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   Applications of Fourier Series to Differential Equations
Fourier theory was initially invented to solve certain differential equations. Therefore, it is of no surprise that Fourier series are widely used for seeking solutions to various ordinary differential equations (ODEs) and partial differential equations (PDEs).

In this section, we consider applications of Fourier series to the solution of ODEs and the most well-known PDEs:
  • the heat equation heat equation
  • the wave equation wave equation
  • Laplace's equation Laplace's equation
   Example 1
Find the Fourier series solution to the differential equation with the boundary conditions .

We will use the Fourier sine series for representation of the nonhomogeneous solution to satisfy the boundary conditions. Using the results of Example 3 on the page Definition of Fourier Series and Typical Examples, we can write the right side of the equation as the series
We assume that the solution has the form
Substituting this into the differential equation, we get
Since the coefficients of each sine mode must be equal to each other, we obtain the algebraic equation
Hence, the solution of the given differential equation is described by the series
   Example 2
Find the periodic solutions of the differential equation , where k is a constant and f (x) is a 2π-periodic function.

Represent the function f (x) on the right-hand side of the equation as a Fourier series:
The complex Fourier coefficients are defined by the formula
Assuming that the solution can be represented as a Fourier series expansion
we find the expression for the derivative:
Substituting this into the differential equation, we get
Since this equation is valid for all n, we obtain
Here cn and k are known numbers. Hence, the solution is given by
   Example 3
Using Fourier series expansion, solve the heat conduction equation in one dimension
with the Dirichlet boundary conditions: T = T1 if x = 0, and T = T2 if x = L. The initial temperature distribution is given by .

First we should define the steady state temperature distribution under the given boundary conditions.
Consider the equation . Integrating, we find the general solution:
Find the coefficients C1 and C2 from the boundary conditions: T0 (0) = T1, T0 (L) = T2.

As a result, we have
Now we construct the time-dependent temperature solution. Introduce the new variable
The boundary conditions for y (x,t) become
The initial condition can be written in the form
Taking into account the new boundary conditions it's natural to apply the Fourier sine series expansion. Then
where the coefficients bn are defined by the formula
(We assume that these coefficients are known.)

We will search the general solution in the form of the series with time-dependent coefficients cn (t).
Obviously, the boundary conditions y (0,t) = 0 and y (L,t) = 0 are satisfied for all times t > 0.
The initial conditions for cn (t) are
Substitute these expressions into the heat conduction equation . Then
Multiply both sides of the last expression by and integrate on the interval [0, L] using the orthogonality relations
Then we get
Solving this ODE for cn (t), we obtain
Since in this case m = n, then we can write
where is a constant depending on the initial conditions.
Taking into account that cn (0) = bn, we get the solution for cn (t) in the form
Hence, the final solution for the heat equation is expressed through the formula
      solution for the heat equation
   Example 4
Find the solution of wave equation for a fixed string
      wave equation for a fixed string
with the boundary conditions u (0,t) = u (L,t) = 0 (the string is fixed at the endponts). The initial displacement and velocity are given by
where f (x) and g (x) are some functions defined by the user, such that

We will look for all periodic solutions in which the variables x and t are separated, i.e. in the form
Substituting this into the wave equation, we obtain
Here the function on the left-hand side depends only on x, whereas the function on the right-hand side depends only on t. This can happen if both sides of the equation are constant. Hence,
If the constant α is positive, we can put to get
with the general solution
Such solution cannot produce periodic functions in t. Therefore, we get that the constant α is negative: . Then our wave equation can be split into two ODEs:
Solving the first equation, we find that
where C1 and C2 are constants of integration.

Considering the boundary conditions for the fixed string, we set
By setting C2 ≠ 0 (otherwise we would get the trivial solution X ≡ 0), we find that λL = πn (n is an integer).

Thus, the so-called eigenvalues are
The corresponding eigenfunctions are written as
For λ = λn the second equation yields
Thus, we can write:
Here n is a positive integer, An and Bn are arbitrary constants depending on the initial conditions.

Now we can combine the general solution of the wave equation as a linear combination of the particular solutions:
Assuming that the series is differentiable, we find that
Determine the constants An and Bn using the initial conditions:
As seen, we should expand the functions f (x) and g (x) into the series based on the orthogonal system .
By the formulas for Fourier coefficients,
Thus, the solution to the wave equation with the given boundary and initial conditions is given by the infinite series
      solution to the wave equation
where the coefficients An and Bn are defined by the formulas given above.

The first term u1(x,t) of the series is called the fundamental tone, the other terms un(x,t) are called harmonics. The period and frequency of a harmonic are given by the formulas
      period and frequency of a harmonic
   Example 5
Find the solution to Laplace's equation
on a circle with the boundary condition

We will find the solution in polar coordinates (r,φ). The relationship between cartesian and polar coordinates is given by the standard formulas (Figure 1):
In polar coordinates, the function u(x,y) becomes u(r,φ) depending on the variables r and φ. Obviously u(r,φ) is a 2π-periodic function in φ. Similarly, the boundary function f (x,y) becomes the function f (φ) depending only on φ.

The Laplace's equation in polar coordinates is
      Laplace's equation in polar coordinates
We will seek the solution u(r,φ) as a Fourier series expansion:
where the Fourier coefficients an (r) and bn (r) depend on the radius r.

Assuming that the function u(r,φ) is smooth enough, so we may differentiate it twice with respect to r and φ, we obtain
Substituting this into the Laplace's equation yields
Since this expression is equal to zero for all r and φ, we conclude that
Thus, we have got the system of ODEs instead of the original PDE. (This method was proposed by Joseph Fourier in 1822.) It is important, that each ODE in the system can be solved independently.

We can check directly that the functions
satisfy the last differential equations. Here the constants an (1) and bn (1) must be found from initial conditions to the ODEs. To get these initial conditions, we expand into Fourier series the function defining the boundary conditions for the Laplace's equation. This yields:
Equating coefficients of cos and sin in this expression, we obtain that
Hence, the system of ODEs has solutions
Thus, the solution to the Laplace's equation is given by
where αn, βn are known numbers defined by the boundary conditions.

We can simplify this answer. Substitute the explicit expressions for the coefficients αn, βn:
Notice that
One can show that using the formula the term in the square brackets reduces to the infinite geometric series whose sum is
Then the answer is given by
      Poisson's integral
The last expression is called Poisson's integral for the unit circle.

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