PY4066 Partial Differential Equations · PY4066 Partial Differential Equations II Recommended texts...

PY4066 Partial Differential Equations IIRecommended texts

• Partial differential equations for scientists and engineersS.J. Farlow, Dover (1993) S-LEN 515.353 M23

• Partial differential equations for scientists and engineersG. Stephenson, Imperial College Press (1996) 515.353 N8

• Elements of soliton theoryG.L. Lamb, Wiley (1980)515.353 M05

Course Outline

• 1 Computer modelling of fluids– Fluid types, transport, conservation laws, wave equations

• 2 Nonlinear wave equations– Solitary waves and solitons, KdV equation, NLS equation

• 3 Solution of PDE’s using integral transforms– Fourier, sine, cosine and Laplace transforms

• 4 Solution of PDE’s using variational principles– Euler-Lagrange equations, method of Ritz, finite element method

1 Computer Modelling of Fluids1.1 Introduction

• Fluid types– Quantum fluids, classical fluids

• Modes of transport– Densities (number, charge, energy, …)– Currents (flows of mass, charge, energy)– Continuity equation relates current and density– Advection, convection, diffusion

• Conservation laws– Mass, momentum, energy, …

• Wave equations

•Fick’s law (diffusive transport)t),( D- t),( rrj ρ∇=

1.2 Diffusive, Convective and Advective Transport

•Diffusion equation

t),f( t),( D- t

t),( 2 rrr =∇∂

∂ ρρ

•Substitute Fick’s law into continuity equation to obtain diffusion equation

•Continuity equation with sources

t),f( t

t),( t),(. rrrj =∂

∂+∇ ρ

Diffusive, Convective and Advective Transport

• Advection (advective transport)

–Molecular motion irrelevantt),v(t),( t),( rrrj ρ=

• Advection-diffusion equation

t),f( t)],t)v(,([ . t),( D- t

t),( 2 rrrrr =∇+∇∂

∂ ρρρ

• Add divergence of advective flux term (when both types of transport are present) to obtain advection-diffusion

A solute (e.g. pollutant) added to a flowing river is subject to advective (flow with river) and diffusive transport (diffusion in water).

Diffusive, Convective and Advective Transport

• Convection occurs when an element of fluid is subjected to an external force owing to an intrinsic property of the fluid element

• For example, if a fluid is heated from below in a gravitational field, a layer of cooler and denser fluid overlies a warmer and less dense fluid. This instability relaxes either bydiffusion of heat (without any transport of the fluid) or by convection, in which case heat is transported along with the fluid

• The Lorenz equations* are a good example

of convection in a simple system *To go to this webpage, click on the link when in ‘show’ mode (F5 k )

1.3 Wave motion in channels

•We will derive coupled, partial differential equations representing conservation of mass and conservation of momentum for water in a channel of width, b.

•The purpose of this is to see where pde’s come from

•Approximate forms for the coupled, nonlinear pde’s which represent the conservation laws can be chosen which lead to wave equations.

Wave motion in channelsConservation laws

•Momentum conservation (Newton’s laws)

•Mass conservation (continuity equation)

•Energy conservation (equations of state)

•These generally result in sets of coupled, nonlinear pde’s

•Under particular circumstances these may result in wave motion, turbulence, etc.

Wave motion in channelsConservation of mass

• Continuity equation is applied to water in the channel with width, b, height, h(x,t), thickness, ∆x, moving parallel to x direction at velocity u(x,t). Water density is ρ

Wave motion in channelsConservation of mass

• Mass of slice = volume . density = b ∆x h ρSubstitute areal density into continuity equation

( ) ( )

( )

equation mass of onConservati 0 xuh

xhu

th

0 t)u(x, t)h(x, x

t)h(x,t

0 t)u(x, ρ t)h(x,∆x bx

ρ t)h(x,∆x bt

=∂∂

+∂∂

+∂∂

=∂∂

+∂∂

=∂∂

+∂∂

Wave motion in channelsConservation of momentum

• Consider a slice where the height of the slice varies within the slice. There is a pressure difference across the slice which tends to accelerate the slice laterally• Compute the pressure difference across the slice

• Recognize that this corresponds to a force, F(x,t)

• Substitute this force into Newton’s 3rd Law (conservation of momentum)

t)F(x, momentum)(dtd =

t)h(x,

x1x 2x

Wave motion in channelsConservation of momentum

dx t)u(x, t)h(x, b ρ

(t)x

(t)x

2

1

∫

• Instantaneous momentum of the slice

x2(t) – x1(t) = ∆x

• Differentiate wrt time using Leibniz’ rule

t),u(x t),(t)h(xx - t),u(x t),h(x (t)x

dx )t)u(x, t)(h(x,t

dx t)u(x, t)h(x, dtd

111

.

222

.

(t)x

(t)x

(t)x

(t)x

2

1

2

1

+∂∂

= ∫∫

Wave motion in channelsConservation of momentum

2

2

x)O( x

t),h(x ) x- (x t),h(x t),h(x

x)O( x

t),u(x ) x- (x t),u(x t),u(x (t)x

t),u(x (t)x

11212

112122

.

11

.

∆+∂

∂+=

∆+∂

∂+==

=

steps) of detailsfor sheet Tutorial (see

x)O( x xuu

tuh b dx t)u(x, t)h(x, b

dtd

for the find werelations thesengsubstituti

2

(t)x

(t)x

2

1

∆+∆

∂∂

+∂∂

=∫ ρρ

momentum of change of rate

Wave motion in channelsConservation of momentum

•Water pressure at depth, h – z, measured from the water surface is ρg (h – z)

•Pressure force averaged over a slice at (x1, t) is

•Force due to gradient of water height

( ) b dzz- t),h(xρg t),F(x

t),h(x

011

1

∫=

t)h(x,

x1x 2x

Wave motion in channelsConservation of momentum

• Difference in pressure force across a slice, of width ∆x, by Taylor series expansion is

( )

2

2t),h(x

0

1

2t),h(x

0112

x)O(x bxhh ρg

x)O( ∆x b dzx

t),h(xρg

x)O( ∆x b dzz- t),h(xdxdρg t),F(x - t),F(x

1

1

∆∆∂∂

=

∆+

∂

∂=

∆+

=

∫

∫

Wave motion in channelsConservation of momentum

• Net force on the slice in the positive x direction is

2x)O( ∆x xhgρbh - ∆+∂∂

• Rate of change of momentum of the slice

2x)O( ∆x xuu

tuρbh ∆+

∂∂

+∂∂

• Equate these and divide by common factors

0 xhg

xuu

tu =

∂∂

+∂∂

+∂∂

Wave motion in channelsResulting equations

Mass ofon Conservati 0 xuh

xhu

th

Momentum ofon Conservati 0 xhg

xuu

tu

=∂∂

+∂∂

+∂∂

=∂∂

+∂∂

+∂∂

A pair of coupled, nonlinear partial differential equations

Wave motion in channelsDerivation of wave equations

• Let the water height h be where h is the mean depthand h is the local wave height

• In the limit:– water is deep compared to the wave height – these equations reduce to the usual, linear wave equation

• In the limits: and– wavelength is long compared to water depth ( )– a nonlinear wave equation with solitary wave solutions

can be derived (see Lamb pp 169 ff for details)

η+=hh

h<<η

h>>λ h<<ηh>>λ

Wave motion in channelsDerivation of wave equations

• If we can neglect and in conservation equations

h<<ηxuu∂∂

xu∂∂η

• For these becomeh<<η

Mass of onConservati

Momentum of onConservati

0 xuh

t

0 x

g tu

=∂∂

+∂∂

=∂∂

+∂∂

η

η

Wave motion in channelsDerivation of wave equations

( )0

th

1x

0 tx

uh t

0 xuh

tt

0 x

g tx

u 0 x

g tu

x

2

2

22

2

2

2

2

2

22

=∂∂

−∂∂

=∂∂

∂+

∂∂

⇒=

∂∂

+∂∂

∂∂

=∂∂

+∂∂

∂⇒=

∂∂

+∂∂

∂∂

ηη

ηη

ηη

g

• Solutions are waves with velocity hg

1.4 Solution of coupled nonlinear pde’s

• Apply leapfrog method for solving pde’s in more than one variable

• Apply forward and/or central finite difference schemes

Solution of coupled nonlinear pde’sLeapfrog methods for coupled equations

• Solve for u and h in alternate timesteps

Solution of coupled nonlinear pde’sCentral difference leapfrog scheme

xg

tu

∂∂

+∂∂ η

onConservati Momentum

0 x2

hhg

t2

uu i1-j i

1j1i

j 1ij =

∆+

∆

−+−−+

( ) hhxtg - uu i

1-j i1j 1i

j 1ij −+

−+∆∆

=

Solution of coupled nonlinear pde’sCentral difference leapfrog scheme

xuh

t ∂∂

+∂∂η

onConservati Mass

0 x2

uuh

t2

hh i1-j i

1j1i

j 1ij _

=∆

+∆

−+−−+

( ) uuxth - hh i

1-j i1j 1i

j 1ij

_−+

−+∆∆

=

Solution of coupled nonlinear pde’sForward/Central difference leapfrog scheme

xg

tu

∂∂

+∂∂ η

onConservati Momentum

0 x2

hhg

t

uu 1i1-j 1i

1jij 1i

j =∆

+∆

+−++−+

( ) hhx2tg - uu 1i

1-j 1i1j i

j 1ij

+−++

+∆∆

=

Solution of coupled nonlinear pde’sForward/Central difference leapfrog scheme

xuh

t ∂∂

+∂∂η

onConservati Mass

0 x2

uuh

t

hh i1-j i

1jij 1i

j _=

∆+

∆

−+−+

( )i1-j i

1j ij 1i

j uux2th - hh

_−+

+∆∆

=

Solution of coupled nonlinear pde’sForward/central difference leapfrog scheme

Momentum ofon Conservati 0 xhg

xuu

tu =

∂∂

+∂∂

+∂∂

0 x2

hhg

x2

uuu

t

uu 1i1-j 1i

1ji

1-j i1ji

j

ij 1i

j =∆

+∆

+∆

+−++−+−+

( ) ( )[ ] hhg uuux2t - uu 1i

1-j 1i1j

i1-j i

1jij i

j 1ij

+−++−+

+ +∆∆

=

Solution of coupled nonlinear pde’sForward/central difference leapfrog scheme

Mass ofon Conservati 0 xuh

xhu

th

=∂∂

+∂∂

+∂∂

0 x2

uuh

x2

hhu

t

hh i1-j i

1jij

i1-j i

1jij

ij 1i

j =∆

+∆

+∆

−+−+−+

( ) ( )[ ]i1-j i

1jij

i1-j i

1jij i

j 1ij uuhhhu

x2t - hh −+−+

+ +∆∆

=