First Order Partial Differential Equations, Part - 1...
Transcript of 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.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics 3 / 50
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.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics 4 / 50
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.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics 5 / 50
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.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics 6 / 50
Example 3
uy = 0 (6)
General solution in D = R2
u = f (x), fεC1(R). (7)
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics 7 / 50
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)
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics 8 / 50
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)
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics 9 / 50
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).
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics11 /50
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.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics12 /50
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.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics13 /50
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)
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics14 /50
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.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics15 /50
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.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics16 /50
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).
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics17 /50
The Cauchy Problem contd..
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics18 /50
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)
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics19 /50
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)}.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics20 /50
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′
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics21 /50
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
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics22 /50
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)
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics23 /50
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?
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics24 /50
Example 8: Characteristic Cauchy problem
2β − 3α = 0⇒ datum curve is a characteristic curveChoose α = 2, β = 3⇒ x = 2η, y = 3η
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics25 /50
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.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics26 /50
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”.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics27 /50
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)
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics28 /50
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.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics29 /50
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)
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics30 /50
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)
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics31 /50
Method of solution of a Cauchy problem contd..
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics32 /50
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).
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics33 /50
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
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics34 /50
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−.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics35 /50
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
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics36 /50
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).
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics37 /50
Figure: Solution is determined in a wedged shaped regionin (x, y)-plane including the lines y = 0 and y = x+ 1.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics38 /50
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 .
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics39 /50
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.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics40 /50
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.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics41 /50
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.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics42 /50
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)
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics43 /50
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.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics45 /50
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)
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics46 /50
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.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics48 /50
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.
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics49 /50
Thank You!
A Model Lession FD PDE Part 1 P. Prasad Department of Mathematics50 /50