Example 2:
In this example, we will see how Fourier series is used to solve a partial differential equation on heat conduction of a circular ring. Again, you will realize the importance of Fourier series. In this problem we study the heat conduction in a circular ring of 1 𝜋 diameter (circumstance is then 1.) that is insulated along its lateral sides. The temperature in the ring is governed by the heat equation ∂u ∂t = 2 ∂ 2 𝑢 ∂x 2 , where c is a constant. We define ( ) the initial temperature of the circular ring and ( , ) the temperature at position x and time t . Here, x represents the arc length along the ring so that the problem becomes one- dimensional although the circular shape is two-dimensional. As we can see, if we go around the ring by an arc length of 1, we come back to the same position as we first started. This means that due to this special shape of a circle, periodicity in space enters the problem. The temperature here is the same as the temperature at the same place, given the time is the same. Thus we can say that temperature is periodic as a function of x: ( ) = ( + 1) ( , ) = ( + 1, ) Since we have already known how to write a Fourier expression, we can write down the Fourier series. This time is a bit different from example one, we use the complex notation to make it easier: ( , ) = ∑ 𝐶 𝑘 2𝜋𝑖𝑘𝑥 ∞ −∞ ( 𝐿 = 1 2 ) Here, the time dependence is in the coefficient 𝐶 𝑘 . We can also write the formula as ( , ) = ∑ 𝐶( ) 2𝜋𝑖𝑘𝑥 ∞ −∞ . Then we plug the formula into the heat equation which is a partial differential equation. 𝜕 𝜕 = ∑𝐶 ′ ( ) 2𝜋𝑖𝑘𝑥 ∞ −∞ ∂ 2 ∂x 2 = ∑𝐶( )(2 𝑖 ) 2 2𝜋𝑖𝑘𝑥 ∞ −∞ The partial differential equation then becomes:
∑𝐶 ′ ( ) 2𝜋𝑖𝑘𝑥 ∞ −∞
= 2 ∑𝐶( )(2 𝑖 ) 2 2𝜋𝑖𝑘𝑥 ∞ −∞
After equating two sides of the equation:
𝐶 ′ ( ) = 2 𝐶( )(2 𝑖 ) 2
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