# 数学代写|偏微分方程代写partial difference equations代考|AMATH353

## 数学代写|偏微分方程代写partial difference equations代考|Term by Term Integration

A Fourier series of a piecewise smooth function $f(x)$ can always be integrated term by term and the result is a convergent infinite series that always converges to $\int_{-L}^L f(x) d x$ even if the original series has jumps.
Example 13
The Fourier series of $f(x)=1$ is
$$1 \sim \frac{4}{\pi} \sum_{n=1,3, \ldots} \frac{1}{n} \sin \frac{n \pi}{L} x$$
Integrate term by term from 0 to $x$ gives
\begin{aligned} x-0 & \sim-\left.\frac{4}{\pi} \sum_{n=1,3, \ldots} \frac{1}{n} \frac{L}{n \pi} \cos \frac{n \pi}{L} x\right|0 ^x \ &=-\frac{4}{\pi} \sum{n=1,3, \ldots} \frac{L}{n^2 \pi} \cos \frac{n \pi}{L} x+\frac{4}{\pi}\left[\frac{L}{1^2 \pi}+\frac{L}{3^2 \pi}+\frac{L}{5^2 \pi}+\cdots\right] \end{aligned}
The last sum is the constant term $(n=0)$ of the Fourier cosine series of $f(x)=x$, which is
$$\frac{1}{L} \int_0^L x d x=\frac{L}{2} .$$
Therefore
$$x \sim \frac{L}{2}-\frac{4 L}{\pi^2} \sum_{n=1,3, \ldots} \frac{1}{n^2} \cos \frac{n \pi}{L} x .$$
We have also found, as a by-product, the sum of the following infinite series
$$\frac{4 L}{\pi^2}\left[\frac{1}{1^2}+\frac{1}{3^2}+\cdots\right]=\frac{L}{2}$$
or
$$\frac{1}{1^2}+\frac{1}{3^2}+\cdots=\frac{\pi^2}{8}$$

## 数学代写|偏微分方程代写partial difference equations代考|Full solution of Several Problems

In this section we give the Fourier coefficients for each of the solutions in the previous chapter.
Example 14
$$\begin{gathered} u_t=k u_{x x}, \ u(0, t)=0, \ u(L, t)=0, \ u(x, 0)=f(x) . \end{gathered}$$
The solution given in the previous chapter is
$$u(x, t)=\sum_{n=1}^{\infty} b_n e^{-k\left(\frac{n \pi}{L}\right)^2 t} \sin \frac{n \pi}{L} x .$$
Upon substituting $t=0$ in (5.9.5) and using (5.9.4) we find that
$$f(x)=\sum_{n=1}^{\infty} b_n \sin \frac{n \pi}{L} x,$$
that is $b_n$ are the coefficients of the expansion of $f(x)$ into Fourier sine series. Therefore
$$b_n=\frac{2}{L} \int_0^L f(x) \sin \frac{n \pi}{L} x d x .$$
Example 15
$$\begin{gathered} u_t=k u_{x x}, \ u(0, t)=u(L, t), \ u_x(0, t)=u_x(L, t), \ u(x, 0)=f(x) . \end{gathered}$$
The solution found in the previous chapter is
$$u(x, t)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n \cos \frac{2 n \pi}{L} x+b_n \sin \frac{2 n \pi}{L} x\right) e^{-k\left(\frac{2 n \pi}{L}\right)^2 t},$$
As in the previous example, we take $t=0$ in (5.9.12) and compare with (5.9.11) we find that
$$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n \cos \frac{2 n \pi}{L} x+b_n \sin \frac{2 n \pi}{L} x\right) .$$

# 偏微分方程代考

## 数学代写|偏微分方程代写partial difference equations代考|Term by Term Integration

$$1 \sim \frac{4}{\pi} \sum_{n=1,3 \ldots} \frac{1}{n} \sin \frac{n \pi}{L} x$$

$$x-0 \sim-\frac{4}{\pi} \sum_{n=1,3, \ldots} \frac{1}{n} \frac{L}{n \pi} \cos \frac{n \pi}{L} x \mid 0^x \quad=-\frac{4}{\pi} \sum n=1,3, \ldots \frac{L}{n^2 \pi} \cos \frac{n \pi}{L} x+\frac{4}{\pi}\left[\frac{L}{1^2 \pi}+\frac{L}{3^2 \pi}\right.$$

$$\frac{1}{L} \int_0^L x d x=\frac{L}{2} .$$

$$x \sim \frac{L}{2}-\frac{4 L}{\pi^2} \sum_{n=1,3, \ldots} \frac{1}{n^2} \cos \frac{n \pi}{L} x .$$

$$\frac{4 L}{\pi^2}\left[\frac{1}{1^2}+\frac{1}{3^2}+\cdots\right]=\frac{L}{2}$$

$$\frac{1}{1^2}+\frac{1}{3^2}+\cdots=\frac{\pi^2}{8}$$

## 数学代写|偏微分方程代写partial difference equations代考|Full solution of Several Problems

$$u_t=k u_{x x}, u(0, t)=0, u(L, t)=0, u(x, 0)=f(x) .$$

$$u(x, t)=\sum_{n=1}^{\infty} b_n e^{-k\left(\frac{n \pi}{L}\right)^2 t} \sin \frac{n \pi}{L} x .$$

$$f(x)=\sum_{n=1}^{\infty} b_n \sin \frac{n \pi}{L} x,$$

$$b_n=\frac{2}{L} \int_0^L f(x) \sin \frac{n \pi}{L} x d x .$$

$$u_t=k u_{x x}, u(0, t)=u(L, t), u_x(0, t)=u_x(L, t), u(x, 0)=f(x) .$$

$$u(x, t)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n \cos \frac{2 n \pi}{L} x+b_n \sin \frac{2 n \pi}{L} x\right) e^{-k\left(\frac{2 n \pi}{L}\right)^2 t}$$

$$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n \cos \frac{2 n \pi}{L} x+b_n \sin \frac{2 n \pi}{L} x\right) .$$

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