Applications of Fourier Series to Differential Equations

Solved Problems

Solution.

First we should define the steady state temperature distribution under the given boundary conditions.

Consider the equation $k\frac{{\partial }^{2}T}{\partial x^2}=0.$ Integrating, we find the general solution:

Find the coefficients $C_1$ and $C_2$ from the boundary conditions: $T_0\left(0\right)=T_1,$ $T_0\left(L\right)=T_2.$ As a result, we have

Now we construct the time-dependent temperature solution. Introduce the new variable

The boundary conditions for $y\left(x,t\right)$ 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 $b_n$ 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 $c_n\left(t\right).$

Obviously, the boundary conditions $y\left(0,t\right)=0$ and $y\left(L,t\right)=0$ are satisfied for all times $t>0.$ The initial conditions for $c_n\left(t\right)$ are

Substitute these expressions into the heat conduction equation $$ Then

Multiply both sides of the last expression by $$ and integrate on the interval $$ using the orthogonality relations

Then we get

or

Solving this ODE for $$ we obtain

Since in this case $$ we can write

where $$ is a constant depending on the initial conditions.

Taking into account that $$ we get the solution for $$ in the form

Consequently, the final solution for the heat equation is expressed through the formula

Solution.

We will look for all periodic solutions in which the variables $$ and $$ are separated, that is in the form

Then

Substituting this into the wave equation, we obtain

Here the function on the left-hand side depends only on $$ whereas the function on the right-hand side depends only on $$ This can happen if both sides of the equation are constant. Hence,

If the constant $$ is positive, we can put $$ to get

with the general solution

Such solution cannot produce periodic functions in $$ Therefore, we get that the constant $$ is negative: $$ Then our wave equation can be split into two ODEs:

Solving the first equation, we find that

where $$ and $$ are constants of integration.

Considering the boundary conditions for the fixed string, we set

Then

By setting $$ (otherwise we would get the trivial solution $$), we find that $$ ($$ is an integer).

Thus, the so-called eigenvalues are

The corresponding eigenfunctions are written as

For $$ the second equation yields

Thus, we can write:

Here $$ is a positive integer, $$ and $$ 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 $$ and $$ using the initial conditions:

As one can see, we should expand the functions $$ and $$ 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

where the coefficients $$ and $$ are defined by the formulas given above.

The first term $$ of the series is called the fundamental tone, the other terms $$ are called harmonics. The period and frequency of a harmonic are given by the formulas

Solution.

We will find the solution in polar coordinates $$ The relationship between cartesian and polar coordinates is given by the standard formulas (Figure $$):

In polar coordinates, the function $$ becomes $$ depending on the variables $$ and $$ Obviously, $$ is a $$-periodic function in $$ Similarly, the boundary function $$ becomes the function $$ depending only on the angle $$

The Laplace's equation in polar coordinates is

We will seek the solution $$ as a Fourier series expansion:

where the Fourier coefficients $$ and $$ depend on the radius $$

Assuming that the function $$ is smooth enough, so we may differentiate it twice with respect to $$ and $$ we obtain

Substituting this into the Laplace's equation yields

Since this expression is equal to zero for all $$ and $$ we conclude that

Thus, we have got the system of ODEs instead of the original PDE. (This method was proposed by Joseph Fourier in $$). 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 $$ and $$ 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 $$ and $$ 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 $$ $$ are known numbers defined by the boundary conditions.

We can simplify this answer. Substitute the explicit expressions for the coefficients $$ $$

Notice that

Therefore,

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

The last expression is called Poisson's integral for the unit circle.