A Revealed-Preference Activity Rule for Quasi-Linear Utilities with Budget Constraints
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Transcript of 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)
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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)
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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
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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)
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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.
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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).
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The Cauchy Problem contd..
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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)
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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
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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)
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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?
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Example 8: Characteristic Cauchy problem
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Example 8: Characteristic Cauchy problem
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 problem
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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.
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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 ”.
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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)
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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 .
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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)
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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)
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Method of solution of a Cauchy problem contd..
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y p
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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).
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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
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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− .
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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
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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 .
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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.
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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.
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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.
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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.
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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)
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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.
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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)
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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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