2.1 Stationary waves
ˆ t
∂u ∂u
= 0 =⇒ 0 = (s, x) ds = u(t, x) = u(0, x) =⇒ u(t, x) = u(0, x)(:= f (x))
∂t 0 ∂t
Watch the domain!
2.2 Transport and traveling waves
Uniform transport
∂u ∂u
+c =0
∂t ∂x
Define the variables ξ = x − ct and η = x + ct such that u(t, x) = v(ξ, η). Then use the chain rule:
∂u ∂v ∂ξ ∂v ∂η ∂u ∂v ∂ξ ∂v ∂η
= + , = +
∂x ∂ξ ∂x ∂η ∂x ∂t ∂ξ ∂t ∂η ∂t
Then
∂u ∂u ∂v ∂ξ ∂v ∂η ∂v ∂ξ ∂v ∂η
+c = + +c +
∂t ∂x ∂ξ ∂t ∂η ∂t ∂ξ ∂x ∂η ∂x
∂v ∂v ∂v ∂v
= (−c) + (c) + c +
∂ξ ∂η ∂ξ ∂η
∂v
=2 =0
∂η
∂v
=0
∂η
Which is then a stationary wave equation. Hence, the general solution for uniform transport is
w(ξ) = u(x − ct)
If we have some initial value u(0, x) = p(x) with general solution w(x), then the solution to the initial
value problem is given by p(x − ct)
Transport with decay
Consider the PDE
∂u ∂u
+c + au = 0, u(0, x) = f (x)
∂t ∂x
This is solved by u(t, x) = f (x − ct)e−at
Nonuniform transport
Consider the PDE
∂u ∂u
+ c(x) =0
∂t ∂t
Define h(t) = u(t, x(t)). We define
ˆ
1
β(x) := dx = t + k =⇒ x(t) = β −1 (t + k)
c(x)
By defining ξ = β(x) − t, we see that the general solution is
u(t, x) = v(β(x) − t)
If we have the initial condition u(0, x) = f (x), then we can find the solution to the initial value problem
by
u(t, x) = f ◦ β −1 (β(x) − t)
1
,2.4 The wave equation: d’Alembert’s Formula
One-dimensional wave equation:
∂2u ∂2u ∂u
2
= c2 2 , u(x, 0) = f (x), (0, x) = g(x)
∂t ∂x ∂t
How to solve? First write as a linear operator:
L = ∂t2 − c2 ∂x2 =⇒ [u] = 0
Now we factor our operator:
L = (∂t − c∂x ) (∂t + x∂x )
Hence we see that both ∂t − c∂x = 0 and ∂t + c∂x = 0 are solutions. Hence, we have that p(x − ct) and
q(x + ct) are solutions. Then by the superposition principle, we see that the full solution to the wave
equation is
u(t, x) = p(x − ct) + q(x + ct)
Theorem 2.15 The solution to the initial value problem
∂2u ∂2u ∂u
2
= c2 2 , u(0, x) = f (x), (0, x) = g(x), −∞ < x < ∞
∂t ∂x ∂t
is given by ˆ x+ct
f (x − ct) + f (x + ct) 1
u(t, x) = + g(z) dz
2 2c x−ct
3.1 Eigensolutions of Linear Evolution Equations
Heat equation:
∂u ∂2u
=
∂t ∂x2
Linear evolutionary form:
∂u
= L[u]
∂t
How to solve the heat equation? Consider 3 cases:
λ Eigenfunctions v(x) Eigensolutions u(t, x) = eλt v(x)
2 2
λ = −ω 2 < 0 cos(ωx), sin(ωx) e−ω t cos(ωx), e−ω t sin(ωx)
λ=0 1, x 1 ,x
2 2
λ = ω2 > 0 e−ωx , eωx eω t−ωx
, eω t+ωx
Any finite linear combination of the above is also a solution.
3.2 Fourier series
Definition 3.2 The Fourier series of a function f (x) defined on −π ≤ x ≤ π is
∞
a0 X
f (x) ∼ + [ak cos(kx) + bk sin(kx)]
2
k=1
whose coefficients are given by the inner product formulae
ˆ
1 π
ak = ⟨f, cos(kx)⟩ = f (x) cos(kx) dx k = 0, 1, 2, 3, . . .
π −π
ˆ
1 π
bk = ⟨f, sin(kx)⟩ = f (x) sin(kx) dx k = 1, 2, 3, . . .
π −π
2
, Periodic extensions
Lemma 3.4 If f (x) is any function defined for −π ≤ x ≤ π, then there is a unique 2π-periodic function
f˜, known as the 2π-periodic extension of f , that satisfies f˜(x) = f (x) for all π < x < π.
Piecewise continuous functions
Definition 3.6 A function f (x) is said to be piecewise continuous on an interval [a, b] if it is defined
and continuous except possibly at a finite number of points a ≤ x1 < 2x < · · · < xn ≤ b. Furthermore,
at each point of discontinuity we require that the left- and right-handed limits
f (x− +
k ) = lim f (x) f (xk ) = lim f (x)
x→x−
k x→x+
k
exist. At the endpoints, only one is required to exist (namely f (a+ ) and f (b− )).
−
We define βk = f (x+
k ) − f (xk ) the magnitude of the jump.
Definition 3.7 A function f (x) is called piecewise C 1 on an interval [a, b] if it is defined, continuous
and continuously differentiable except at a finite number of points a ≤ x1 < · · · < xn ≤ b. At each
exceptional point, the left-and right-hand limits of both the function and its derivative exist.
The convergence theorem
Theorem 3.8 If f˜(x) is a 2π-periodic, piecewise C 1 function, then, at any x ∈ R, its Fourier series
converges to
f˜(x), if f˜ is continuous at x
1 ˜ +
[f (x ) + f˜(x− )], if x is a jump discontinuity
2
Even and odd functions
Definitions and lemma’s:
• A function is called even if f (−x) = f (x)
• A function is called odd if f (−x) = −f (x)
• The sum of two even functions is even
• The sum of two odd functions is odd
• the product of two even or two odd functions is even
• the product of an even and an odd function is odd
´a
• If f is odd and integrable on the symmetric interval [−a, a], then −a
f (x)dx = 0
´a ´a
• If g is even and integrable on the symmetric interval [−a, a], then −a
g(x)dx = 2 −a
g(x)dx
Proposition 3.14 If f (x) is even, then f (x) can be represented by a Fourier cosine series (i.e. bk = 0
for all k). If f (x) is odd, then f (x) can be represented by a Fourier sine series (i.e. ak = 0 for all k).
Conversely, a convergent Fourier cosine series always represents an even function, while a convergent sine
series always represents an odd function.
3
ˆ t
∂u ∂u
= 0 =⇒ 0 = (s, x) ds = u(t, x) = u(0, x) =⇒ u(t, x) = u(0, x)(:= f (x))
∂t 0 ∂t
Watch the domain!
2.2 Transport and traveling waves
Uniform transport
∂u ∂u
+c =0
∂t ∂x
Define the variables ξ = x − ct and η = x + ct such that u(t, x) = v(ξ, η). Then use the chain rule:
∂u ∂v ∂ξ ∂v ∂η ∂u ∂v ∂ξ ∂v ∂η
= + , = +
∂x ∂ξ ∂x ∂η ∂x ∂t ∂ξ ∂t ∂η ∂t
Then
∂u ∂u ∂v ∂ξ ∂v ∂η ∂v ∂ξ ∂v ∂η
+c = + +c +
∂t ∂x ∂ξ ∂t ∂η ∂t ∂ξ ∂x ∂η ∂x
∂v ∂v ∂v ∂v
= (−c) + (c) + c +
∂ξ ∂η ∂ξ ∂η
∂v
=2 =0
∂η
∂v
=0
∂η
Which is then a stationary wave equation. Hence, the general solution for uniform transport is
w(ξ) = u(x − ct)
If we have some initial value u(0, x) = p(x) with general solution w(x), then the solution to the initial
value problem is given by p(x − ct)
Transport with decay
Consider the PDE
∂u ∂u
+c + au = 0, u(0, x) = f (x)
∂t ∂x
This is solved by u(t, x) = f (x − ct)e−at
Nonuniform transport
Consider the PDE
∂u ∂u
+ c(x) =0
∂t ∂t
Define h(t) = u(t, x(t)). We define
ˆ
1
β(x) := dx = t + k =⇒ x(t) = β −1 (t + k)
c(x)
By defining ξ = β(x) − t, we see that the general solution is
u(t, x) = v(β(x) − t)
If we have the initial condition u(0, x) = f (x), then we can find the solution to the initial value problem
by
u(t, x) = f ◦ β −1 (β(x) − t)
1
,2.4 The wave equation: d’Alembert’s Formula
One-dimensional wave equation:
∂2u ∂2u ∂u
2
= c2 2 , u(x, 0) = f (x), (0, x) = g(x)
∂t ∂x ∂t
How to solve? First write as a linear operator:
L = ∂t2 − c2 ∂x2 =⇒ [u] = 0
Now we factor our operator:
L = (∂t − c∂x ) (∂t + x∂x )
Hence we see that both ∂t − c∂x = 0 and ∂t + c∂x = 0 are solutions. Hence, we have that p(x − ct) and
q(x + ct) are solutions. Then by the superposition principle, we see that the full solution to the wave
equation is
u(t, x) = p(x − ct) + q(x + ct)
Theorem 2.15 The solution to the initial value problem
∂2u ∂2u ∂u
2
= c2 2 , u(0, x) = f (x), (0, x) = g(x), −∞ < x < ∞
∂t ∂x ∂t
is given by ˆ x+ct
f (x − ct) + f (x + ct) 1
u(t, x) = + g(z) dz
2 2c x−ct
3.1 Eigensolutions of Linear Evolution Equations
Heat equation:
∂u ∂2u
=
∂t ∂x2
Linear evolutionary form:
∂u
= L[u]
∂t
How to solve the heat equation? Consider 3 cases:
λ Eigenfunctions v(x) Eigensolutions u(t, x) = eλt v(x)
2 2
λ = −ω 2 < 0 cos(ωx), sin(ωx) e−ω t cos(ωx), e−ω t sin(ωx)
λ=0 1, x 1 ,x
2 2
λ = ω2 > 0 e−ωx , eωx eω t−ωx
, eω t+ωx
Any finite linear combination of the above is also a solution.
3.2 Fourier series
Definition 3.2 The Fourier series of a function f (x) defined on −π ≤ x ≤ π is
∞
a0 X
f (x) ∼ + [ak cos(kx) + bk sin(kx)]
2
k=1
whose coefficients are given by the inner product formulae
ˆ
1 π
ak = ⟨f, cos(kx)⟩ = f (x) cos(kx) dx k = 0, 1, 2, 3, . . .
π −π
ˆ
1 π
bk = ⟨f, sin(kx)⟩ = f (x) sin(kx) dx k = 1, 2, 3, . . .
π −π
2
, Periodic extensions
Lemma 3.4 If f (x) is any function defined for −π ≤ x ≤ π, then there is a unique 2π-periodic function
f˜, known as the 2π-periodic extension of f , that satisfies f˜(x) = f (x) for all π < x < π.
Piecewise continuous functions
Definition 3.6 A function f (x) is said to be piecewise continuous on an interval [a, b] if it is defined
and continuous except possibly at a finite number of points a ≤ x1 < 2x < · · · < xn ≤ b. Furthermore,
at each point of discontinuity we require that the left- and right-handed limits
f (x− +
k ) = lim f (x) f (xk ) = lim f (x)
x→x−
k x→x+
k
exist. At the endpoints, only one is required to exist (namely f (a+ ) and f (b− )).
−
We define βk = f (x+
k ) − f (xk ) the magnitude of the jump.
Definition 3.7 A function f (x) is called piecewise C 1 on an interval [a, b] if it is defined, continuous
and continuously differentiable except at a finite number of points a ≤ x1 < · · · < xn ≤ b. At each
exceptional point, the left-and right-hand limits of both the function and its derivative exist.
The convergence theorem
Theorem 3.8 If f˜(x) is a 2π-periodic, piecewise C 1 function, then, at any x ∈ R, its Fourier series
converges to
f˜(x), if f˜ is continuous at x
1 ˜ +
[f (x ) + f˜(x− )], if x is a jump discontinuity
2
Even and odd functions
Definitions and lemma’s:
• A function is called even if f (−x) = f (x)
• A function is called odd if f (−x) = −f (x)
• The sum of two even functions is even
• The sum of two odd functions is odd
• the product of two even or two odd functions is even
• the product of an even and an odd function is odd
´a
• If f is odd and integrable on the symmetric interval [−a, a], then −a
f (x)dx = 0
´a ´a
• If g is even and integrable on the symmetric interval [−a, a], then −a
g(x)dx = 2 −a
g(x)dx
Proposition 3.14 If f (x) is even, then f (x) can be represented by a Fourier cosine series (i.e. bk = 0
for all k). If f (x) is odd, then f (x) can be represented by a Fourier sine series (i.e. ak = 0 for all k).
Conversely, a convergent Fourier cosine series always represents an even function, while a convergent sine
series always represents an odd function.
3
