# Dirac Delta and Fourier Series

#### rannasquaer

##### New member
A beam of length L with fixed ends, has a concentrated force P applied in the center exactly in L / 2.

In the differential equation:

$\frac{d^4y(x)}{dx^4}=\frac{1}{\text{EI}}q(x)$

In which

$q(x)= P \delta(x-\frac{L}{2})$

P represents an infinitely concentrated charge distribution

The problem can be solved through developments in Fourier sine series, suppose that

$y(x)=\sum_{n=1}^{\infty} b_n \sin (\frac{n \pi x}{\text{L}})$

Demonstrate and explain step by step to obtain the equation below

$\delta(x-\frac{\text{L}}{2})= \frac{2}{\text{L}} \sum_{n=1}^{\infty} \sin (\frac{n \pi}{2}) \sin (\frac{n \pi x}{\text{L}})$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hi rannasquaer, welcome to MHB!

The Fourier sine series says that we can write an odd function $f(x)$ with period $L$ as:
$$f(x)=\sum_{n=1}^\infty B_n \sin\frac{n\pi x}L\quad\text{with}\quad B_n = \frac 2L\int_0^L f(x) \sin\frac{n\pi x}L\,dx \quad\quad (1)$$

Substitute $f(x)=\delta(x-\frac L2)$ to find:
$B_n = \frac 2L\int_0^L \delta\big(x-\frac L2\big) \sin\frac{n\pi x}L\,dx$

To evaluate this, we use the property of the Dirac $\delta$ function that if $a<c<b$ then $\int_a^b \delta(x-c)g(x)\,dx = g(c)$.
So
$B_n = \frac 2L\int_0^L \delta\big(x-\frac L2\big) \sin\big(\frac{n\pi x}L\big)\,dx = \frac 2L\, \sin\big(\frac{n\pi}L\frac L2\big) = \frac 2L\, \sin \frac{n\pi}2$

Substitute in $(1)$ and find:
$\delta\big(x-\frac L2\big) = \sum_{n=1}^\infty \frac 2L\, \sin \frac{n\pi}2 \sin\frac{n\pi x}L$

Last edited:
• rannasquaer

#### rannasquaer

##### New member
Thank you so much, I was having trouble understanding what to use as f(x), I thought I should use q(x), and everything was going wrong.

Thank you!

• Klaas van Aarsen