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4/10/2020 http://numericalmethods.eng.usf.edu
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What is a Partial Differential
Equation ?

Ordinary Differential Equations have only one independent
 variable
3 dy

5 y
2 
3 e
 x
, y ( 0 )

5 dx
Partial Differential Equations have more than one independent variable
3
 2
 x u



2 u
 x
2  y
2
2 y
2 subject to certain conditions: where u is the dependent variable, and
x and y are the independent variables.

Example of an Ordinary
Differential Equation
Spherical
Ball
Hot Water

 hA
    a

 mC d
 dt
Assumption: Ball is a lumped system.
Number of Independent variables: One (t)

Example of an Partial
Differential Equation
Spherical
Ball
Hot Water k r
2

 r r
2

T
 r
 r
2 k sin


  sin


T
 
 k r
2 sin
2

 2
T
 
2
 
C

T
 t
, t

0 , T ( r ,

,

, 0 )

T a


Assumption: Ball is not a lumped system.
Number of Independent variables: Four (r, θ , φ ,t)

Classification of 2 nd Order
Linear PDE’s
A
 2 u
 x
2

B
 2 u
 x
 y

C
 2 u
 y
2

D

0
A , B , and C x and
x y u
  u
, , and , .
x y
D

Classification of 2 nd Order
Linear PDE’s
A
 2 u
 x
2

B
 2 u
 x
 y

C
 2 u
 y
2

D

0




Classification of 2 nd Order
Linear PDE’s: Elliptic
A
 2 u
 x
2

B
 2 u
 x
 y

C
 2 u
 y
2

D

0
B

4 AC is elliptic.

0

Classification of 2 nd Order
Linear PDE’s: Elliptic
A
 2 u
 x
2

B
 2 u
 x
 y

C
 2 u
 y
2

D

0
Example:
 2
T
 x
2

 2
T
 y
2

0
B
2 
4 AC

0

4 ( 1 )( 1 )
 
4

0 therefore the equation is elliptic.

Classification of 2 nd Order
Linear PDE’s: Parabolic
A
 2 u
 x
2

B
 2 u
 x
 y

C
 2 u
 y
2

D

0
B

4 AC

0 equation is parabolic.

Classification of 2 nd Order
Linear PDE’s: Parabolic
A
 2 u
 x
2

B
 2 u
 x
 y

C
 2 u
 y
2

D

0
Example:

T
 t
 k
 2
T
 x
2
B
2 
4 AC

0

4 ( 0 )( k )

0 therefore the equation is parabolic.

Classification of 2 nd Order
Linear PDE’s: Hyperbolic
A
 2 u
 x
2

B
 2 u
 x
 y

C
 2 u
 y
2

D

0
B

4 AC

0 equation is hyperbolic.

Classification of 2 nd Order
Linear PDE’s: Hyperbolic
A
 2 u
 x
2

B
 2 u
 x
 y

C
 2 u
 y
2

D

0
Example:  2 y
 x
2

1 c
2
 2 y
 t
2 where, giving c
2
B
2 
4 AC

0

4 ( 1 )(

2
1
)
 c
4
2

0 c therefore the equation is hyperbolic.

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Acknowledgement
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This material is based upon work supported by the National
Science Foundation under Grant # 0717624. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the
National Science Foundation.