Partial Differential Equations - Background
description
Transcript of Partial Differential Equations - Background
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
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.
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Examples
2
2
x
TC
t
T:)D1(EquationConductionHeat
2
2
2
2
y
T
x
TC
t
T:)D2(EquationConductionHeat
0yx
:EquationLaplaceD22
2
2
22
)y,x(fyx
:EquationPoissonD22
2
2
22
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Examples (contd.)
2
2
2
2
x
uC
t
u:EquationWaveD1
2
2
2
2
2
2
y
u
x
uC
t
u
:)membranevibrating(EquationWaveD2
)plateVibrating(02
t
u2
h4
y
u4
2y
2x
u4
24
x
u4
D
)beamVibrating(04
x
u4
C2
t
u2
:EquationsthOrder4
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
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
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
(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)
2
2
x
utu
xu
tu
xsinx
uu
tu
3
3
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
(2) Number of Independent Variables
Examples:
(2 variables: x and t)
(3 variables: r, , and t)
2
2
x
utu
2
2
22
2 u
r
1ru
r1
r
utu
(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.
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Examples: (Linear)
(Non-linear)
(Linear)
(Non-linear)
(Non-linear)
(Linear)
(Non-linear)
tsinx
ue
t
u2
2t
2
2
0y
uy
x
u2
2
2
2
0tu
x
uu
2
2
0uyu
yxu
x 2
y2
2
2
eusinxu
x
u
xsiny
uyx
u2
x
u2
22
2
2
1yu
uxu 2
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
(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)
)y,x(fy
u
x
u2
2
2
2
0tu
x
u2
2
ux
u
t
u2
2
2
2
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Examples
(Non-homogeneous)
(Homogeneous)
5utu
xu
5ut
)5u(x
)5u(
(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.
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Examples
(Variable coefficients)
(C constant; constant coefficients)
0y
ux
x
u2
22
2
2
0t
uC
x
u2
2
2
2
(6) Canonical forms for 2nd order PDEs (Linear)
(Standard Form)GFuyu
Exu
Dy
uC
yxu
Bx
uA
2
22
2
2
where A, B, C, D, E, F, and G are either real constants or real-valued functions of x and/or y.
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
PDE is Elliptic 0AC4B2
PDE is Hyperbolic 0AC4B2
PDE is Parabolic 0AC4B2
Parabolic PDE solution “propagates” or diffuses
Hyperbolic PDE solution propagates as a wave
Elliptic PDE equilibrium
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
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.
0FEyDxCyBxyAx 22 0AC4B2
0AC4B2
0AC4B2
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Examples
(a) Here, A=1, B=0, C=2, D=E=F=G=0 B2-4AC = 0 - 4(1)(2) = -8 < 0 this equation is elliptic.
(b)
Here, A=1, B=0, C=-2, D=E=F=G=0 B2-4AC = 0 - 4(1)(-2) = 8 > 0 this equation is hyperbolic.
(c)
Here, A=1, B=0, E=-2, C=D=F=G=0 B2-4AC = 0 - 4(1)(0) = 0 this equation is parabolic.
0y
u2
x
u2
2
2
2
0y
u2
x
u2
2
2
2
0yu
2x
u2
2
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Examples
(d) Here, A=1, B=-4, C=1, D=E=F=G=0 B2-4AC = 16 - 4(1)(1) = 12 > 0 this equation is hyperbolic.
(e)
Here, A=3, B=-4, C=-5, D=E=F=G=0 B2-4AC = 16 - 4(3)(-5) = 76 > 0 this equation is hyperbolic.
(f) Here, A=3, B=-4, C=-5, D=8, E=-9, F=6, G=27exy B2-4AC = 16 - 4(3)(-5) > 0 this equation is hyperbolic.
0y
uyx
u4
x
u2
22
2
2
xy2
22
2
2
e27u6yu
9xu
8y
u5
yxu
4x
u3
0y
u5
yxu
4x
u3
2
22
2
2
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Examples
(g)
Here, A=y, B=0, C=-1, D=E=F=G=0 B2-4AC = 0 - 4(y)(-1) = 4y for y>0, this equation is hyperbolic; for y=0, this equation is parabolic; for y<0, this equation is elliptic.
0y
u
x
uy
2
2
2
2
Hyperbolic
x
y
Elliptic Parabolic
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Examples
(h)
Here, A=1, B=2x, C=1-y2, D=E=F=G=0 B2-4AC = 4x2 - 4(1)(1-y2) = 4x2+4y2-4 or x2+y2 >,=,< 0
0y
u)y1(
yxu
x2x
u2
22
2
2
2
Elliptic
Hyperbolic
x
y
Parabolic onsurface of circle
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Examples
(i)
Here, A=1, B=-y, C=0, D=E=F=G=0 B2-4AC = y2 for y=0, this equation is parabolic; for y0, this equation is hyperbolic.
0uyu
yxu
xyx
uy
x
u 2
2
2
Hyperbolic
Hyperbolic
x
y
Parabolic
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Example
(j)
Here, A=sin2x, B=sin2x, C=cos2x, D=E=F=G=0 B2-4AC = sin22x-4sin2xcos2x = 4sin2xcos2x-4sin2xcos2x = 0 this equation is parabolic everywhere.
xy
u)x(cos
yxu
)x2(sinx
u)x(sin
2
22
2
2
22
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Example
(k)
This must be converted to 2nd order form first:
and
subtracting,
Now, A=1, B=0, C=-1, D=E=F=G=0 B2-4AC=4 > 0
Hyperbolic.
0xu
tu
0xtu
t
u 2
2
2
0
x
utx
u2
22
0x
u
t
u2
2
2
2
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Example
(l)
Again, convert to 2nd order form first:
and
adding,
Again, A=1, B=0, C=-1, D=E=F=G=0 B2-4AC = 4 > 0 Hyperbolic.
0xu
tu
0xtu
t
u 2
2
2
0
x
utx
u2
22
0x
u
t
u2
2
2
2
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Wave equation (hyperbolic)
The wave equation has the form:
This equation can be factored as follows:
This implies that x+t and x-t define characteristic directions, i.e. directions along which the PDE will collapse into an ODE.
0t
u
x
u2
2
2
2
utxtxt
uxu
txt
u
x
u2
2
2
2
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Wave equation (contd.)
Let =x+t and =x-t
Similarly,
Thus, and
xxx
ttt
2tx
2tx
0u
4u22t
u
x
u 2
2
2
2
2
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Wave equation (contd.)
Thus, and
and
or, by integrating again,
0u
0u
)(*fu
)(*g
u
direction0xtheintraveling.e.i,wavetraveling
rightwardrepresents
direction0xtheintraveling.e.i,wavetraveling
leftwardrepresents
)tx(g)tx(f)(g)(fu
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
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).
)x(Vxm2t
i2
22
1
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
We now return to the case study problem for adiabatic ( ), frictionless, quasi-1D flow:
For simplicity, we have assumed that ne=0 (non-ionized flow) for now.
0Q
0uAP2u
)1(RT
xA1
2u
)1(RT
t
xP
x)Au(
A1
t)u(
0x
)uA(A1
t
22
2
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
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 M<1, and hyperbolic character for M>1.•Thus, M=1, the choking point, exists to delineate the elliptic and hyperbolic regions of the flow.
RTu
M
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
• 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.
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
Conservation equations for quasi-1D, isothermal flow
At steady state,
• An analytical solution can be obtained for this case.
RTP
ttanconsT
0xu
uxP
ttanconsmuA
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CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering
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:
RTP
ttanconsTxP
x)Au(
A1
t)u(
0x
)uA(A1
t2