introduction to partial differential...
TRANSCRIPT
9/14/2011 http://numericalmethods.eng.usf.edu 1
Introduction to Partial Differential Equations
http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM Undergraduates
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 is the dependent variable, and x and y are the independent variables.
5)0(,353 2 ==+ − yeydxdy x
222
2
2
2
3 yxyu
xu
+=∂∂
+∂∂
Example of an Ordinary Differential Equation
Assumption: Ball is a lumped system. Number of Independent variables:
One (t)
Hot Water
Spherical Ball
( )dtdmChA aθθθ =−
Example of an Partial Differential Equation
Assumption: Ball is not a lumped system. Number of Independent variables:
Four (r, θ,φ,t)
Hot Water
Spherical Ball
aTrTttTCT
rkT
rk
rTr
rrk
=≥∂∂
=∂∂
+
∂∂
∂∂
+
∂∂
∂∂ )0,,,(,0,
sinsin
sin 2
2
2222
2 φθρφθθ
θθθ
Classification of 2nd Order Linear PDE’s
where are functions of ,and is a function of
02
22
2
2
=+∂∂
+∂∂
∂+
∂∂ D
yuC
yxuB
xuA
CBA and,,yx and D
, , and , .u ux y ux y∂ ∂∂ ∂
Classification of 2nd Order Linear PDE’s
Can Be: Elliptic Parabolic Hyperbolic
02
22
2
2
=+∂∂
+∂∂
∂+
∂∂ D
yuC
yxuB
xuA
Classification of 2nd Order Linear PDE’s: Elliptic
02
22
2
2
=+∂∂
+∂∂
∂+
∂∂ D
yuC
yxuB
xuA
042 <− ACBIf ,then equation is elliptic.
Classification of 2nd Order Linear PDE’s: Elliptic
02
22
2
2
=+∂∂
+∂∂
∂+
∂∂ D
yuC
yxuB
xuA
Example:
where, giving
therefore the equation is elliptic.
02
2
2
2
=∂∂
+∂∂
yT
xT
1,0,1 === CBA
04)1)(1(4042 <−=−=− ACB
Classification of 2nd Order Linear PDE’s: Parabolic
02
22
2
2
=+∂∂
+∂∂
∂+
∂∂ D
yuC
yxuB
xuA
2 4 0B AC− =If ,then the equation is parabolic.
Classification of 2nd Order Linear PDE’s: Parabolic
02
22
2
2
=+∂∂
+∂∂
∂+
∂∂ D
yuC
yxuB
xuA
Example:
where, giving
therefore the equation is parabolic.
2
2
xTk
tT
∂∂
=∂∂
1,0,0, −==== DCBkAACB 42 − ))(0(40 k−= 0=
Classification of 2nd Order Linear PDE’s: Hyperbolic
02
22
2
2
=+∂∂
+∂∂
∂+
∂∂ D
yuC
yxuB
xuA
2 4 0B AC− >If ,then the equation is hyperbolic.
Classification of 2nd Order Linear PDE’s: Hyperbolic
02
22
2
2
=+∂∂
+∂∂
∂+
∂∂ D
yuC
yxuB
xuA
Example:
where, giving
therefore the equation is hyperbolic.
2
2
22
2 1ty
cxy
∂∂
=∂∂
2
1,0,1c
CBA −===
)1)(1(404 22
cACB −
−=− 042 >=
c
THE END