### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - Jilong Bi

Pages: [1]
1
##### Term Test 2 / Re: TT2--P3
« on: March 23, 2018, 09:42:24 AM »
First separate of variable $$u(x,y) =X(x)Y(y),$$
take derivative
$$X''Y + XY'' = -\lambda XY$$
$$\implies \frac{X''}{X}+\frac{Y''}{Y}= -\lambda$$
By the given condition
$$X''+\lambda_1X = 0 , X'(0)+X'(a) = 0$$
This is Neumann boundary condition
As n = 0,$$\lambda_1 = 0 ,X_0 = \frac{1}{2}$$
As n = 1,2,...,$$\lambda_1 = \frac{n^2\pi ^2}{a^2},X_1 = \cos\frac{n\pi x}{a}$$
For Y, this is Dirichlet boundary condition  $$Y''+\lambda_2Y= 0 , Y(0)+Y(b) = 0$$
As m = 1,2,...,$$\lambda_2 = \frac{m^2\pi ^2}{b^2},Y_1 = \sin\frac{m\pi y}{b}$$
$\lambda =\lambda_1 + \lambda_2$, for n,m = 1,2,....
$$\implies \lambda = \pi^2(\frac{m^2}{b^2}+\frac{n^2}{a^2})$$
for n,m = 1,2,....
$$u(x,y) =\cos\frac{n\pi x}{a} \sin\frac{m\pi y}{b}$$
For n = 0, $\lambda$  = 0 and u(x,y) = 0

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

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
\begin{gather*}
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).
\end{gather*}
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
\end{gather*}
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
\end{gather*}

3
##### Quiz-5 / Quiz5 T0101
« on: March 08, 2018, 01:23:53 PM »
http://www.math.toronto.edu/courses/apm346h1/20181/PDE-textbook/Chapter5/S5.2.P.html 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 $$sin{\beta}x =\frac{ e^{i{\beta}x}- e^{-i{\beta}x}}{2i}$$
$$\implies \widehat {g}(k) = \frac{1}{2i\sqrt{2{\pi}a}}[e^{\frac{-(k-{\beta})^2}{2a}}-e^{\frac{-(k+{\beta})^2}{2a}}]$$

4
##### Term Test 1 / Re: P3
« on: February 15, 2018, 10:16:26 PM »
should $\phi(t) = 3cos(-\frac{1}{3}t) + constant$ ? And the answer for -t<x<2t is 3$cos(\frac{x-2t}{3}) -3$

5
##### Term Test 1 / Re: P5 Night
« on: February 15, 2018, 10:03:55 PM »
We know k = 1 ,so by formular:  $$\frac{1}{\sqrt{4{\pi}t}}\int_{-1}^1 e^{-\frac{(x-y)^2}{4t}}ydy$$  Let$$w = {\frac{(y-x)}{\sqrt{4t}}} \Rightarrow y =\sqrt{4t}w+x \Rightarrow dy = \sqrt{4t} dw$$ then the formular becomes  $$\frac{1}{\sqrt{4{\pi}t}}\int_{b}^a e^{-w^2}(\sqrt{4t}w+x) \sqrt{4t} dw \Rightarrow \frac{1}{\sqrt{{\pi}}}\int_{b}^a e^{-w^2}(\sqrt{4t}w+x) dw \Rightarrow \frac{1}{\sqrt{{\pi}}}\int_{b}^a \sqrt{4t}we^{-w^2} dw+ \frac{1}{\sqrt{{\pi}}}\int_{b}^a xe^{-w^2} dw$$  $$\frac{2\sqrt{t}}{\sqrt{{\pi}}}(-\frac{1}{2}e^{-a^2}+\frac{1}{2}e^{-b^2}) +\frac{x}{2}[erf(a)-erf(b)]$$ $$\Rightarrow\frac{\sqrt{t}}{\sqrt{{\pi}}} (\frac{1}{2}e^{-b^2}-\frac{1}{2}e^{-a^2} +\frac{x}{2}[erf(a)-erf(b)]$$ $$a = {\frac{1 -x}{\sqrt{4t}}}$$ $$b ={\frac{-1-x}{\sqrt{4t}}}$$

6
##### Term Test 1 / Re: P1
« on: February 15, 2018, 09:23:12 PM »
Characteristic Equation :$$\frac{dt}{1} = \frac{dx}{3(t^2-1)} = \frac{du}{6t^2}$$
a). From $$\frac{dt}{1} = \frac{dx}{3(t^2-1)} \\ (3t^2-3)dt = dx \\ \Rightarrow t^3 -3t +D =x \\ \Rightarrow D = x -t^3 -3t$$ b).From :$$\frac{dt}{1} = \frac{du}{6t^2} \\ 6t^2dt = du \\ \Rightarrow 2t^3+A = u\\ \Rightarrow u(x,t) = 2t^3+\phi(x-t^3-3t)$$ c). $$u(x,0) = \phi(x) = x \\ \Rightarrow u(x,t) = 2t^3+x-t^3-3t \\ \Rightarrow u(x,t) = t^3+x-3t$$

7
##### Term Test 1 / Re: P4
« on: February 15, 2018, 09:11:09 PM »
fist take derivative with respect to t,$$\frac{1}{2}\int_0^L(2u_tu_{tt}+c^22u_xu_{xt}+2auu_t)dx+\frac{ac^2}{2}2u_t(0,t)u_{tt}(0,t)+\frac{bc^2}{2}2u_{t}(L,t)u_{tt}(L,t)$$ By the boundary condition given, $$u_{tt} = c^2u_{xx}-au$$ $$\Rightarrow \int_0^L(c^2u_tu_{xx}-auu_t+c^2u_xu_{xt}+auu_t)dx+ac^2u_t(0,t)u_{tt}(0,t)+bc^2u_{t}(L,t)u_{tt}(L,t)$$ $$\Rightarrow c^2\int_0^L(u_tu_{xx}+u_xu_{xt})dx+ac^2u_t(0,t)u_{tt}(0,t)+bc^2u_{t}(L,t)u_{tt}(L,t)$$ Since $$u_tu_x$$ take derivative with respect to x is equal to $$u_tu_{xx}+u_xu_{xt}$$ $$\Rightarrow c^2 u_t(L,t)u_x(L,t) - c^2u_t(0,t)u_x(0,t)+ac^2u_t(0,t)u_{tt}(0,t)+bc^2u_{t}(L,t)u_{tt}(L,t)$$ By another boundary condition $$u_x(0,t) = au_{tt}(0,t)$$ $$\Rightarrow c^2 u_t(L,t)u_x(L,t) - c^2au_t(0,t)u_{tt}(0,t)+ac^2u_t(0,t)u_{tt}(0,t)+bc^2u_{t}(L,t)u_{tt}(L,t)$$ And for the last boundary condition $$-u_x(L,t)=bu_{tt}(L,t)$$$$\Rightarrow -c^2b u_t(L,t)u_{tt}(L,t) - c^2au_t(0,t)u_{tt}(0,t)+ac^2u_t(0,t)u_{tt}(0,t)+bc^2u_{t}(L,t)u_{tt}(L,t)$$  last step: $$-c^2b u_t(L,t)u_{tt}(L,t) +cb^2u_{t}(L,t)u_{tt}(L,t)- c^2au_t(0,t)u_{tt}(0,t)+ca^2u_t(0,t)u_{tt}(0,t)$$  All terms cancel ,so the equation equal to 0 and the energy does not depend on t.

8
##### Quiz-2 / Re: Q2-T0101
« on: February 03, 2018, 01:13:09 PM »
For the general solution :    u(x,t) =\frac{1}{2}[g(x+ct)+g(x-ct)]+\frac{1}{2c}
\int_{x-ct}^{x+ct}h(y)dy

$$\\$$for part(a):

u(x,t) =\frac{1}{2}[g(x+ct)+g(x-ct)]

$$u(x,t) = \begin{cases} 0, & \text{x-ct >1 , x+ct \geq1} \\ \frac{1}{2}, & \text{\vert x-ct \vert <1 , x+ct \geq1}\\ 1, & \text{\vert x-ct \vert <1 , \vert x+ct \vert<1}\\ 1, & \text{x-ct <-1 , x+ct \geq1}\\ \end{cases}$$for part(b):

u(x,t) =\frac{1}{2c}
\int_{x-ct}^{x+ct}h(y)dy

$$u(x,t) = \begin{cases} 0, & \text{x-ct >\frac{\pi}{2} , x+ct >\frac{\pi}{2}} \\ \frac{1}{2c}(1-\sin(x-ct), & \text{\vert x-ct \vert <\frac{\pi}{2} ,x+ct \geq\frac{\pi}{2}}\\ \frac{1}{2c}(\sin(x+ct)-\sin(x-ct), & \text{\vert x-ct \vert <\frac{\pi}{2} , \vert x+ct \vert<\frac{\pi}{2}}\\ \frac{1}{c}, & \text{x-ct <-\frac{\pi}{2} , x+ct \geq\frac{\pi}{2} }\\ \end{cases}$$

Pages: [1]