Applications of Fourier Series to Differential Equations
Solved Problems
Example 3.
Using Fourier series expansion, solve the heat conduction equation in one dimension \[\frac{{\partial T}}{{\partial t}} = k\frac{{{\partial ^2}T}}{{\partial {x^2}}}\] with the Dirichlet boundary conditions: \(T = {T_1}\) if \(x = 0\) and \(T = {T_2}\) if \(x = L.\) The initial temperature distribution is given by \(T\left( {x,0} \right) = f\left( x \right).\)
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:
\[{T_0}\left( x \right) = {C_1} + {C_2}x.\]
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
Obviously, the boundary conditions \(y\left( {0,t} \right) = 0\) and \(y\left( {L,t} \right) = 0\) are satisfied for all times \(t \gt 0.\) The initial conditions for \({c_n}\left( t \right)\) are
\[{c_n}\left( 0 \right) = {b_n},\;\; n = 0,1,2, \ldots\]
Substitute these expressions into the heat conduction equation \(k \frac{{{\partial ^2}y}}{{\partial {x^2}}} = \frac{{\partial y}}{{\partial t}}.\) Then
Multiply both sides of the last expression by \({\sin {\frac{{m\pi x}}{L}}}\) and integrate on the interval \(\left[ {0,L} \right]\) using the orthogonality relations
\[
\int\limits_0^L {\sin \frac{{n\pi x}}{L}\sin \frac{{m\pi x}}{L}dx} =
\begin{cases}
0, & \text{if} & m \ne n \\
\frac{L}{2}, & \text{if} & m = n
\end{cases}.\]
with the boundary conditions \(u\left( {0,t} \right) = u\left( {L,t} \right) = 0\) (the string is fixed at the endponts). The initial displacement and velocity are given by
\[u\left( {x,0} \right) = f\left( x \right),\; \frac{{\partial u\left( {x,0} \right)}}{{\partial t}} = g\left( x \right),\]
where \(f\left( x \right)\) and \(g\left( x \right)\) are some functions defined by the user, such that
\[f\left( 0 \right) = f\left( L \right) = g\left( 0 \right) = g\left( L \right) = 0.\]
Solution.
We will look for all periodic solutions in which the variables \(x\) and \(t\) are separated, that is in the form
\[u\left( {x,t} \right) = X\left( x \right) T\left( t \right).\]
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,
Such solution cannot produce periodic functions in \(t.\) Therefore, we get that the constant \(\alpha\) is negative: \(\alpha = - {\lambda ^2}.\) Then our wave equation can be split into two ODEs:
As one can see, we should expand the functions \(f\left( x \right)\) and \(g\left( x \right)\) into the series based on the orthogonal system \(\left\{ {\sin \frac{{\pi nx}}{L}} \right\}.\) By the formulas for Fourier coefficients,
\[{A_n} = \frac{2}{L}\int\limits_0^L {f\left( x \right)\sin \frac{{\pi nx}}{L}dx} ,\;\; {B_n} = \frac{2}{{a\pi n}}\int\limits_0^L {g\left( x \right)\sin \frac{{\pi nx}}{L}dx} ,\;\; n = 1,2,3, \ldots\]
Thus, the solution to the wave equation with the given boundary and initial conditions is given by the infinite series
where the coefficients \({A_n}\) and \({B_n}\) are defined by the formulas given above.
The first term \({u_1}\left( {x,t} \right)\) of the series is called the fundamental tone, the other terms \({u_n}\left( {x,t} \right)\) are called harmonics. The period and frequency of a harmonic are given by the formulas
We will find the solution in polar coordinates \(\left( {r,\varphi } \right).\) The relationship between cartesian and polar coordinates is given by the standard formulas (Figure \(1\)):
In polar coordinates, the function \(u\left( {x,y} \right)\) becomes \(u\left( {r,\varphi } \right)\) depending on the variables \(r\) and \(\varphi.\) Obviously, \(u\left( {r,\varphi } \right)\) is a \(2\pi\)-periodic function in \(\varphi.\) Similarly, the boundary function \(f\left( {x,y} \right)\) becomes the function \(f\left( \varphi \right)\) depending only on the angle \(\varphi.\)
We will seek the solution \(u\left( {r,\varphi } \right)\) as a Fourier series expansion:
\[u\left( {r,\varphi } \right) = \frac{{{a_0}\left( r \right)}}{2} + \sum\limits_{n = 1}^\infty {\left[ {{a_n}\left( r \right)\cos n\varphi + {b_n}\left( r \right)\sin n\varphi } \right]} ,\]
where the Fourier coefficients \({{a_n}\left( r \right)}\) and \({{b_n}\left( r \right)}\) depend on the radius \(r.\)
Assuming that the function \(u\left( {r,\varphi } \right)\) is smooth enough, so we may differentiate it twice with respect to \(r\) and \(\varphi,\) we obtain
\[\frac{{\partial u}}{{\partial r}} = \frac{{{a'_0} \left( r \right)}}{2} + \sum\limits_{n = 1}^\infty {\left[ {{a'_n} \left( r \right)\cos n\varphi + {b'_n} \left( r \right)\sin n\varphi } \right]} ,\]
\[\frac{{{\partial ^2}u}}{{\partial {r^2}}} = \frac{{{a^{\prime\prime}_0} \left( r \right)}}{2} + \sum\limits_{n = 1}^\infty {\left[ {{a^{\prime\prime}_n}\left( r \right)\cos n\varphi + {b^{\prime\prime}_n}\left( r \right)\sin n\varphi } \right]} ,\]
Substituting this into the Laplace's equation yields
\[\frac{{{r^2}{a^{\prime\prime}_0}\left( r \right)}}{2} + \frac{{r{a'_0} \left( r \right)}}{2}
+ \sum\limits_{n = 1}^\infty {\Big[ {\Big( {{r^2}{a^{\prime\prime}_n}\left( r \right) + r{a'_n} \left( r \right) - {n^2}{a_n}\left( r \right)} \Big)\cos n\varphi }}
+ {\Big( {{r^2}{b^{\prime\prime}_n}\left( r \right) + r{b'_n} \left( r \right) - {n^2}{b_n}\left( r \right)} \Big)\sin n\varphi } \Big] = 0.\]
Since this expression is equal to zero for all \(r\) and \(\varphi,\) we conclude that
\[{r^2}{a^{\prime\prime}_n}\left( r \right) + r{a'_n} \left( r \right) - {n^2}{a_n}\left( r \right) = 0\;\; \text{for}\;\;n = 0,1,2,3, \ldots\]
\[{r^2}{b^{\prime\prime}_n}\left( r \right) + r{b'_n} \left( r \right) - {n^2}{b_n}\left( r \right) = 0\;\; \text{for}\;\;n = 1,2,3, \ldots \]
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
\[{a_n}\left( r \right) = {a_n}\left( 1 \right){r^n},\;\; {b_n}\left( r \right) = {b_n}\left( 1 \right){r^n}\]
satisfy the last differential equations.
Here the constants \({a_n}\left( 1 \right)\) and \({b_n}\left( 1 \right)\) must be found from initial conditions to the ODEs. To get these initial conditions, we expand into Fourier series the function \(f\left( \varphi \right) = u\left( {1,\varphi } \right)\) defining the boundary conditions for the Laplace's equation. This yields:
One can show that using the formula \(\cos x = {\frac{{{e^{ix}} + {e^{ - ix}}}}{2}},\) the term in the square brackets reduces to the infinite geometric series whose sum is