Editing Fourier transform (section)

=== Analysis of differential equations ===
Perhaps the most important use of the Fourier transformation is to solve [[partial differential equation]]s.
Many of the equations of the mathematical physics of the nineteenth century can be treated this way. Fourier studied the heat equation, which in one dimension and in dimensionless units is
<math display="block">\frac{\partial^2 y(x, t)}{\partial^2 x} = \frac{\partial y(x, t)}{\partial t}.</math>
The example we will give, a slightly more difficult one, is the wave equation in one dimension,
<math display="block">\frac{\partial^2y(x, t)}{\partial^2 x} = \frac{\partial^2y(x, t)}{\partial^2t}.</math>

As usual, the problem is not to find a solution: there are infinitely many. The problem is that of the so-called "boundary problem": find a solution which satisfies the "boundary conditions"
<math display="block">y(x, 0) = f(x),\qquad \frac{\partial y(x, 0)}{\partial t} = g(x).</math>

Here, {{mvar|f}} and {{mvar|g}} are given functions. For the heat equation, only one boundary condition can be required (usually the first one). But for the wave equation, there are still infinitely many solutions {{mvar|y}} which satisfy the first boundary condition. But when one imposes both conditions, there is only one possible solution.

It is easier to find the Fourier transform {{mvar|ŷ}} of the solution than to find the solution directly. This is because the Fourier transformation takes differentiation into multiplication by the Fourier-dual variable, and so a partial differential equation applied to the original function is transformed into multiplication by polynomial functions of the dual variables applied to the transformed function. After {{mvar|ŷ}} is determined, we can apply the inverse Fourier transformation to find {{mvar|y}}.

Fourier's method is as follows. First, note that any function of the forms
<math display="block"> \cos\bigl(2\pi\xi(x\pm t)\bigr) \text{ or } \sin\bigl(2\pi\xi(x \pm t)\bigr)</math>
satisfies the wave equation. These are called the elementary solutions.

Second, note that therefore any integral
<math display="block">\begin{align}
  y(x, t) = \int_{0}^{\infty} d\xi \Bigl[ &a_+(\xi)\cos\bigl(2\pi\xi(x + t)\bigr) + a_-(\xi)\cos\bigl(2\pi\xi(x - t)\bigr) +{} \\
                                     &b_+(\xi)\sin\bigl(2\pi\xi(x + t)\bigr) + b_-(\xi)\sin\left(2\pi\xi(x - t)\right) \Bigr]
\end{align}</math>
satisfies the wave equation for arbitrary {{math|''a''<sub>+</sub>, ''a''<sub>−</sub>, ''b''<sub>+</sub>, ''b''<sub>−</sub>}}. This integral may be interpreted as a continuous linear combination of solutions for the linear equation.

Now this resembles the formula for the Fourier synthesis of a function. In fact, this is the real inverse Fourier transform of {{math|''a''<sub>±</sub>}} and {{math|''b''<sub>±</sub>}} in the variable {{mvar|x}}.

The third step is to examine how to find the specific unknown coefficient functions {{math|''a''<sub>±</sub>}} and {{math|''b''<sub>±</sub>}} that will lead to {{mvar|y}} satisfying the boundary conditions. We are interested in the values of these solutions at {{math|1=''t'' = 0}}. So we will set {{math|1=''t'' = 0}}. Assuming that the conditions needed for Fourier inversion are satisfied, we can then find the Fourier sine and cosine transforms (in the variable {{mvar|x}}) of both sides and obtain
<math display="block"> 2\int_{-\infty}^\infty y(x,0) \cos(2\pi\xi x) \, dx = a_+ + a_-</math>
and
<math display="block">2\int_{-\infty}^\infty y(x,0) \sin(2\pi\xi x) \, dx = b_+ + b_-.</math>

Similarly, taking the derivative of {{mvar|y}} with respect to {{mvar|t}} and then applying the Fourier sine and cosine transformations yields
<math display="block">2\int_{-\infty}^\infty \frac{\partial y(u,0)}{\partial t} \sin (2\pi\xi x) \, dx = (2\pi\xi)\left(-a_+ + a_-\right)</math>
and
<math display="block">2\int_{-\infty}^\infty \frac{\partial y(u,0)}{\partial t} \cos (2\pi\xi x) \, dx = (2\pi\xi)\left(b_+ - b_-\right).</math>

These are four linear equations for the four unknowns {{math|''a''<sub>±</sub>}} and {{math|''b''<sub>±</sub>}}, in terms of the Fourier sine and cosine transforms of the boundary conditions, which are easily solved by elementary algebra, provided that these transforms can be found.

In summary, we chose a set of elementary solutions, parametrized by {{mvar|ξ}}, of which the general solution would be a (continuous) linear combination in the form of an integral over the parameter {{mvar|ξ}}. But this integral was in the form of a Fourier integral. The next step was to express the boundary conditions in terms of these integrals, and set them equal to the given functions {{mvar|f}} and {{mvar|g}}. But these expressions also took the form of a Fourier integral because of the properties of the Fourier transform of a derivative. The last step was to exploit Fourier inversion by applying the Fourier transformation to both sides, thus obtaining expressions for the coefficient functions {{math|''a''<sub>±</sub>}} and {{math|''b''<sub>±</sub>}} in terms of the given boundary conditions {{mvar|f}} and {{mvar|g}}.

From a higher point of view, Fourier's procedure can be reformulated more conceptually. Since there are two variables, we will use the Fourier transformation in both {{mvar|x}} and {{mvar|t}} rather than operate as Fourier did, who only transformed in the spatial variables. Note that {{mvar|ŷ}} must be considered in the sense of a distribution since {{math|''y''(''x'', ''t'')}} is not going to be {{math|''L''<sup>1</sup>}}: as a wave, it will persist through time and thus is not a transient phenomenon. But it will be bounded and so its Fourier transform can be defined as a distribution. The operational properties of the Fourier transformation that are relevant to this equation are that it takes differentiation in {{mvar|x}} to multiplication by {{math|''i''2π''ξ''}} and differentiation with respect to {{mvar|t}} to multiplication by {{math|''i''2π''f''}} where {{mvar|f}} is the frequency. Then the wave equation becomes an algebraic equation in {{mvar|ŷ}}:
<math display="block">\xi^2 \hat y (\xi, f) = f^2 \hat y (\xi, f).</math>
This is equivalent to requiring {{math|1=''ŷ''(''ξ'', ''f'') = 0}} unless {{math|1=''ξ'' = ±''f''}}. Right away, this explains why the choice of elementary solutions we made earlier worked so well: obviously {{math|1=''f̂'' = ''δ''(''ξ'' ± ''f'')}} will be solutions. Applying Fourier inversion to these delta functions, we obtain the elementary solutions we picked earlier. But from the higher point of view, one does not pick elementary solutions, but rather considers the space of all distributions which are supported on the (degenerate) conic {{math|1=''ξ''{{isup|2}} − ''f''{{isup|2}} = 0}}.

We may as well consider the distributions supported on the conic that are given by distributions of one variable on the line {{math|1=''ξ'' = ''f''}} plus distributions on the line {{math|''ξ'' {{=}} −''f''}} as follows: if {{mvar|Φ}} is any test function,
<math display="block">\iint \hat y \phi(\xi,f) \, d\xi \, df = \int s_+ \phi(\xi,\xi) \, d\xi + \int s_- \phi(\xi,-\xi) \, d\xi,</math>
where {{math|''s''<sub>+</sub>}}, and {{math|''s''<sub>−</sub>}}, are distributions of one variable.

Then Fourier inversion gives, for the boundary conditions, something very similar to what we had more concretely above (put {{math|1=''Φ''(''ξ'', ''f'') = ''e''<sup>''i''2π(''xξ''+''tf'')</sup>}}, which is clearly of polynomial growth):
<math display="block"> y(x,0) = \int\bigl\{s_+(\xi) + s_-(\xi)\bigr\} e^{i 2\pi \xi x+0} \, d\xi </math>
and
<math display="block"> \frac{\partial y(x,0)}{\partial t} = \int\bigl\{s_+(\xi) - s_-(\xi)\bigr\} i 2\pi \xi e^{i 2\pi\xi x+0} \, d\xi.</math>

Now, as before, applying the one-variable Fourier transformation in the variable {{mvar|x}} to these functions of {{mvar|x}} yields two equations in the two unknown distributions {{math|''s''<sub>±</sub>}} (which can be taken to be ordinary functions if the boundary conditions are {{math|''L''<sup>1</sup>}} or {{math|''L''<sup>2</sup>}}).

From a calculational point of view, the drawback of course is that one must first calculate the Fourier transforms of the boundary conditions, then assemble the solution from these, and then calculate an inverse Fourier transform. Closed form formulas are rare, except when there is some geometric symmetry that can be exploited, and the numerical calculations are difficult because of the oscillatory nature of the integrals, which makes convergence slow and hard to estimate. For practical calculations, other methods are often used.

The twentieth century has seen the extension of these methods to all linear partial differential equations with polynomial coefficients, and by extending the notion of Fourier transformation to include Fourier integral operators, some non-linear equations as well.