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Topics - Jilong Bi

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Quiz-6 / Quiz 6 T0101
« on: March 16, 2018, 01:44:27 AM »
& \Delta u:=u_{xx}+u_{yy}=0&& \text{in } r<a\\[3pt]
& u_r|_{r=a}=f(\theta).
where we use polar coordinates $(r,\theta)$ and $f(\theta)=\left\{\begin{aligned}
&1 &&0<\theta<\pi,\\
-&1 &&\pi<\theta<2\pi.

The expected answer: solution as a series.

Given the condition $r<a$. We can discard the logr and $r^{-n}$ part ,so the general solution is
u(r,\theta) =\frac{1}{2}A_0+\sum_{n=1}^\infty r^{n} (A_n {\cos}(n\theta)+B_n {\sin}(n\theta))\\
\implies u_r(r,\theta)=\sum_{n=1}^\infty nr^{n-1} (A_n {\cos}(n\theta)+B_n {\sin}(n\theta))\\
\implies u_r(a,\theta)=\sum_{n=1}^\infty na^{n-1} (A_n {\cos}(n\theta)+B_n {\sin}(n\theta)) = f(\theta).
For $n  = 0,1,2,3\ldots$
$$ A_n = \frac{1}{{\pi}na^{n-1}}\int_{-\pi}^{\pi} f(\theta){\cos}(n\theta)d\theta $$
For $n  = 1,2,3\ldots$
$$ B_n = \frac{1}{{\pi}na^{n-1}}\int_{-\pi}^{\pi} f(\theta){\sin}(n\theta)d\theta $$
Since $f(\theta)$ is odd function $\implies$ $A_n$ = 0 and
\begin{gather*}B_n = \frac{2}{{\pi}na^{n-1}}\int_{0}^{\pi} f(\theta){\sin}(n\theta)d\theta  \\
\implies B_n = \frac{2}{{\pi}na^{n-1}}\int_{0}^{\pi}{\sin}(n\theta)d\theta
We integral this get $-\frac{1}{n}\cos(n\theta)$ from 0 to $\pi$
$$\implies B_n = \frac{2}{{\pi}n^{2}a^{n-1}}[1-(-1)^{n}] $$
If we write n = 2m, it becomes 0, so we consider n = 2m+1
\begin{gather*}\implies B_n = \frac{4}{{\pi}(2m+1)^{2}a^{2m}}\\
u(r,\theta) =  \frac{4}{{\pi}}\sum_{m=0}^\infty \frac{r^{2m+1}}{(2m+1)^{2}a^{2m}} \sin(2m+1)\theta 

Quiz-5 / Quiz5 T0101
« on: March 08, 2018, 01:23:53 PM » problem4 1.
given $$f(x) =e^{\frac{-ax^2}{2}}$$
$$\implies\widehat {f}(k) = (\frac{1}{\sqrt{2{\pi}a}})e^{\frac{-k^2}{2a}}$$
By theorem $$g(x) = f(x)e^{i{\beta}x} \implies \widehat {g}(k) =  \widehat {f}(k-{\beta})$$
$$cos{\beta}x =\frac{ e^{i{\beta}x}+ e^{-i{\beta}x}}{2}$$
$$\implies \widehat {g}(k)  = \frac{1}{2}\widehat {f}(k-{\beta})+\frac{1}{2}\widehat {f}(k+{\beta})$$
$$\implies \widehat {g}(k)  =  \frac{1}{2\sqrt{2{\pi}a}}[e^{\frac{-(k-{\beta})^2}{2a}}+e^{\frac{-(k+{\beta})^2}{2a}}]$$
same reson for $$sin{\beta}x =\frac{ e^{i{\beta}x}- e^{-i{\beta}x}}{2i}$$
$$\implies \widehat {g}(k)  =  \frac{1}{2i\sqrt{2{\pi}a}}[e^{\frac{-(k-{\beta})^2}{2a}}-e^{\frac{-(k+{\beta})^2}{2a}}]$$

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