First Order Partial Differential Equation, Part - 2:Non-linear Equation
PHOOLAN PRASAD
DEPARTMENT OF MATHEMATICSINDIAN INSTITUTE OF SCIENCE, BANGALORE
First order non-linear equation
F (x, y, u, p, q) = 0, p = ux, q = uy
F ∈ C2(D3), domain D3 ⊂ R5 (1)
No directional derivative of u in (x, y)- plane for
a general F .
Take a known solution u(x, y) ∈ C2(D), D ⊂ R2
F (x, y, u(x, y), p(x, y), q(x, y) = 0 in D. (2)
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Charpit Equqtions
Taking x derivative
Fx + Fuux + Fppx + Fqqx = 0
Using qx = (uy)x = (ux)y = py
Fppx + Fqpy = −Fx − pFu(3)
Beautiful, p is differentiated in the direction (Fp, Fq).Similarly
Fpqx + Fqqy = −Fy − qFu (4)
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Charpit Equations contd..
Along one parameter family of curves in (x, y)- planegiven by
dx
dσ= Fp,
dy
dσ= Fq (5)
we havedp
dσ= −Fx − pFu (6)
dq
dσ= −Fy − qFu (7)
Further
du
dσ= ux
dx
dσ+ uy
dy
dσ= pFp + qFq (8)
Derived for a given solution u = u(x, y).A Model Lession FD PDE Part 2 P. Prasad Department of Mathematics 4 / 28
Charpit Equations contd..
These, 5 equations for 5 quantities x, y, u, p, q arecomplete irrespective of the solution u(x, y).They are Charpit equations.
Given values (u0, p0, q0) at (x0, y0), such that(x0, y0, u0, p0, q0) ∈ D3
⇒ local unique solution of Charpit Equations with(x0, y0, u0, p0, q0) at σ = 0.
Autonomous system⇒ 4 parameter family of solutions.(x, y, u, p, q) = (x, y, u, p, q)(σ, c1, c2, c3, c4).
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Charpit Equations contd..
Theorem. The function F is constant for everysolution of the Charpit’s equations i.e.F (x(σ), y(σ), u(σ), p(σ), q(σ)) = C(c1, c2, c3, c4) isindependent of σ.Proof. Simple,
dF
dσ=dx
dσFx +
dy
dσFy +
du
dσFu +
dp
dσFp +
dq
dσFq = 0 (9)
when we use Charpit’s equations. In order thatsolution of Charpit’s equations satisfies the relationF = 0, choose c1, c2, c3 and c4 such that
C(c1, c2, c3, c4) = 0⇒ c4 = c4(c1, c2, c3) (10)
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Monge strip and characteristic curves
Monge strip is a solution of the Charpit’s equationssatisfying
F (x(σ), y(σ), u(σ), p(σ), q(σ) = 0. (11)
It is a 3 parameter family of functions
(x, y, u, p, q)(σ, c1, c2, c3) = 0 (12)
Characteristic curves: From the Monge strips, thecurves x(σ, c1, c2, c3), y(σ, c1, c2, c3) in (x, y)- plane are3-parameter family of characteristic curves.
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Cauchy data
u(x, y) : D → Rγ : (x = x0(η), y = y0(η)) is curve in D.u0(η) = u(x0(η), y0(η))
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Solution of a Cauchy problem
Value of u is carried along a characteristic notindependently but together with values of p and q.
We need values of x0, y0, u0, p0 and q0 at P0 on datumcurve as initial data for Charpit’s equations.
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Solution of a Cauchy problem contd..
How to get values of p0 and q0 from Cauchy data?
u = u0(η) on γ : x = x0(η), y = y0(η) (13)
First we have
F (x0(η), y0(η), u0(η), p0(η), q0(η)) = 0. (14)
Differentiating u0(η) = u(x0(η), y0(η)) wrt η, we get
u′0(η) = p0(η)x′0(η) + q0y′0(η). (15)
Solve now p0(η) and q0(η). Cauchy data for Charpit’s ODEs
(x(σ), y(σ), u(σ), p(σ), q(σ)) |σ=0= (x0, y0, u0, p0, q0)(η) (16)
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Solution of a Cauchy problem contd..
Solve the Charpit’s equations
dx
dσ= Fp,
dy
dσ= Fq,
du
dσ= pFp + qFq (17)
dp
dσ= −(Fx + pFu),
dq
dσ= −(Fy + qFu) (18)
with above initial conditions:
x = x(σ, η), y = y(σ, η) (19)
u = u(σ, η), p = p(σ, η), q = q(σ, η) (20)
Then solve σ = σ(x, y), η = η(x, y) from (20) and getu(x, y) = u(σ(x, y), η(x, y)).We can also get ux = p(x, y), uy = q(x, y)
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Solution of a Cauchy problem contd..
Theorem
F (x, y, u, p, q) ∈ C2(D3), domain D3 ⊂ R5
x0(η), y0(η), u0(η) ∈ C2(I), η ∈ I ⊂ R(x0(η), y0(η), u0(η)p0(η), q0(η)) ∈ D3, η ∈ Ip0(η), q0(η) ∈ C1(I), η ∈ Idx0dη Fq(x0, y0, u0, p0, q0)− dy0
dη Fp(x0, y0, u0, p0, q0) 6= 0
⇒ There exist a unique solution of the Cauchyproblem such that
u(x, y)|γ = u0, p(x, y)|γ = p0, q(x, y)|γ = q0 (21)
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Solution of a Cauchy problem contd..
Important point for the existence and uniqueness ofthe Cauchy problem is that the datum curve γ is nowhere tangential to a characteristic curve.
If γ is a characteristic curve, the data u0(η) is to berestricted (i.e., the equations of p0 and q0 are alsosatisfied) and when this restriction is imposed, thesolution of the Cauchy problem is non-unique -infinity of solutions exist.
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Isotropic wave motion with constant velocity
Consider an isotropic wave moving into a uniform medium withconstant velocity c (like light wave)
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Isotropic wave motion with constant velocitycontd..
Let a wavefront in such a wave be u(x, y) = ct.c(t+ δt) = u(x+ δx, y + δy)Taylor expansion of u up to first order terms (using ct = u(x, y))
c√u2x + u2y
δt =ux√u2x + u2y
δx+uy√u2x + u2y
δy = n1δx+ n2δy = δn
(22)
c√u2x + u2y
=δn = normal displacement
δt= c (23)
⇒ p2 + q2 = 1; p = ux, q = uy (24)
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Isotropic wave motion with constant velocitycontd..Problem. Find successive positions of the wavefront u(x, y) = ct whenthe initial position is
αx+ βy = 0, α2 + β2 = 1, where u = 0 (25)
Cauchy problemF ≡ p2 + q2 = 1
γ : x0 = βη, y0 = −αηcauchy data
u0 = 0
Values of p0 and q0
p20 + q20 = 1, βp0 − αq0 = 0
⇒ p0 = ±α, q0 = ±β(26)
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Isotropic wave motion with constant velocitycontd..
Solution of Charpit’s equations
dx
dσ= 2p,
dy
dσ= 2q,
du
dσ= 2(p2 + q2) = 2,
dp
dσ= 0,
dq
dσ= 0
(27)
Solution are
x = ±2ασ + βη, y = ±2βσ − αηu = 2σ, p = ±α, q = ±β
⇒ σ = ±1
2(αx+ βy)
⇒ u = ±(αx+ βy)
(28)
Two solutions of 2 problems, but uniqueness theorem not violated.
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Isotropic wave motion with constant velocitycontd..
Wavefronts
αx + βy = ±ct (29)
+ sign for forward propagating wavefront
− sign for backward propagating wavefront
Normal distance at time t from the initialposition = ±ct
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We have presented the theory of characteristics of firstorder PDEs briefly. It is based on the existence ofcharacteristics curves in the (x, y)-plane.
Along each of these characteristics we derive a numberof compatibility conditions, which are transportequations and which are sufficient to carry all necessaryinformation from the datum curve in the Cauchyproblem into a domain in which solution is determined.
In this sense every first order PDE is a hyperbolicequation.
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We have omitted a special class of solutions known ascomplete integral, for which any standard text may beconsulted.
Every solution of the PDE (1) can be obtained from acomplete integral.
We can also solve a Cauchy problem with its help.
Complete integral plays an important role in Physics.
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So far we have discussed only a genuine solution, whichis valid locally.
We have seen that characteristic carry informationabout the solution. Characteristic curves are theonly curve which can sustain discontinuities ofcertain types in the solution. For a linear equationthe discontinuities can be in the solution and itsderivatives, for a quasilinear equation thediscontinuities can be in the first and higher orderderivatives and for nonlinear equations thediscontinuities can be in second and higher orderderivatives.
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It is possible to develop a theory of first order PDEstarting from the definition of characteristic curvesas the curves which carry above type ofdiscontinuities. See P. Prasad, A theory of firstorder PDE through propagation of discontinuities,RMS Mathematics News letter, 2000, 10, 89-103.
Absence of real characteristic curves of the equationux + iuy = 0 in (x, y)-plane shows that its solutioncan not have discontinuities of any type along acurve in (x, y)-plane. This is related to theregularity of a solution of an elliptic PDE
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There is a fairly complete theory of weak solutions ofHamilton-Jacobi equations, a particular case of thenonlinear equation (1). Generally the domain ofvalidity of a weak solution with Cauchy data on thex-axis is at least half of the (x, y)-plane.
Theory of a single conservation law, a first orderequation, is particularly interesting not only from thepoint of view of theory but also from the point of viewof applications (Prasad, 2001).
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1 Consider the partial differential equation
F ≡ u(p2 + q2)− 1 = 0.
(i) Show that the general solution of the Charpit’s equations is a fourparameter family of strips represented by
x = x0 +2
3u0(2σ)
32 cos θ, y = y0 +
2
3u0(2σ)
32 sin θ,
u = 2u0σ, p =cos θ√
2σ, q =
sin θ√2σ
where x0, y0, u0 and θ are the parameters.(ii) Find the three parameter sub-family representing the totality of all
Monge strips.(iii) Show that the characteristic curves consist of all straight lines in
the (x, y)-plane.
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2 Solve the following Cauchy problems:(i) 1
2 (p2 + q2) = u with Cauchy data prescribed on the circlex2 + y2 = 1 by
u(cos θ, sin θ) = 1, 0 ≤ θ ≤ 2π
(ii) p2 + q2 +(p− 1
2x) (q − 1
2y)− u = 0 with Cauchy data prescribed on
the x-axis byu(x, 0) = 0
(iii) 2pq − u = 0 with Cauchy data prescribed on the y-axis by
u(0, y) =1
2y2
(iv) 2p2x+ qy − u = 0 with Cauchy data on x-axis
u(x, 1) = −1
2x.
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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.
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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 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.
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Thank You!
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