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Solutions to Partial Differential Equations Using Fourier Series 📂Partial Differential Equations

Solutions to Partial Differential Equations Using Fourier Series

Definition

For the function f∈L2[−π,π]f \in \mathcal{L}^{2} [- \pi , \pi] in the Hilbert space, and for ak=1π∫−ππf(x)cos⁡kxdx\displaystyle a_{k} = {{1} \over {\pi}} \int_{- \pi}^{\pi} f(x) \cos kx dx and bk=1π∫−ππf(x)sin⁡kxdx\displaystyle b_{k} = {{1} \over {\pi}} \int_{- \pi}^{\pi} f(x) \sin kx dx,

f(x)∼a02+∑k=1∞(akcos⁡kx+bxsin⁡kx) f(x) \sim {{a_{0}} \over {2}} + \sum_{k=1}^{\infty} \left( a_{k} \cos kx + b_{x} \sin kx \right)

is called the Fourier series of ff.

Description

Unlike the Taylor series, which approximates a function with polynomials, the Fourier series approximates it with trigonometric polynomials. This complex form of the Fourier series can be applied to a variety of functions that are difficult to approach with the conventional Taylor series, making it useful. However, there is no guarantee that the Fourier series will converge, and even if it does converge, whether it will precisely converge to ff is a separate issue. For undergraduate-level solutions to partial differential equations, these issues are set aside for the time being, but one should be aware of these flaws.

For convenience, from now on, we will use == instead of ∼\sim. Let’s take a look at why the Fourier series is defined in such a form. Since L2\mathcal{L}^2 is in the case where Lp\mathcal{L}^{p} to p=2p=2, it is possible to define the inner product and norm for f,g:[−π,π]→Rf,g : [-\pi, \pi] \to \mathbb{R} as follows.

<f,g>:=1π∫−ππf(x)g(x)dx∥f∥:=<f,f>:=1π∫−ππ∣f(x)∣2dx \left< f, g \right> := {{1} \over{\pi}} \int_{-\pi}^{\pi} f(x) g(x) dx \left\| f \right\| : = \left< f, f \right> := \sqrt{ {{1} \over{\pi}} \int_{-\pi}^{\pi} \left| f(x) \right| ^2 dx }

For the even function sin⁡kx\sin k x and the odd function cos⁡kx\cos k x, there is <cos⁡kx,cos⁡lx>=<sin⁡kx,cos⁡lx>=0\displaystyle \left< \cos kx , \cos lx \right> = \left< \sin kx , \cos lx \right> = 0 for k≠lk \ne l. Of course, <cos⁡lx,sin⁡lx>=<sin⁡lx,cos⁡lx>=0\displaystyle \left< \cos lx , \sin lx \right> = \left< \sin lx , \cos lx \right> = 0 also holds for k=lk = l. The only case where 00 does not occur in the inner product of trigonometric functions is <cos⁡lx,cos⁡lx>≠0≠<sin⁡lx,sin⁡lx>\displaystyle \left< \cos lx , \cos lx \right> \ne 0 \ne \left< \sin lx , \sin lx \right>.

By using the above facts, calculating <f,cos⁡lx>f\left< f, \cos lx \right> f, we obtain

<f,cos⁡lx>=a02<1,cos⁡lx>+∑k=1∞(ak<cos⁡kx,cos⁡lx>+bx<sin⁡kx,cos⁡lx>)=al \left< f, \cos lx \right> = {{a_{0}} \over {2}} \left< 1 , \cos lx \right> + \sum_{k=1}^{\infty} \left( a_{k} \left< \cos kx , \cos lx \right> + b_{x} \left< \sin kx , \cos lx \right> \right) = a_{l}

Similarly, we can obtain <f,sin⁡lx>=bl\left< f , \sin lx \right> = b_{l}, and

<f,1>=a02<1,1>=a02∥1∥2=a02 \left< f , 1 \right> = {{a_{0}} \over {2}} \left< 1, 1 \right> = {{a_{0}} \over {2}} \left\| 1 \right\| ^2 = {{a_{0}} \over {2}}

it is.

Approximating ff with trigonometric polynomials means projecting ff into the vector space of trigonometric functions. From the perspective of linear algebra, thinking about each term means that it looks like either <f(x),cos⁡nx>cos⁡nx\left< f(x) , \cos nx \right> \cos nx or <f(x),sin⁡nx>sin⁡nx\left< f(x) , \sin nx \right> \sin nx, which can be seen as projecting the function ff onto the basis {cos⁡nx,sin⁡nx ∣ n∈N}\left\{ \cos nx , \sin nx \ | \ n \in \mathbb{N} \right\}.

Using the properties of odd and even functions, not only this but also the following formula can significantly reduce calculations.

If f∈L2[−π,π]f \in \mathcal{L}^{2} [-\pi , \pi ] is an odd function, for ak=2π∫0πf(x)cos⁡kxdx\displaystyle a_{k} = {{2} \over {\pi}} \int_{0} \pi f(x) \cos kx dx

f(x)∼a02+∑k=1∞akcos⁡kx f(x) \sim {{a_{0 } } \over {2 }} + \sum_{k=1}^{\infty} a_{k} \cos kx

If f∈L2[−π,π]f \in \mathcal{L}^{2} [-\pi , \pi ] is an even function, for bk=2π∫0πf(x)sin⁡kxdx\displaystyle b_{k} = {{2} \over {\pi}} \int_{0} \pi f(x) \sin kx dx

f(x)∼∑k=1∞bksin⁡kx f(x) \sim \sum_{k=1}^{\infty} b_{k} \sin kx

Meanwhile, by introducing the following definition for f,g:[−π,π]→Cf,g : [-\pi, \pi] \to \mathbb{C}, a generalization to complex numbers is possible.

<f,g>:=12π∫−ππf(x)g(x)‾dx \left< f, g \right> := {{1} \over{2 \pi}} \int_{-\pi}^{\pi} f(x) \overline{ g(x) } dx

∥f∥:=<f,f>:=12π∫−ππ∣f(x)∣2dx \left\| f \right\| : = \left< f, f \right> := \sqrt{ {{1} \over{2 \pi}} \int_{-\pi}^{\pi} \left| f(x) \right| ^2 dx }

f(x)∼∑k∈Zckeikx=⋯+c−2e−i2x+c−1e−ix+c0+c1eix+c2ei2x+⋯ f(x) \sim \sum_{k \in \mathbb{Z}} c_{k} e^{ikx} = \cdots +c_{-2}e^{-i 2 x } + c_{-1}e^{-i x } + c_{0} +c_{1}e^{i x } + c_{2}e^{i 2 x } + \cdots

ck=<f,eikx>=12π∫−ππf(x)e−ikxdx c_{k} = \left<f , e^{ikx} \right> = {{1} \over {2 \pi}} \int_{-\pi}^{\pi} f(x) e^{-ikx} dx

From Euler’s formula, since there is cos⁡kx=eikx+e−ikx2\displaystyle \cos kx = {{e^{ikx} + e^{-ikx} } \over {2}} and sin⁡kx=eikx−e−ikx2i\displaystyle \sin kx = {{e^{ikx} - e^{-ikx} } \over {2i}},

ck+c−k=12π∫−ππf(x)(eikx+e−ikx)dx=12π∫−ππf(x)⋅2cos⁡kxdx=1π∫−ππf(x)cos⁡kxdx=ak \begin{align*} c_{k} + c_{-k} =& {{1 } \over {2 \pi }} \int_{-\pi}^{\pi} f(x) \left(e^{ikx} + e^{-ikx} \right) dx \\ =& {{1 } \over { 2 \pi }} \int_{-\pi}^{\pi} f(x) \cdot 2 \cos kx dx \\ =& {{1 } \over { \pi }} \int_{-\pi}^{\pi} f(x) \cos kx dx \\ =& a_{k} \end{align*}

Similarly, we can obtain bk=i(ck−c−k)b_{k} = i (c_{k} - c_{-k}).

Therefore, although it seems that the standardization constant has changed at first glance, in reality, it properly covers f,g:[−π,π]→Rf,g : [-\pi, \pi] \to \mathbb{R}. As you can see, to understand these techniques, even roughly, background knowledge in linear algebra, complex analysis, and real analysis is required.

See also