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

First Order Partial Differential Equation, Part - 2: Non-linear Equation PHOOLAN PRASAD DEPARTMENT OF MATHEMATICS INDIAN INSTITUTE OF SCIENCE, BANGAL...
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First Order Partial Differential Equation, Part - 2: Non-linear Equation PHOOLAN PRASAD

DEPARTMENT OF MATHEMATICS INDIAN INSTITUTE OF SCIENCE, BANGALORE

First order non-linear equation

F (x, y, u, p, q) = 0, p = ux, q = uy F ∈ C 2(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) ∈ C 2(D), D ⊂ R2 F (x, y, u(x, y), p(x, y), q(x, y) = 0 in D.

A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

(2)

2 / 28

Charpit Equqtions Taking x derivative Fx + Fu ux + Fp px + Fq qx = 0 Using qx = (uy )x = (ux )y = py Fp px + Fq py = −Fx − pFu

(3)

Beautiful, p is differentiated in the direction (Fp , Fq ). Similarly Fp qx + Fq qy = −Fy − qFu (4)

A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

3 / 28

Charpit Equations contd.. Along one parameter family of curves in (x, y)- plane given by dx dy = Fp , = Fq (5) dσ dσ we have dp = −Fx − pFu (6) dσ dq = −Fy − qFu dσ

(7)

Further dx dy du = ux + uy = pFp + qFq dσ dσ dσ Derived for a given solution u = u(x, y). A Model Lession FD PDE Part 2

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Department of Mathematics

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Charpit Equations contd.. These, 5 equations for 5 quantities x, y, u, p, q are complete 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 ).

A Model Lession FD PDE Part 2

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Department of Mathematics

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Charpit Equations contd.. Theorem. The function F is constant for every solution of the Charpit’s equations i.e. F (x(σ), y(σ), u(σ), p(σ), q(σ)) = C(c1 , c2 , c3 , c4 ) is independent of σ. Proof. Simple, dF dx dy du dp dq = Fx + Fy + Fu + Fp + Fq = 0 (9) dσ dσ dσ dσ dσ dσ when we use Charpit’s equations. In order that solution of Charpit’s equations satisfies the relation F = 0, choose c1 , c2 , c3 and c4 such that C(c1 , c2 , c3 , c4 ) = 0 ⇒ c4 = c4 (c1 , c2 , c3 ) A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

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Monge strip and characteristic curves Monge strip is a solution of the Charpit’s equations satisfying 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, the curves x(σ, c1 , c2 , c3 ), y(σ, c1 , c2 , c3 ) in (x, y)- plane are 3-parameter family of characteristic curves.

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Department of Mathematics

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Cauchy data u(x, y) : D → R γ : (x = x0 (η), y = y0 (η)) is curve in D. u0 (η) = u(x0 (η), y0 (η))

A Model Lession FD PDE Part 2

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Department of Mathematics

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Solution of a Cauchy problem

Value of u is carried along a characteristic not independently but together with values of p and q. We need values of x0 , y0 , u0 , p0 and q0 at P0 on datum curve as initial data for Charpit’s equations.

A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

9 / 28

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)

F (x0 (η), y0 (η), u0 (η), p0 (η), q0 (η)) = 0.

(14)

First we have

Differentiating u0 (η) = u(x0 (η), y0 (η)) wrt η, we get u00 (η) = p0 (η)x00 (η) + q0 y00 (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)

A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

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Solution of a Cauchy problem contd.. Solve the Charpit’s equations dx dy du = Fp , = Fq , = pFp + qFq dσ dσ dσ dp dq = −(Fx + pFu ), = −(Fy + qFu ) dσ dσ with above initial conditions:

(17)

(18)

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 get u(x, y) = u(σ(x, y), η(x, y)). We can also get ux = p(x, y), uy = q(x, y) A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

11 / 28

Solution of a Cauchy problem contd.. Theorem F (x, y, u, p, q) ∈ C 2 (D3 ), domain D3 ⊂ R5 x0 (η), y0 (η), u0 (η) ∈ C 2 (I), η ∈ I ⊂ R (x0 (η), y0 (η), u0 (η)p0 (η), q0 (η)) ∈ D3 , η ∈ I p0 (η), q0 (η) ∈ C 1 (I), η ∈ I dy0 dx0 dη Fq (x0 , y0 , u0 , p0 , q0 ) − dη Fp (x0 , y0 , u0 , p0 , q0 ) 6= 0 ⇒ There exist a unique solution of the Cauchy problem such that u(x, y)|γ = u0 , p(x, y)|γ = p0 , q(x, y)|γ = q0

A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

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12 / 28

Solution of a Cauchy problem contd..

Important point for the existence and uniqueness of the Cauchy problem is that the datum curve γ is no where tangential to a characteristic curve. If γ is a characteristic curve, the data u0 (η) is to be restricted (i.e., the equations of p0 and q0 are also satisfied) and when this restriction is imposed, the solution of the Cauchy problem is non-unique infinity of solutions exist.

A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

13 / 28

Isotropic wave motion with constant velocity Consider an isotropic wave moving into a uniform medium with constant velocity c (like light wave)

A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

14 / 28

Isotropic wave motion with constant velocity contd.. 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)) ux uy c p 2 δt = p 2 δx + p 2 δy = n1 δx + n2 δy = δn 2 2 ux + uy ux + uy ux + u2y (22) c δn = normal displacement p 2 = =c 2 δt ux + uy

(23)

⇒ p2 + q 2 = 1; p = ux , q = uy

(24)

A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

15 / 28

Isotropic wave motion with constant velocity contd.. Problem. Find successive positions of the wavefront u(x, y) = ct when the initial position is αx + βy = 0, α2 + β 2 = 1, where u = 0

(25)

Cauchy problem F ≡ p2 + q 2 = 1 γ : x0 = βη, y0 = −αη cauchy data u0 = 0 Values of p0 and q0 p20 + q02 = 1, βp0 − αq0 = 0 ⇒ p0 = ±α, q0 = ±β A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

(26) 16 / 28

Isotropic wave motion with constant velocity contd.. Solution of Charpit’s equations dx dy = 2p, = 2q, dσ dσ du dp dq = 2(p2 + q 2 ) = 2, = 0, =0 dσ dσ dσ

(27)

x = ±2ασ + βη, y = ±2βσ − αη u = 2σ, p = ±α, q = ±β 1 ⇒ σ = ± (αx + βy) 2 ⇒ u = ±(αx + βy)

(28)

Solution are

Two solutions of 2 problems, but uniqueness theorem not violated. A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

17 / 28

Isotropic wave motion with constant velocity contd..

Wavefronts αx + βy = ±ct

(29)

+ sign for forward propagating wavefront − sign for backward propagating wavefront Normal distance at time t from the initial position = ±ct

A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

18 / 28

We have presented the theory of characteristics of first order PDEs briefly. It is based on the existence of characteristics curves in the (x, y)-plane. Along each of these characteristics we derive a number of compatibility conditions, which are transport equations and which are sufficient to carry all necessary information from the datum curve in the Cauchy problem into a domain in which solution is determined. In this sense every first order PDE is a hyperbolic equation.

A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

19 / 28

We have omitted a special class of solutions known as complete integral, for which any standard text may be consulted. Every solution of the PDE (1) can be obtained from a complete integral. We can also solve a Cauchy problem with its help. Complete integral plays an important role in Physics.

A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

20 / 28

So far we have discussed only a genuine solution, which is valid locally. We have seen that characteristic carry information about the solution. Characteristic curves are the only curve which can sustain discontinuities of certain types in the solution. For a linear equation the discontinuities can be in the solution and its derivatives, for a quasilinear equation the discontinuities can be in the first and higher order derivatives and for nonlinear equations the discontinuities can be in second and higher order derivatives. A Model Lession FD PDE Part 2

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Department of Mathematics

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It is possible to develop a theory of first order PDE starting from the definition of characteristic curves as the curves which carry above type of discontinuities. See P. Prasad, A theory of first order PDE through propagation of discontinuities, RMS Mathematics News letter, 2000, 10, 89-103. Absence of real characteristic curves of the equation ux + iuy = 0 in (x, y)-plane shows that its solution can not have discontinuities of any type along a curve in (x, y)-plane. This is related to the regularity of a solution of an elliptic PDE

A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

22 / 28

There is a fairly complete theory of weak solutions of Hamilton-Jacobi equations, a particular case of the nonlinear equation (1). Generally the domain of validity of a weak solution with Cauchy data on the x-axis is at least half of the (x, y)-plane. Theory of a single conservation law, a first order equation, is particularly interesting not only from the point of view of theory but also from the point of view of applications (Prasad, 2001).

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Department of Mathematics

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1 Consider the partial differential equation F ≡ u(p2 + q 2 ) − 1 = 0. (i) Show that the general solution of the Charpit’s equations is a four parameter family of strips represented by 3 3 2 2 x = x0 + u0 (2σ) 2 cos θ, y = y0 + u0 (2σ) 2 sin θ, 3 3 cos θ sin θ u = 2u0 σ, p = √ , q = √ 2σ 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.

A Model Lession FD PDE Part 2

P. Prasad

Department of Mathematics

24 / 28

2 Solve the following Cauchy problems: (i)

1 2 2 (p 2

+ q 2 ) = u with Cauchy data prescribed on the circle x + y 2 = 1 by u(cos θ, sin θ) = 1, 0 ≤ θ ≤ 2π

(ii) p2 + q 2 + p − 12 x the x-axis by



 q − 21 y − u = 0 with Cauchy data prescribed on u(x, 0) = 0

(iii) 2pq − u = 0 with Cauchy data prescribed on the y-axis by u(0, y) =

1 2 y 2

(iv) 2p2 x + qy − u = 0 with Cauchy data on x-axis 1 u(x, 1) = − x. 2

A Model Lession FD PDE Part 2

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Department of Mathematics

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1

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

2

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

3

F. John. Partial Differential Equations. Springer-Verlag, New York, 1982.

4

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

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Department of Mathematics

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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 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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Department of Mathematics

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Thank You!

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Department of Mathematics

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