4/3/2013 1
Introduction to Partial
Differential Equations
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Partial Differential Equations - Numerical Methods - Lecture Slides, Slides of Mathematical Methods for Numerical Analysis and Optimization
The main points are: Parabolic Partial Differential Equations, Independent Variables, One Dependent Variable, Heat-Conduction Equation, Physical Example, Internal Temperature, Schematic Diagram, Interior Nodes, Explicit Method
Typology: Slides
2012/2013
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4/3/2013 1
Introduction to Partial
Differential Equations
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What is a Partial Differential Equation
- Ordinary Differential Equations have only one independent variable
- Partial Differential Equations have more than one independent variable
subject to certain conditions: where u is the dependent variable, and x and y are the independent variables.
3 5 3 , ( 0 ) 5
2
- = =
− y e y dx
dy (^) x
2 2 2
2
2
2 3 x y y
u
x
u = + ∂
∂
∂
∂
Example of an Ordinary Differential
Equation
- Assumption: Ball is a lumped system.
- Number of Independent variables: One (t)
Hot Water
Spherical Ball
dt
d hA (^) a mC
θ θ −θ =
Classification of 2
nd
Order Linear
PDE’s
where are
functions of ,and is
a function of
2
2 2
2
2
- = ∂
D
y
u C x y
u B x
u A
A , B , and C
x and y D
, , and ,.
u u x y u x y
Classification of 2
nd
Order Linear
PDE’s
can be:
- Elliptic
- Parabolic
- Hyperbolic
2
2 2
2
2
- = ∂
D
y
u C x y
u B x
u A
Classification of 2
nd
Order Linear
PDE’s: Elliptic
Example:
where, giving
therefore the equation is elliptic.
2
2 2
2
2
- = ∂
D
y
u C x y
u B x
u A
2 0
2
2
2
∂
∂
∂
∂
y
T
x
T
A = 1 , B = 0 , C = 1
4 0 4 ( 1 )( 1 ) 4 0
2 B − AC = − = − <
Classification of 2
nd
Order Linear
PDE’s: Parabolic
If ,then the
equation is parabolic.
2
2 2
2
2
- = ∂
D
y
u C x y
u B x
u A
2 B − 4 AC = 0
Classification of 2
nd
Order Linear
PDE’s: Hyperbolic
If ,then the
equation is hyperbolic.
2
2 2
2
2
- = ∂
D
y
u C x y
u B x
u A
2 B − 4 AC > 0
Classification of 2
nd
Order Linear
PDE’s: Hyperbolic
Example:
where, giving
therefore the equation is hyperbolic.
2
2 2
2
2
- = ∂
D
y
u C x y
u B x
u A
2
2 2 2
(^2 )
t
y x c
y ∂
∂
∂
∂
2
1 1 , 0 , c
A = B = C = −
)
1 4 0 4 ( 1 )( 2
2
c
B AC
− − = −^0
4 = 2 > c