Applications of Partial Differential Equations
P. Sam Johnson
March 6, 2020
P. Sam Johnson Applications of Partial Differential Equations March 6, 2020 1/233
,Overview
Partial differential equations are classified in many ways. One such way is
classifying them as linear and non-linear. Linear equations have an
algebraic nature in their solution sets; in the sense that these solutions can
be superimposed. Nonlinear equations do not share this property.
Equally important in classification schemes of a PDE is the specific nature
of the physical phenomenon that it describes; for example, a PDE can be
classified as wave-like, diffusion like, or static, depending upon whether it
models wave propagation, a diffusion process, or an equilibrium state,
respectively. For example, Laplace’s equation is a linear equilibrium
equation; the heat equation is a linear diffusion equation because the heat
flow is a diffusion process. In three lectures, we discuss some physical
examples and methods for solving them using PDE as a tool.
P. Sam Johnson Applications of Partial Differential Equations March 6, 2020 2/233
,Classification of Partial Differential Equations of Second
order
The general form of second order partial differential equation is given by
∂2z ∂2z ∂2z ∂z ∂z
A 2
+ B + C 2
+D +E + Fz = G (x, y )
∂x ∂x∂y ∂y ∂x ∂y
where A, B, C , D, E , F and G are functions of x and y or constants.
Now consider the term
B 2 − 4AC .
(i) If B 2 − 4AC < 0, then the equation is Elliptic
(ii) If B 2 − 4AC > 0, then the equation is Hyperbolic
(iii) If B 2 − 4AC = 0, then the equation is Parabolic.
P. Sam Johnson Applications of Partial Differential Equations March 6, 2020 3/233
, Examples
Exercise 1.
Classify the following equations
1. uxx − 3uxy + uyy = 0
Solution. Here A = 1, B = −3 and C = 1. Therefore
B 2 − 4AC = (−3)2 − 4(1)(1) = 9 − 4 = 5 > 0
Hence the given equation is hyperbolic.
2. 4uxx − 7uxy + 3uyy = 0
Solution. Here A = 4, B = −7 and C = 3. Therefore
B 2 − 4AC = (−7)2 − 4(4)(3) = 49 − 48 = 1 > 0
Hence the given equation is hyperbolic.
P. Sam Johnson Applications of Partial Differential Equations March 6, 2020 4/233
P. Sam Johnson
March 6, 2020
P. Sam Johnson Applications of Partial Differential Equations March 6, 2020 1/233
,Overview
Partial differential equations are classified in many ways. One such way is
classifying them as linear and non-linear. Linear equations have an
algebraic nature in their solution sets; in the sense that these solutions can
be superimposed. Nonlinear equations do not share this property.
Equally important in classification schemes of a PDE is the specific nature
of the physical phenomenon that it describes; for example, a PDE can be
classified as wave-like, diffusion like, or static, depending upon whether it
models wave propagation, a diffusion process, or an equilibrium state,
respectively. For example, Laplace’s equation is a linear equilibrium
equation; the heat equation is a linear diffusion equation because the heat
flow is a diffusion process. In three lectures, we discuss some physical
examples and methods for solving them using PDE as a tool.
P. Sam Johnson Applications of Partial Differential Equations March 6, 2020 2/233
,Classification of Partial Differential Equations of Second
order
The general form of second order partial differential equation is given by
∂2z ∂2z ∂2z ∂z ∂z
A 2
+ B + C 2
+D +E + Fz = G (x, y )
∂x ∂x∂y ∂y ∂x ∂y
where A, B, C , D, E , F and G are functions of x and y or constants.
Now consider the term
B 2 − 4AC .
(i) If B 2 − 4AC < 0, then the equation is Elliptic
(ii) If B 2 − 4AC > 0, then the equation is Hyperbolic
(iii) If B 2 − 4AC = 0, then the equation is Parabolic.
P. Sam Johnson Applications of Partial Differential Equations March 6, 2020 3/233
, Examples
Exercise 1.
Classify the following equations
1. uxx − 3uxy + uyy = 0
Solution. Here A = 1, B = −3 and C = 1. Therefore
B 2 − 4AC = (−3)2 − 4(1)(1) = 9 − 4 = 5 > 0
Hence the given equation is hyperbolic.
2. 4uxx − 7uxy + 3uyy = 0
Solution. Here A = 4, B = −7 and C = 3. Therefore
B 2 − 4AC = (−7)2 − 4(4)(3) = 49 − 48 = 1 > 0
Hence the given equation is hyperbolic.
P. Sam Johnson Applications of Partial Differential Equations March 6, 2020 4/233
