First Order Partial Differential Equation, Part - 2: Non-linear

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)

A Model Lession FD PDE Part 2 P. Prasad Department of Mathematics 2 / 28

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)


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


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).



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)


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.


Cauchy data

u(x, y) : D → Rγ : (x = x0(η), y = y0(η)) is curve in D.u0(η) = u(x0(η), y0(η))


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.


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)

A Model Lession FD PDE Part 2 P. Prasad Department of Mathematics10 /28


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)



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)



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.


Isotropic wave motion with constant velocity

Consider an isotropic wave moving into a uniform medium withconstant velocity c (like light wave)


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)


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)



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.



Wavefronts

αx + βy = ±ct (29)

+ sign for forward propagating wavefront

− sign for backward propagating wavefront

Normal distance at time t from the initialposition = ±ct


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.


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.


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.


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


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).


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.


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.


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.


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First Order Partial Differential Equation, Part - 2: Non-linear

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