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Partial Differential Equations - Background

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CIS888.11V/EE894R/ME894V A Case Study in ... (Non-homogeneous) (Homogeneous) (Homogenous) ... (Non-homogeneous) (Homogeneous) (5) Types of Coefficients ... – PowerPoint PPT presentation

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Title: Partial Differential Equations - Background


1
Partial Differential Equations - Background
  • Physical problems are governed by many PDEs
  • Some are governed by first order PDEs
  • Numerous problems are governed by second order
    PDEs
  • A few problems are governed by fourth-order PDEs.

2
Examples
3
Examples (contd.)
4
Classification of Partial Differential Equations
(PDEs)
  • There are 6 basic classifications
  • (1) Order of PDE
  • (2) Number of independent variables
  • (3) Linearity
  • (4) Homogeneity
  • (5) Types of coefficients
  • (6) Canonical forms for 2nd order PDEs

5
(1) Order of PDEs
  • The order of a PDE is the order of the highest
    partial derivative in the equation.
  • Examples
  • (2nd order)
  • (1st order)
  • (3rd order)

6
(2) Number of Independent Variables
  • Examples
  • (2 variables x and t)
  • (3 variables r, q, and t)

(3) Linearity
PDEs can be linear or non-linear. A PDE is
linear if the dependent variable and all its
derivatives appear in a linear fashion (i.e. they
are not multiplied together or squared for
example.
7
  • Examples (Linear)
  • (Non-linear)
  • (Linear)
  • (Non-linear)
  • (Non-linear)
  • (Linear)
  • (Non-linear)

8
(4) Homogeneity
  • A PDE is called homogenous if after writing the
    terms in order, the right hand side is zero.
  • Examples
  • (Non-homogeneous)
  • (Homogeneous)
  • (Homogenous)

9
Examples
  • (Non-homogeneous)
  • (Homogeneous)

(5) Types of Coefficients
If the coefficients in front of each term
involving the dependent variable and its
derivatives are independent of the variables
(dependent or independent), then that PDE is one
with constant coefficients.
10
Examples
  • (Variable coefficients)
  • (C constant constant coefficients)

(6) Canonical forms for 2nd order PDEs (Linear)
(Standard Form)
where A, B, C, D, E, F, and G are either real
constants or real-valued functions of x and/or y.
11
  • PDE is Elliptic

PDE is Parabolic
PDE is Hyperbolic
Parabolic PDE ? solution propagates or
diffuses Hyperbolic PDE ? solution propagates as
a wave Elliptic PDE ? equilibrium
12
  • This terminology of elliptic, parabolic, and
    hyperbolic, reflect the analogy between the
    standard form for the linear, 2nd order PDE and
    conic sections encountered in analytical
    geometry
  • for which when one obtains the equation for an
    ellipse, when one obtains the equation for a
    parabola, and when
  • one gets the equation for a hyperbola.

13
Examples
  • (a) Here, A1, B0, C2, DEFG0 ? B2-4AC
    0 - 4(1)(2) -8 lt 0 ? this equation is elliptic.
  • (b)
  • Here, A1, B0, C-2, DEFG0 ? B2-4AC 0 -
    4(1)(-2) 8 gt 0 ? this equation is hyperbolic.
  • (c)
  • Here, A1, B0, E-2, CDFG0 ? B2-4AC 0 -
    4(1)(0) 0 ? this equation is parabolic.

14
Examples
  • (d) Here, A1, B-4, C1, DEFG0 ? B2-4AC
    16 - 4(1)(1) 12 gt 0 ? this equation is
    hyperbolic.
  • (e)
  • Here, A3, B-4, C-5, DEFG0 ? B2-4AC 16 -
    4(3)(-5) 76 gt 0 ? this equation is hyperbolic.
  • (f) Here, A3, B-4, C-5, D8, E-9,
    F6, G27exy ? B2-4AC 16 - 4(3)(-5) gt 0 ? this
    equation is hyperbolic.

15
Examples
  • (g)
  • Here, Ay, B0, C-1, DEFG0 ? B2-4AC 0 -
    4(y)(-1) 4y ? for ygt0, this equation is
    hyperbolic for y0, this equation is parabolic
    for ylt0, this equation is elliptic.

16
Examples
  • (h)
  • Here, A1, B2x, C1-y2, DEFG0 ? B2-4AC 4x2
    - 4(1)(1-y2) 4x24y2-4 or x2y2 gt,,lt 0

17
Examples
  • (i)
  • Here, A1, B-y, C0, DEFG0 ? B2-4AC y2 ?
    for y0, this equation is parabolic for y?0,
    this equation is hyperbolic.

18
Example
  • (j)
  • Here, Asin2x, Bsin2x, Ccos2x, DEFG0 ?
    B2-4AC sin22x-4sin2xcos2x 4sin2xcos2x-4sin2xco
    s2x 0 ? this equation is parabolic everywhere.

19
Example
  • (k)
  • This must be converted to 2nd order form first
  • and
  • subtracting,
  • Now, A1, B0, C-1, DEFG0 ? B2-4AC4 gt 0
  • ? Hyperbolic.

20
Example
  • (l)
  • Again, convert to 2nd order form first
  • and
  • adding,
  • Again, A1, B0, C-1, DEFG0 ? B2-4AC 4 gt 0
    ? Hyperbolic.

21
Wave equation (hyperbolic)
  • The wave equation has the form
  • This equation can be factored as follows
  • This implies that xt and x-t define
    characteristic directions, i.e. directions along
    which the PDE will collapse into an ODE.

22
Wave equation (contd.)
  • Let xxt and hx-t
  • ?
  • Similarly,
  • Thus, and
  • ??

23
Wave equation (contd.)
  • Thus, and
  • ? and
  • or, by integrating again,

24
  • Note that there are other important equations in
    mathematical physics, such as
  • Schroedinger eq. 1-D
  • which is a wave equation by virtue of the
    imaginary constant i?. Note that but for the i
    ( ), this equation would be parabolic.
    However, the i makes all the difference and
    this is a wave equation (hyperbolic).

25
  • We now return to the case study problem for
    adiabatic ( ), frictionless, quasi-1D flow
  • For simplicity, we have assumed that ne0
    (non-ionized flow) for now.

26
  • Recall that for steady flow, we found that M 1
    when dA/dx 0 (choking or sonic condition),
  • where
  • It turns out that this system of equations
    exhibits elliptic character for Mlt1, and
    hyperbolic character for Mgt1.
  • Thus, M1, the choking point, exists to delineate
    the elliptic and hyperbolic regions of the flow.

27
  • If we are interested in transient, i.e.
    time-dependent solutions, analytical solutions do
    not exist (except for drastic simplifications)
    and the governing equations must be solved
    numerically.
  • If we are interested in obtaining steady state
    solutions, they can be obtained numerically as
    well (in quasi-1D flow, analytical solutions were
    obtained), in one of two ways
  • Direct solution of steady state equations, or
  • Time-marching from an arbitrary initial state to
    the steady state.

28
Conservation equations for quasi-1D, isothermal
flow
  • At steady state,
  • An analytical solution can be obtained for this
    case.

29
Conservation equations for quasi-1D, isothermal
flow (contd.)
  • A steady state solution can also be obtained by
    time-marching, i.e. solving the unsteady
    equations
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