Math, asked by nageshp767, 9 months ago

a
Reduce the equation Uxx - x´uyy = 0 to a
canonical form​

Answers

Answered by PhilipTheGreat
0

MA 201: Second Order Linear PDE

Canonical Transformation

Lecture - 6

MA201(2016):PDE

• A second order PDE with two independent variables x and y is

given by

F(x, y, u, ux , uy , uxy , uxx , uyy ) = 0. (1)

What is the linear form?

• The unknown function u(x, y) satisfies an equation:

Auxx + Buxy + Cuyy + Dux + Euy + Fu + G = 0. (2)

Facts:

• The expression Lu ≡ Auxx + Buxy + Cuyy is called the Principal part

of the equation.

• Classification of such PDEs is based on this principal part.

MA201(2016):PDE

• At a point (x, y), the above equation is said to be

Hyperbolic if B

2

(x, y) − 4A(x, y)C(x, y) > 0

Parabolic if B

2

(x, y) − 4A(x, y)C(x, y) = 0

Elliptic if B

2

(x, y) − 4A(x, y)C(x, y) < 0

• Each category relates to specific problems

1. Laplace’s Equation

uxx + uyy = 0.

2. Wave Equation

utt − uxx = 0.

3. Heat Equation

ut = uxx.

MA201(2016):PDE

Methods and Techniques for Solving PDEs

• Change of coordinates: A PDE can be changed to an ODE or to an

easier PDE by changing the coordinates of the problem.

• Separation of variables: A PDE in n independent variables is reduced

to n ODEs.

• Integral transforms: A PDE in n independent variables is reduced to

one in (n − 1) independent variables. Hence, a PDE in two variables

could be changed to an ODE.

• Numerical Methods

MA201(2016):PDE

Change of coordinates: Canonical Transformations

• This can be achieved by introducing new coordinates

ξ = ξ(x, y), η = η(x, y).

• Compute the partial derivatives

MA201(2016):PDE

• Substitute these values into the original equation to obtain a new

form

where the new coefficients are as follows

Whether the form of PDE remains invariant even after

coordinate transformation?

MA201(2016):PDE

• It can be observed that

2A˜ B˜

B˜ 2C˜

=

ξx ξy

ηx ηy

2A B

B 2C

ξx ξy

ηx ηy

t

(3)

• Taking the determinant on both sides gives

B˜ 2 − 4A˜C˜ = (ξxηy − ξy ηx )

2

(B

2 − 4AC) = J

2

(B

2 − 4AC) (4)

• J is the Jacobian of the transformation and we select the

transformation (ξ, η) such that J 6= 0.

• Transformation ξ = ξ(x, y) and η = η(x, y) are called canonical

transformation or characteristics and the reduced form of the PDE is

called canonical form.

MA201(2016):PDE

Canonical Transformations: Hyperbolic PDE

• Hyperbolic PDE’s: B

2 − 4AC > 0

Natural Choice: A˜ = 0 and C˜ = 0

Leads to Algebraic Equations:

2 + Bλ + C = 0, λ1 =

ξx

ξy

, λ2 =

ηx

ηy

.

Observations:

ξ(x, y) = c For ξ solve dy

dx = −λ1(x, y)

⇒ ξx + ξy

dy

dx = 0,

• For η solve dy

dx = −λ2(x, y).

• Canonical form for hyperbolic equation may be read as

uξη = φ(ξ, η, u, uξ, uη).

MA201(2016):PDE

Hyperbolic Equation with Constant Coefficients

• Assume that either A 6= 0 or C 6= 0.

• What can you say when A and C vanish simultaneously?

• For A 6= 0, canonical transformation (ξ, η) takes the form

ξ = y + λ1x, η = y + λ2x, λ1,2 =

−B ±

B2 − 4AC

2A

.

For A 6= 0 and C = 0, we have

ξ = y, η = y −

B

A

x.

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