Chapter 1
Lecture 1 - Introduction to
Partial Differential Equations
ODE - Equations which define functions of a single independent variable
by prescribing a relationship between the values of the function and its
derivatives.
EG:
(1.1)
y (x) + ey(x) = 0.
PDE - Involve multivariable functions u(x, t), u(x, y) that are determined by
prescribing a relationship between the function value and its partial derivatives.
EG 1:
a
∂
∂
u(x, y) + b u(x, y) = c
∂x
∂y
First Order PDE
(1.2)
3
Lecture 1 - Introduction to Partial Differential Equations
EG 2: Some Classic Second Order PDEs:
Quadric
Classification
T = X2
Parabolic
X2 + Y 2 = k
Elliptic
T 2 − c2 X 2 = k
Hyperbolic
Eq.
∂
∂t u(x, t)
∂2
∂x2 u(x, y)
+
∂2
∂t2 u(x, t)
=
Name
∂2
∂x2 u(x, t)
∂2
∂y 2 u(x, y)
= f (x, y)
2
∂
− c2 ∂x
2 u(x, t) = 0
Heat Equation or Diffusion Eq
Poisson Eq f ≡
0
Laplace Eq f = 0
The Wave Eq
By analogy with quadric surfaces aX 2 + 2bXY + c2 Y 2 + · · · = k that can be
reduced to a standard form by coordinate rotation, the most general linear
2nd order PDE
(1.3)
auxx + 2buxy + cuyy + · · ·
can be reduced by a transformation of coordinates to one of the Heat,
Laplace or the Wave Eq.
1.1
Modeling and Derivation of PDE:
1D Conservation Law: Traffic flow on a highway.
Consider the traffic flow on a highway and let u(x, t) be the density of
cars at x at time t.
[u] = # of cars/unit length.
(1.4)
Let q(x, t) be the flux of cars at x at time t.
[q] = # of cars/unit time.
4
(1.5)
1.1. MODELING AND DERIVATION OF PDE:
{u(x, t + Δt) − u(x, t)}Δx {q(x, t) − q(x + Δx, t)}Δt
(1.6)
Let Δx → 0 and Δt → 0:
∂u ∂q
+
=0
∂t
∂x
(1.7)
• conservation of cars
• conservation of heat
• conservation of chemicals.
How does q change with u?
Convection - and the first order Wave Equation: Assume that q
varies linearly with u, i.e., q = cu from which it follows that
∂u
∂u
+c
=0
∂t
∂x
(1.8)
But this is just a wave equation. To see this consider the following moving
coordinate system.
x = x − ct transformation of coordinates
(1.9)
Guess:
solves ut + cux = 0
u(x, t) = f (x − ct)
ut = −cf
ux = f (1.10)
Therefore ut + cux = −cf + cf = 0.
Thus ut + cux = 0 has solutions of the form u(x, t) = f (x − ct) which
represents a right moving wave. What happens if q = −cu in which case
∂u
∂u
−c
=0
∂t
∂x
(1.11)
5
Lecture 1 - Introduction to Partial Differential Equations
Exercise: Show that (1.11) has a solution of the form u(x, t) = f (x + ct)
which represents a left moving wave.
Note:
∂
∂2u
∂
∂
∂
∂2u
+c
−c
u(x, t) = 2 − c2 2 = 0
(1.12)
∂t
∂x
∂t
∂x
∂t
∂x
is the 2nd order wave equation that has both left and right moving wave
solutions.
Fourier’s Law: Heat flows from hotter regions to colder ones?
q = −α2
∂u
∂x
(1.13)
In this case the conservation law reduces to the form:
∂2u
∂u
= α2 2
∂t
∂x
2D Heat Equation:
6
∂u
= α2
∂t
The Heat Equation
∂2u ∂2u
+ 2
∂x2
∂y
(1.14)
(1.15)
1.2. THE WAVE EQUATION:
1.2
The Wave Equation:
Consider an elastic rod having a density ρ and cross-sectional area A, and let
σ(x, t) be the pressure in the rod at x at time t and u(x, t) the displacement
of the rod from equilibrium.
Balance of Linear Momentum F = M a.
σ(x + Δx, t)A − σ(x, t)A = ρAΔx
σ(x + Δx, t) − σ(x, t)
Δx
∂σ
Δx → 0
∂x
∂2u
∂t2
∂2u
= ρ 2
∂t
∂2u
∂t2
= ρ
(1.16)
(BLM) Balance of Linear Momentum
Hooke’s Law
∂u
∂x
Plug into (BLM) to obtain the 2nd order wave equation.
2
2
E ∂ u
∂2u
E
2∂ u
=
=c
, where c =
∂t2
ρ ∂x2
∂x2
ρ
σ=E
(1.17)
(1.18)
7
Lecture 1 - Introduction to Partial Differential Equations
1.3
The Drunkard’s Walk - The Heat Equation:
Let u(x, t) be the density of fruit-flies at point x at time t. Find an equation
for the density of flies at t + Δt.
u(x, t + Δt) = pu(x + Δx, t) + (1 − 2p)u(x, t) + pu(x − Δx, t)
[u(x + Δx, t) − u(x, t)] [u(x, t) − u(x − Δx, t)]
= u(x, t) + pΔx
−
Δx
Δx
∂u
∂u
(x,
t)
−
(x
−
Δx,
t)
∂x
(1.19)
u(x, t) + pΔx2 ∂x
Δx
∂2u
u(x, t) + pΔx2 2
∂x
2
u(x, t + Δt) − u(x, t)
Δx2 ∂ 2 u
∂u
2∂ u
p
=
α
⇒
The Heat Eq.
Δt
Δt ∂x2
∂t
∂x2
What is the Mean Absolute Deviation of the Drunkard?
xN
x2N
sj
= ±Δx
tj
= jΔt
= s1 + s2 + · · · + sN ∼ 0 Expected Value
2
= (s1 + · · · + sN ) =
s21
+ ··· +
2
s2N
+ 2(s1 s2 + · · · + sN −1 sN )
∼ N Δx
Therefore
8
tN
Δx2 = k 2 tN
Δt
√
|xN | ∼ k tN
x2N
(1.20)
1.3. THE DRUNKARD’S WALK - THE HEAT EQUATION:
Drunkard Walk
5
4
3
2
x
1
0
−1
−2
−3
−4
0
2
4
6
8
10
t
12
14
16
Figure 1.1: Simulation with N = 1000 trajectories for 200 steps along with
the mean absolute deviation envelopes shown in red
9
18
20