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 9Cauchy problem

ux + uy = uu(x, 0) = 1⇒

x0 = η, y0 = 0, u0 = 1 (49)

Step 1. Characteristic curves

dx

dσ= 1⇒ x = σ + η

dy

dσ= 1⇒ y = σ (50)

Step 2. Therefore σ = y, η = x− yStep 3. Compatibility condition

du

dσ= u⇒ u = u0(η)eσ = eσ

Step 4. Solution u = ey

exists on D = R2


Example 10

Cauchy problem

ux + uy = u2

u(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

du

dσ= u2 ⇒ u =

1

u0(η)− σ

Step 3. Solution u = 11−y

exists on the domain D = y < 1 and u→ +∞ as y → 1−.


Example 11

Cauchy problem

uux + uy = 0u(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

dx

dσ= u,

dy

dσ= 1,

du

dσ= 0 (53)

Step 2. Quasilinear equations, characteristics dependon the solution

u = η


Example 11 contd..

Step 3.x = η(σ + 1), y = σ

Step 4. From solution of characteristic equationsσ = y and η = x

y+1

Step 5. Solution is u = xy+1 , but D =?

Step 6. Characteristic curves are straight lines

x

y + 1= η, 0 ≤ η ≤ |

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


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


Example 12

Cauchy problem

uux + uy = 0u(x, 0) = 1

2 , 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 =1

2.

Step 3. The characteristic curves are

y − 2x = −2η, 0 ≤ η ≤ 1. (55)

on which solution has the same value u = 12 .


Example 12 contd..

Step 4. The solution u = 12 of the Cauchy problem is determined in an

infinite 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

General solution contains an arbitrary function.

Theorem : If φ(x, y, u) = C1 and ψ(x, y, u) = C2 betwo independent first integrals of the ODEs

dx

a(x, y, u)=

dy

b(x, y, u)=

du

c(x, y, u)(56)

and φ2u + ψ2

u 6= 0, the general solution of the PDEaux + buy = c is given by

h(φ(x, y, u), ψ(x, y, u)) = 0 (57)

where h is an arbitrary function.


Example 13

uux + uy = 0 (58)

dx

u=dy

1=du

0(59)

Note 0 appearing in a denominator to be properly interpreted

⇒ u = C1

x− C1y = C2

⇒ x− uy = C2 ⇒ (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 14Consider the differential equation

(y + 2ux)ux − (x+ 2uy)uy =1

2(x2 − y2) (62)

The characteristic equations and the compatibility condition are

dx

y + 2ux=

dy

−(x+ 2uy)=

du12(x2 − y2)

(63)

To get one first integral we derive from these

xdx+ ydy

2u(x2 − y2)=

2du

x2 − y2(64)

which immediately leads to

ϕ(x, y, u) ≡ x2 + y2 − 4u2 = C1 (65)


Example 14 contd..

For another independent first integral we derive asecond combination

ydx+ xdy

y2 − x2=

2du

x2 − y2(66)

which leads to

ψ(x, y, u) ≡ xy + 2u = C2 (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 functions of their arguments.A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics

44 /50

Example 14 contd..

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

⇒ x = η, y = η, u = 0

From (58) and (60) we get 2η2 = C1 and η2 = C2 whichgives C1 = 2C2. 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.


Exercise

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

(2x+ u)ux + (2y + 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

uux + uy = 0, y > 0, −∞ < x <∞with Cauchy data

u(x, 0) =

{α2 − x2 for |x| ≤ α

0 for |x| > α.(70)


Exercise contd..

3. Find the solution of the differential equation(1− m

ru)ux −mMuy = 0

satisfying

u(0, y) =My

ρ− ywhere m, r, ρ,M are constants, in a neighourhood ofthe point x = 0, y = 0.

4. Find the general integral of the equation

(2x− y)y2ux + 8(y − 2x)x2uy = 2(4x2 + y2)u

and deduce the solution of the Cauchy problem whenthe u(x, 0) = 1

2x on a portion of the x-axis.A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics

47 /50

1 R. Courant and D. Hilbert. Methods of Mathematical Physics,vol 2: Partial Differential Equations. Interscience Publishers, NewYork, 1962.

2 L. C. Evans. Partial Differential Equations. Graduate Studies inMathematics, Vol 19, American Mathematical Society, 1999.

3 F. John. Partial Differential Equations. Springer-Verlag, NewYork, 1982.

4 P. Prasad. (1997) Nonlinearity, Conservation Law and Shocks,Part I: Genuine Nonlinearity and Discontinuous Solutions,RESONANCE- Journal of Science Education by Indian Academyof Sciences, Bangalore, Vol-2, No.2, 8-18.P. Prasad. (1997) Nonlinearity, Conservation Law and Shocks,Part II: Stability Consideration and Examples, RESONANCE-Journal of Science Education by Indian Academy of Sciences,Bangalore, Vol-2, No.7, 8-19.


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 first order PDE through propagation ofdiscontinuities. 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.


Thank You!


first order partial differential equations, part - 1...

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