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Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic Equations David J. Willis March 11, 2015 David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic Equations March 11, 2015 1 / 20

Video 15: PDE Classi cation: Elliptic, Parabolic and ...faculty.uml.edu/dwillis/MECH5200/PDFs/2015_22520... · References and Acknowledgements The following materials were used in

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Video 15: PDE Classification: Elliptic, Parabolic andHyperbolic Equations

David J. Willis

March 11, 2015

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 1 / 20

Table of contents

1 PDE ClassificationEquation ClassificationPDE Classification

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 2 / 20

References and Acknowledgements

The following materials were used in the preparation of this lecture:

1 Handbook of Grid Generation, Thompson, Soni and Weatherhill.

2 MIT Open Courseware 16.920 Course

The author of these slides wishes to thank these sources for making thecurrent lecture.

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 3 / 20

Table of contents

1 PDE ClassificationEquation ClassificationPDE Classification

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 4 / 20

PDE Classification Equation Classification

Equation Classification

Consider the following equation:

Ax2 + Bxy + Cy 2 = f

or

A

(x

y

)2

+ B

(x

y

)+ C =

f

y 2

Az2 + Bz + C = f ′

What does this equation represent?

Let’s plot it for several different A,B and C values.

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 5 / 20

PDE Classification Equation Classification

Equation Classification

A = 2, B = 1, C = 2→ B2 − 4AC < 0

Paraboloid Surface – Forms and Ellipse

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 6 / 20

PDE Classification Equation Classification

Equation Classification

A = 1, B = 2, C = 1→ B2 − 4AC = 0

Pure Parabolic Surface

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 7 / 20

PDE Classification Equation Classification

Equation Classification

A = 1, B = 2, C = 1→ B2 − 4AC > 0

Hyperbolic Surface

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 8 / 20

PDE Classification Equation Classification

Summary

The value of the discriminant: B2 − 4AC defines the equation

Contour shapes and the intersection of the shape with the x − yplane is:

B2 − 4AC < 0 : Circles and Ellipse (parabaloid)B2 − 4AC = 0 : Parallel lines (pure parabola)B2 − 4AC < 0 : X-lines (hyperbola)

a) Elliptic b) Parabolic c) Hyperbolic

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 9 / 20

PDE Classification Equation Classification

Table of contents

1 PDE ClassificationEquation ClassificationPDE Classification

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 10 / 20

PDE Classification PDE Classification

PDE Classification

Consider a general second order PDE:

A∂2u

∂x2+ B

∂2u

∂xy+ C

∂2u

∂y 2+ D

∂u

∂x+ E

∂u

∂y+ F = 0 (1)

This is identical in form to the equation for a conic section ( a generalform of the equation we considered earlier):

Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0 (2)

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 11 / 20

PDE Classification PDE Classification

PDE Classification

Aside: We can show that the characteristic equation:

A∂2u

∂x2+ B

∂2u

∂xy+ C

∂2u

∂y 2= 0 (3)

Analysed using a Fourier series expansion:

u =1

4π2

∑j

∑k

ujk expi(σx )j x expi(σy )k y (4)

Yields:

A

(σx

σy

)2

+ B

(σx

σy

)+ C = 0 (5)

Let’s have a look at how this equation applies to PDEs.

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 12 / 20

PDE Classification PDE Classification

PDE Classification: Elliptic PDE

Start with Laplace’s or Poisson’s Equation

∂2φ

∂x2+∂2φ

∂y 2= 1 · ∂

∂x2+ 1 · ∂

∂y 2= 0

So, A = C = 1 and B = 0, so B2 − 4AC = −4 < 0

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 13 / 20

PDE Classification PDE Classification

PDE Classification: Example

Let’s say you solve he membrane problem from last unit with a singlepoint force applied transverse to the membrane:

The point load is distributed throughout the membrane.

The point has base influence throughout the membrane.

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 14 / 20

PDE Classification PDE Classification

PDE Classification: Example

Let’s add another point load:

The point load has influence throughout the membrane.

Deflection at any point is dependent on the forcing conditions at allpoints on the membrane.

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 15 / 20

PDE Classification PDE Classification

PDE Classification: Parabolic PDE

Next, let’s consider the heat equation:

∂T

∂t− ∂2T

∂x2= f (6)

A = 1,C = 0,B = 0. Therefore, B2 − 4AC = 0

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 16 / 20

PDE Classification PDE Classification

PDE Classification: Example

Consider the unsteady heat equation

The temperature of the bar is dependent on the temperature in thetime leading up to the current time.

The temperature in the future will depend on the temperature now.

Ie. If we turn up the heat now, it will only affect the temperature attimes after the current time.

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 17 / 20

PDE Classification PDE Classification

PDE Classification: Hyperbolic PDE

Next, let’s consider the second order wave equation:

∂2u

∂t2− ∂2u

∂x2= f (7)

A = 1,B = 0,C = −1. Therefore, B2 − 4AC > 0

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 18 / 20

PDE Classification PDE Classification

PDE Classification: Example

The wave equation propagates information to well defined zones ofinfluence.

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 19 / 20

PDE Classification PDE Classification

What have we learned?

A second order (partial differential) equation can be classified into 3types:

Elliptic: Laplace and other similar equations. Elliptic equations havezones of influence and dependence that include the whole domain.Parabolic: Unsteady heat, parabolized Navier-Stokes, boundary layers,and other similar equations. Parabolic equations have zones ofinfluence that includes the downstream domain. The zone ofdependence is the upstream domain.Hyperbolic: Wave equation, supersonic Navier-Stokes, and othersimilar equations. Hyperbolic equations have zones of influence anddependence that converge to the point along characteristic lines(upstream & downstream).

David J. Willis Video 15: PDE Classification: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 20 / 20