#19 (11/05/2025)

Laplace Equation

Consider a plate in which the temperature is kept at 0 on three sides (x=0, x=π and y=0) and a distributed temperature distribution is given on one side (y=1).

Inside the body, the temperature is subject to the Laplace equation as

∂² u

∂x²

∂² u

∂y²

0 in D,

(1)

u(0, y)

0 (0 < y < 1)

(2)

u(π, y)

0 (0 < y < 1)

(3)

u(x, 0)

0 (0 < x < π)

(4)

u(x, 1)

f(x) (0 < x < π)

(5)

The Laplace equation can be rewritten as

∂² u

∂x²

= −

∂² u

∂y²

(6)

Because of the homegeneous boundary conditins along the x axis, x can be taken as the main variable and y can be taken as the secondary variable. Hence, u(x, y) can be expressed by the Fourier series on x as

u(x, y) =

∞
∑
n=1

u_n(y) sin n x.

(7)

Substituting Eq. (7) into Eq. (6), we have

∞
∑
n=1

−n² u_n(y) sin n x =

∞
∑
n=1

−

d² u_n(y)

d y²

sin nx,

(8)

d² u_n(y)

d y²

− n² u_n(y) = 0.

(9)

Equation (9) can be solved as

u_n(y) = A_n e^{n y}+ B_n e^{− ny}.

(10)

Therefore, we have

u(x, y) =

∞
∑
n=1

(A_n e^{n y}+ B_n e^{− ny}) sin n x.

(11)

It's time to consider the boundary condition in the y direction:

Along y=0, u(x, 0)=0 so

0 =

∞
∑
n=1

(A_n + B_n ) sin nx,

B_n = − A_n,

which yields

u(x, y) =

∞
∑
n=1

A_n (e^{n y}−e^{− ny}) sin n x.

(12)

Along y=1, u(x, 1)=f(x) so

f(x) =

∞
∑
n=1

A_n (eⁿ − e⁻ⁿ) sin nx,

(13)

Equation (13) implies that A_n (eⁿ − e⁻ⁿ) is the Fourier coefficient of f(x).¹

Therefore,

A_n (eⁿ − e⁻ⁿ) =

⌠
⌡

π

0

f(x) sin n x dx,

(14)

A_n =

(eⁿ − e⁻ⁿ)

⌠
⌡

π

0

f(x) sin n x dx,

(15)

Example:

For f(x)=1, we have

A_n =

(eⁿ − e⁻ⁿ)

1−(−1)ⁿ

(16)

u(x, y) =

∞
∑
n=1

(eⁿ − e⁻ⁿ)

1−(−1)ⁿ

(e^{n y}−e^{− n y}) sin n x.

(17)

Here is an example of the profile of u(x, y) when f(x) = 1.

Fourier Cosine Transform

The discrete cosine transform (DCT) does not use complex numbers or negative frequencies. DCT was derived by K. R. Rao, former UTA professor. To illustrate DCT, Cosine Transform of a continuous function, f(x), is explained below:

A function, f(x), is defined over [0, 2 π]. Extend f(x) to [−2π, 0] and define ~f(x) as

(x) ≡

⎧
⎪
⎨
⎪
⎩

f(x)

0 < x < 2π

f(−x)

−2π < x < 0

(18)

The Fourier series for ~f(x) over [−2π, 2π] is expressed as

(x) =

∞
∑
k=−∞

e^{[(i k x)/2]}.

(19)

The Fourier coefficient, ~F_k, is expressed as

4π

⌠
⌡

2π

−2π

(x) e^{−[(i k x)/2]}dx.

(20)

It is possible to rewrite ~F_k without complex functions as

4π

⌠
⌡

0

−2π

(x) e^{−[(i k x)/2]} dx +

4π

⌠
⌡

2π

0

(x) e^{−[(i k x)/2]} dx

4π

⌠
⌡

2π

0

(−x) e^{[(i k x)/2]} dx +

4π

⌠
⌡

2π

0

(x) e^{−[(i k x)/2]} dx

4π

⌠
⌡

2π

0

(x) (e^{[(i k x)/2]}+e^{[(−i k x)/2]}) dx

2π

⌠
⌡

2π

0

(x) cos

d x.

(21)

Note that

−k

(22)

Therefore, ~f(x) can be also rewritten without complex functions as

(x)

−1
∑
k=−∞

e^{[(i k x)/2]} +

∞
∑
k=1

e^{[(i k x)/2]}

∞
∑
k=1

−k

e^{−[(i k x)/2]} +

∞
∑
k=1

e^{[(i k x)/2]}

∞
∑
k=1

(e^{−[(i k x)/2]}+e^{[(i k x)/2]})

+ 2

∞
∑
k=1

cos

(23)

Continuous Fourier Cosine Transform for f(x) over [0, 2π]

f(x)

F₀ + 2

∞
∑
k=1

F_k cos

(24)

F(k)

2π

⌠
⌡

2π

0

f(x) cos

d x.

(25)

Example:

f(x)=x, 0 < x < 2π

Conventional transform

f(x) = ∞
∑
m=−∞
C_m e^{i m x}, C_m = 1
2π
⌠
⌡ 2 π

0
f(x) e^{−i m x} dx = ⎧
⎪
⎨
⎪
⎩

i
m
,
m ≠ 0

π,
m = 0

f(x) = −1
∑
m=−∞
i
m
e^{i m x}+ π+ ∞
∑
m=1
i
m
e^{i m x}
Cosine transform

F_k = 2 (−1+(−1)^k
k² π
, F₀ = π.

f(x)=π+ 4
π
∞
∑
k=1
−1+(−1)^k
k²
cos k x
2

Discrete Fourier Cosine Transform²

for (f₀, f₁, f₂, …f_N−1)

f_l

F₀

N−1
∑
k=1

F_k cos

k (2 l+1)π

(26)

F(k)

N−1
∑
l = 0

f_l cos

k (2 l+1)π

(27)

Dirac delta functions and Heaviside step functions

Definition of Dirac delta function

The definition of the Dirac delta function is

⌠
⌡

∞

−∞

δ(x) f(x) dx = f(0),

(28)

i.e. δ(x) is defined by its action as an operator and is therefore different from ordinary functions. The Dirac delta function is one of generalized functions (distributions, hyperfunctions) that are used in physics extensively.

The Dirac delta function can be realized by the limit of a sequence of functions defined as

w_k(x) =

⎧
⎪
⎨
⎪
⎩

|x| <

otherwise

(29)

⌠
⌡

∞

−∞

w_k(x) f(x) dx

⌠
⌡

1/k

−1/k

f(x) dx

⌠
⌡

1/k

−1/k

f(x) dx

⌠
⌡

1/k

−1/k

⎛
⎝

f(0)+ f′(0) x +

f"(0)

x² +

f"′(0)

x³ +

f""(0)

x⁴ + …

⎞
⎠

f(0) +

f"(0)

k²

f""(0)

120

k⁴

+ …

∼

f(0) as k → ∞.

(30)

By comparing Eq. (28) with Eq. (30), one can obtain

δ(x) =

lim
k→ ∞

w_k(x).

(31)

In physics, a point force at x=a with a magnitude of F can be expressed as

w = F δ(x−a).

(32)

Derivatives of delta functions

It is also possible to define derivatives of delta functions as distributions. By using integrations by parts, one can start with

⌠
⌡

∞

−∞

δ′(x) f(x) dx

[δ(x) f(x)]_−∞^∞−

⌠
⌡

∞

−∞

δ(x) f′(x) dx

−f′(0).

(33)

So the action of δ′(x) is to (1) differentiate f(x), (2) evaluate f′(x) at x=0 and (3) multiply by (−1).

To understand the physical meaning of δ′(x), consider a sequence of functions, m_k(x), defined as

m_k (x) ≡

⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩

4 k²

−

< x < 0

− 4 k²

0 < x <

otherwise.

(34)

Then,

⌠
⌡

∞

−∞

m_k(x) f(x) dx

⌠
⌡

0

−1/(2k)

4 k² f(x)dx +

⌠
⌡

1/(2k)

0

(−4 k²) f(x)dx

4k²

⌠
⌡

0

−1/(2k)

f(x)dx −4k²

⌠
⌡

1/(2k)

0

f(x)dx

4k²

⌠
⌡

0

−1/(2k)

⎛
⎝

f(0)+ f′(0) x +

f"(0)

x² +

f"′(0)

x³ + …

⎞
⎠

−4 k²

⌠
⌡

1/(2k)

0

⎛
⎝

f(0)+ f′(0) x +

f"(0)

x² +

f"′(0)

x³ + …

⎞
⎠

4k²

⎛
⎝

f(0)

−

f′(0)

8 k²

f"(0)

48 k³

−

f"′(0)

384 k⁴

+…

⎞
⎠

−4k²

⎛
⎝

f(0)

f′(0)

8 k²

f"(0)

48 k³

f"′(0)

384 k⁴

+…

⎞
⎠

−f′(0) −

f"′(0)

k²

− …

∼

−f′(0) as k→ ∞.

(35)

By comparing Eq. (33) with Eq. (35), one finds

δ′(x) =

lim
k→ ∞

m_k(x).

(36)

If we replace the distributed load by the equivalent point force, the figure of m_k(x) can be seen as a moment (couple) as

It is thus seen that a point moment located at x=a with a magnitude of M₁ can be expressed by δ′(x) as

M = M₁ δ′(x−a).

(37)

Higher order derivatives of δ(x) can be evaluated in a similar manner. For instance, the second order derivative of δ(x), δ"(x), is

⌠
⌡

∞

−∞

δ"(x) g(x) dx

⎡
⎣

δ′(x) g(x) − δ(x) g′(x)

⎤
⎦

∞
−∞

⌠
⌡

∞

−∞

δ(x) g"(x) dx

g"(0),

(38)

in general

⌠
⌡

∞

−∞

δ⁽ⁿ⁾(x) g(x) dx = (−1)ⁿ g⁽ⁿ⁾(0).

(39)

Heaviside step function

The Heaviside step function, H(x), is defined as

H(x) =

⎧
⎪
⎨
⎪
⎩

x < 0

x > 0.

(40)

Obviously, H(x) is discontinuous at x=0 and in traditional calculus, the derivative of H(x) at x = 0 does not exist. However, consider

⌠
⌡

∞

−∞

H^′(x) g(x)dx

⎡
⎣

H(x) g(x)

⎤
⎦

∞
−∞

−

⌠
⌡

∞

−∞

H(x) g′(x) dx

g(∞)−0 −

⌠
⌡

∞

0

g′(x) dx

g(∞) −

⎡
⎣

g(x)

⎤
⎦

∞
0

g(0).

(41)

where an integration by parts was used. By comparing Eq. (41) with Eq. (28), one can find

H ′(x) = δ(x).

(42)

which implies that if one includes δ(x) in a group of functions, it is possible to differentiate discontinuous functions as many times as needed.

Examples

g(x) δ(x) = g(0) δ(x).

⌠
⌡

∞

−∞

⎛
⎝

g(x) δ(x)

⎞
⎠

f(x) dx

⌠
⌡

∞

−∞

δ(x)

⎛
⎝

g(x) f(x)

⎞
⎠

g(0) f(0).

(43)

⌠
⌡

∞

−∞

⎛
⎝

g(0) δ(x)

⎞
⎠

f(x) dx

⌠
⌡

∞

−∞

δ(x)

⎛
⎝

g(0) f(x)

⎞
⎠

g(0) f(0).

(44)

x δ′(x) = − δ(x).

⌠
⌡

∞

−∞

⎛
⎝

x δ′(x)

⎞
⎠

f(x) dx

⌠
⌡

∞

−∞

δ′(x)

⎛
⎝

x f(x)

⎞
⎠

−

(x f(x)) |_x=0

− ( f(x)+ x f′(x))|_x=0

− f(0).

(45)

⌠
⌡

∞

−∞

⎛
⎝

−δ(x)

⎞
⎠

f(x) dx = − f(0).

(46)

x² δ"(x) = 2 δ(x).

d⁴

dx⁴

|x|³ = 12 δ(x).

|x| = 2 H(x)−1.

|x| = x H(x) − x H(−x).

(47)

|x|

H(x) + xδ(x) − H(−x) −(−1) x δ(−x)

2H(x)−1 + 2x δ(x),

(48)

H(−x) = 1− H(x),

(49)

x δ(x) = 0,

(50)

⌠
⌡

∞

−∞

⎛
⎝

x δ(x)

⎞
⎠

g(x) dx

⌠
⌡

∞

−∞

δ(x)

⎛
⎝

x g(x)

⎞
⎠

xg(x)|_x=0

(51)

|x| = 2H(x)−1.

(52)

Integration of singularity functions

⌠
⌡

x

−∞

δ(x−a) dx

H(x−a)

(53)

⌠
⌡

x

−∞

H(x−a) dx

(x − a) H(x−a)

(54)

⌠
⌡

x

−∞

(x−a) H(x−a) dx

(x−a)²

H(x−a)

(55)

⌠
⌡

x

−∞

(x−a)² H(x−a) dx

(x−a)³

H(x−a)

(56)

…

(57)

In general,

(x−a)ⁿ H(x−a) =

⎧
⎪
⎨
⎪
⎩

x < a

(x−a)ⁿ

x > a

(58)

i.e. the function, (x−a)ⁿ H(x−a), is identically 0 when the argument, x−a, is negative.

Beam loading

Example

A cantilevered beam is subject to a partially distributed load starting at x=a. The problem is to find the reaction force and the point moment at x=0.

First we can assume that all the unknown quantities, i.e. R₁ and M₁, are positive, i.e. the directions shown in the figure. Once we can express q(x) using the singularity functions, V(x) and M(x) can be automatically expressed by integrating each singular function as

q(x)

M₁ δ′(x) + R₁ δ(x) − w H(x−a),

(59)

V(x)

M₁ δ(x) + R₁ H(x) − w (x−a) H(x),

(60)

M(x)

M₁ H(x) + R₁ x H(x) − w

(x−a)²

H(x−a).

(61)

Substitute x=l+ϵ where ϵ is a small number, i.e. we are looking at a point a little bit right to the end of the beam. Since there is no beam material at x=l+ϵ, both M(x) and V(x) should be zero. So Eqs. (60, 61) become

M₁ ×0 + R₁ ×1 − w (l−a)

(62)

M₁ ×1 + R₁ ×l− w

(l−a)²

(63)

Note that δ(x)|_{x→ l} = 0 and H(x)|_{x→ l} = 1 etc.... Eqs. (62, 63) can be solved to yield

R₁

w (l−a)

(64)

M₁

−

(l+a)(l−a)

(65)

The result of Eq. (65) implies that the correct sign of M₁ is ccw. It is noted that Eq. (62) is the balance-of-force equation while Eq. (63) is the balance-of-moment equation

Footnotes:

f(x) =

∞
∑
n=1

f_n sin n x, f_n =

⌠
⌡

π

0

f(x) sin n x dx.

²Ahmed, N., Natarajan, T. and Rao, K. R. "Discrete Cosine Transform," IEEE Transactions on Computers, C-23 (1): 90-93, doi:10.1109/T-C.1974.

File translated from T_EX by T_TH, version 4.03.
On 04 Nov 2025, 17:24.

Laplace Equation

Fourier Cosine Transform

Example:

Discrete Fourier Cosine Transform2

Dirac delta functions and Heaviside step functions

Definition of Dirac delta function

Derivatives of delta functions

Heaviside step function

Examples

Integration of singularity functions

Beam loading

Footnotes:

Discrete Fourier Cosine Transform²