1
Final Exam / Re: FE Problem 1
« on: April 14, 2015, 11:40:50 PM »
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2
Final Exam / Re: FE Problem 2
« on: April 14, 2015, 11:39:04 PM »
\begin{align}
u(x,t)&=\frac{1}{\sqrt{4k\pi t}}[\int_0 ^\infty e^{-\frac{(x-y)^2}{4kt}}f(y) dy
+\int_0 ^\infty e^{-\frac{(x+y)^2}{4kt}}f(y)]\\
&=\frac{1}{\sqrt{4k\pi t}}[\int_0 ^1 e^{-\frac{(x-y)^2}{4kt}}+ e^{-\frac{(x+y)^2}{4kt}}dy]\\
&=\frac{1}{\sqrt{12\pi t}}[\int_0 ^1 e^{-\frac{(x-y)^2}{12t}}+ e^{-\frac{(x+y)^2}{12t}}dy]\\
\end{align}
Let $ I_1(x) = \int_0 ^1 e^{-\frac{(x-y)^2}{2}}$ then $I_2(x)= I_!(-x)$\\
Let $z_1 = (x-y)$, we get
\begin{align}
I_1(x) &= \int_0 ^1 e^{-\frac{(x-y)^2}{2}}dy\\
&=\int_{x-1} ^{x} e^{-\frac{z^2}{2}}dz\\
&=\int_{0} ^{x-1} e^{-\frac{z^2}{2}}dz+\int_{0} ^{x} e^{-\frac{z^2}{2}}dz\\
&=\sqrt{\frac{\pi}{2}}(erf(x-1)+erf(x)))\\
I_2(x)&=\sqrt{\frac{\pi}{2}}(erf(-x-1)+erf(-x)))
\end{align}
Then, we get
\begin{align}
u(x, t) &= \frac{1}{\sqrt{12\pi t}}e^{\frac{1}{6t}}[\sqrt{\frac{\pi}{2}}(erf(x-1)+erf(x)))+\sqrt{\frac{\pi}{2}}(erf(-x-1)+erf(-x)))]\\
&= \frac{1}{\sqrt{24t}}e^{\frac{1}{6t}}(erf(x-1)+erf(x)+erf(-x-1)+erf(-x)]\\
\end{align}
u(x,t)&=\frac{1}{\sqrt{4k\pi t}}[\int_0 ^\infty e^{-\frac{(x-y)^2}{4kt}}f(y) dy
+\int_0 ^\infty e^{-\frac{(x+y)^2}{4kt}}f(y)]\\
&=\frac{1}{\sqrt{4k\pi t}}[\int_0 ^1 e^{-\frac{(x-y)^2}{4kt}}+ e^{-\frac{(x+y)^2}{4kt}}dy]\\
&=\frac{1}{\sqrt{12\pi t}}[\int_0 ^1 e^{-\frac{(x-y)^2}{12t}}+ e^{-\frac{(x+y)^2}{12t}}dy]\\
\end{align}
Let $ I_1(x) = \int_0 ^1 e^{-\frac{(x-y)^2}{2}}$ then $I_2(x)= I_!(-x)$\\
Let $z_1 = (x-y)$, we get
\begin{align}
I_1(x) &= \int_0 ^1 e^{-\frac{(x-y)^2}{2}}dy\\
&=\int_{x-1} ^{x} e^{-\frac{z^2}{2}}dz\\
&=\int_{0} ^{x-1} e^{-\frac{z^2}{2}}dz+\int_{0} ^{x} e^{-\frac{z^2}{2}}dz\\
&=\sqrt{\frac{\pi}{2}}(erf(x-1)+erf(x)))\\
I_2(x)&=\sqrt{\frac{\pi}{2}}(erf(-x-1)+erf(-x)))
\end{align}
Then, we get
\begin{align}
u(x, t) &= \frac{1}{\sqrt{12\pi t}}e^{\frac{1}{6t}}[\sqrt{\frac{\pi}{2}}(erf(x-1)+erf(x)))+\sqrt{\frac{\pi}{2}}(erf(-x-1)+erf(-x)))]\\
&= \frac{1}{\sqrt{24t}}e^{\frac{1}{6t}}(erf(x-1)+erf(x)+erf(-x-1)+erf(-x)]\\
\end{align}
3
HA10 / Re: HA10 Problem 1
« on: April 02, 2015, 08:43:07 PM »
\begin{align*}
&\Phi(u):=\frac{1}{2}\int_0^2 u_{xx}^2\,dx,\\
&\Psi(u)= \frac{1}{2}\int_0^2 u_{x}^2\,dx =1,\\
&u(0)=u'(0)=u(2)=u'(2)=0.
\end{align*}
a. Write down Euler–Lagrange equation $\delta (\Phi-\lambda \Psi)=0$.
Let $L (x, u, u', u'') = u_{xx}^2$
then
\begin{align}
\delta \Phi&=
\iiint_\Omega\Bigl(\frac{\partial L}{\partial u} - \sum_{1\le j\le n} \frac{\partial\ }{\partial x_j} \frac{\partial L}{\partial u_{x_j}} \Bigr)\delta u \,dx \\
&=\frac{\partial^2}{\partial x^2}2u_{xx}\\
&=2u_{xxxx}
\end{align}
similarly,
\begin{align}
L_2(x,u,u') = u_x^2\\
\delta\Psi &=\frac{\partial L}{\partial u}-\frac{\partial}{\partial x}\frac{\partial L}{\partial u_x}\\
&=-\frac{\partial}{\partial x}2u_x\\
&=-2u_{xx}
\end{align}
Thus
\begin{equation}
\delta(\Phi-\lambda\Psi) = u_{xxxx}+\lambda u_{xx}=0
\end{equation}
b. Under boundary conditions (\ref{eq-10.3}) solve it and find out eigenvalues $\lambda$ for each solution exists.
Characteristic Equation:
\begin{equation}
k^4+\lambda k^2 = 0\\
k = 0,\sqrt{\lambda}i, -\sqrt{\lambda}i
\end{equation}
Let $\lambda = \omega ^2$
Then $u = A + Bx+C\cos(\omega x)+D\sin(\omega x)$
By hint, we have $u(-1)=u'(-1)=u(1)=u'(-1) = 0$
Even eigenfunction is:
\begin{align}
X &= A+C\cos(\omega x)\\
X(1) &= A+C\cos(\omega)=0\\
X'(1) &= -C\sin(\omega) = 0\\
\sin(\omega) &=0 \implies \omega = n\pi\\
\lambda_n &= \omega ^2=(n\pi)^2\\
A+C\cos(\omega) = 0\\
A = 1\\
C = (-1)^{n+1}\\
\end{align}
$X_n = 1+ (-1)^{n+1}cos(nx) $ is the even eigenfunction
Odd function
\begin{align}
X &= Bx + D \sin (\omega x)\\
X(1) &=B+D\cos(\omega ) = 0\\
X'(1) &=B+D\cos(\omega)\omega = 0\\
\cos(\omega)\omega &= \sin(\omega)\\
\omega&=\tan(\omega)\\
\end{align}
Then the graph $\omega$ and $\tan(\omega)$ intersections are the eigenvalues
&\Phi(u):=\frac{1}{2}\int_0^2 u_{xx}^2\,dx,\\
&\Psi(u)= \frac{1}{2}\int_0^2 u_{x}^2\,dx =1,\\
&u(0)=u'(0)=u(2)=u'(2)=0.
\end{align*}
a. Write down Euler–Lagrange equation $\delta (\Phi-\lambda \Psi)=0$.
Let $L (x, u, u', u'') = u_{xx}^2$
then
\begin{align}
\delta \Phi&=
\iiint_\Omega\Bigl(\frac{\partial L}{\partial u} - \sum_{1\le j\le n} \frac{\partial\ }{\partial x_j} \frac{\partial L}{\partial u_{x_j}} \Bigr)\delta u \,dx \\
&=\frac{\partial^2}{\partial x^2}2u_{xx}\\
&=2u_{xxxx}
\end{align}
similarly,
\begin{align}
L_2(x,u,u') = u_x^2\\
\delta\Psi &=\frac{\partial L}{\partial u}-\frac{\partial}{\partial x}\frac{\partial L}{\partial u_x}\\
&=-\frac{\partial}{\partial x}2u_x\\
&=-2u_{xx}
\end{align}
Thus
\begin{equation}
\delta(\Phi-\lambda\Psi) = u_{xxxx}+\lambda u_{xx}=0
\end{equation}
b. Under boundary conditions (\ref{eq-10.3}) solve it and find out eigenvalues $\lambda$ for each solution exists.
Characteristic Equation:
\begin{equation}
k^4+\lambda k^2 = 0\\
k = 0,\sqrt{\lambda}i, -\sqrt{\lambda}i
\end{equation}
Let $\lambda = \omega ^2$
Then $u = A + Bx+C\cos(\omega x)+D\sin(\omega x)$
By hint, we have $u(-1)=u'(-1)=u(1)=u'(-1) = 0$
Even eigenfunction is:
\begin{align}
X &= A+C\cos(\omega x)\\
X(1) &= A+C\cos(\omega)=0\\
X'(1) &= -C\sin(\omega) = 0\\
\sin(\omega) &=0 \implies \omega = n\pi\\
\lambda_n &= \omega ^2=(n\pi)^2\\
A+C\cos(\omega) = 0\\
A = 1\\
C = (-1)^{n+1}\\
\end{align}
$X_n = 1+ (-1)^{n+1}cos(nx) $ is the even eigenfunction
Odd function
\begin{align}
X &= Bx + D \sin (\omega x)\\
X(1) &=B+D\cos(\omega ) = 0\\
X'(1) &=B+D\cos(\omega)\omega = 0\\
\cos(\omega)\omega &= \sin(\omega)\\
\omega&=\tan(\omega)\\
\end{align}
Then the graph $\omega$ and $\tan(\omega)$ intersections are the eigenvalues
4
HA9 / Re: HA9 Problem 1
« on: March 26, 2015, 11:26:42 PM »
Q1
From lecture section 28.1, we have
\begin{equation}
\Delta =\partial_\rho^2 + \frac{2}{\rho}\partial_\rho +
\frac{1}{\rho^2}\Lambda
\end{equation}
with
\begin{equation}
\Lambda:=
\bigl(\partial_{\phi}^2 + \cot(\phi)\partial_\phi \bigr) +\frac{1}{\sin^2(\phi)}\partial_{\theta}^2.
\end{equation}
Plug $u=P(\rho)Y(\phi,\theta)$ into $\Delta u=0$: \begin{equation}
P''(\rho)Y(\phi,\theta) + \frac{2}{\rho}P' (\rho)Y(\phi,\theta) + \frac{1}{\rho^2} P(\rho)\Lambda Y(\phi,\theta)=0
\end{equation}
which could be rewritten as
\begin{equation}
\frac{\rho^2 P''(\rho) + \rho P' (\rho)}{P(\rho)}+
\frac{\Lambda Y(\phi,\theta)}{Y(\phi,\theta)}=0
\end{equation}
and since the first term depends only on $\rho$ and the second only on$\phi, \theta$ we conclude that both are constant:
\begin{align}
\rho^2 P'' +2\rho P' = \lambda P,\\
\Lambda Y(\phi,\theta)=-\lambda Y(\phi,\theta).
\end{align}
The first equation is of Euler type and it has solutions $P:=\rho^l$ iff
$\lambda= l(l+1)$. However if we are considering ball, solution must be infinitely smooth in its center due to some general properties of Laplace equation and this is possible iff $l=0,1,2,\ldots$ and in this case $u$ must be a polynomial of $(x,y,z)$.Which in this case, u depends on $x^2+y^2+z^2$ and $z$
Then suppose that
\begin{align}
u= &z^4 + a (1-x^2-y^2-z^2) + bz^2 (1-x^2-y^2-z^2) + c(1-x^2-y^2-z^2) ^2\\\\
=&z^4 + a (1-x^2-y^2-z^2) + b (z^2-z^2x^2-z^2y^2-z^4) + c(1-x^2-y^2-z^2) ^2
\end{align}
\begin{align}
\Delta u &= uxx+uyy+uzz\\
&=12 z^2 -6a + 2b( 1 -\rho^2 -7 z^2) -6c +20c\rho^2\\\text{Suppose $1-\rho^2-z^2=0$}\\
&=12z^2-6a+2b(-6z^2)-6c+20c(1-z^2)\\
&=12z^2-6a-12bz^2-6c+20c-20cz^2\\
&=(12-12b-20c)z^2-6a-6c=0
\end{align}
Then we have
\begin{align}
12-12b-20c=0\\
-6a-6c=0
\end{align}
Then $a = -c$, $b=1-\frac{5}{3}c$, and c is arbitrary.
From lecture section 28.1, we have
\begin{equation}
\Delta =\partial_\rho^2 + \frac{2}{\rho}\partial_\rho +
\frac{1}{\rho^2}\Lambda
\end{equation}
with
\begin{equation}
\Lambda:=
\bigl(\partial_{\phi}^2 + \cot(\phi)\partial_\phi \bigr) +\frac{1}{\sin^2(\phi)}\partial_{\theta}^2.
\end{equation}
Plug $u=P(\rho)Y(\phi,\theta)$ into $\Delta u=0$: \begin{equation}
P''(\rho)Y(\phi,\theta) + \frac{2}{\rho}P' (\rho)Y(\phi,\theta) + \frac{1}{\rho^2} P(\rho)\Lambda Y(\phi,\theta)=0
\end{equation}
which could be rewritten as
\begin{equation}
\frac{\rho^2 P''(\rho) + \rho P' (\rho)}{P(\rho)}+
\frac{\Lambda Y(\phi,\theta)}{Y(\phi,\theta)}=0
\end{equation}
and since the first term depends only on $\rho$ and the second only on$\phi, \theta$ we conclude that both are constant:
\begin{align}
\rho^2 P'' +2\rho P' = \lambda P,\\
\Lambda Y(\phi,\theta)=-\lambda Y(\phi,\theta).
\end{align}
The first equation is of Euler type and it has solutions $P:=\rho^l$ iff
$\lambda= l(l+1)$. However if we are considering ball, solution must be infinitely smooth in its center due to some general properties of Laplace equation and this is possible iff $l=0,1,2,\ldots$ and in this case $u$ must be a polynomial of $(x,y,z)$.Which in this case, u depends on $x^2+y^2+z^2$ and $z$
Then suppose that
\begin{align}
u= &z^4 + a (1-x^2-y^2-z^2) + bz^2 (1-x^2-y^2-z^2) + c(1-x^2-y^2-z^2) ^2\\\\
=&z^4 + a (1-x^2-y^2-z^2) + b (z^2-z^2x^2-z^2y^2-z^4) + c(1-x^2-y^2-z^2) ^2
\end{align}
\begin{align}
\Delta u &= uxx+uyy+uzz\\
&=12 z^2 -6a + 2b( 1 -\rho^2 -7 z^2) -6c +20c\rho^2\\\text{Suppose $1-\rho^2-z^2=0$}\\
&=12z^2-6a+2b(-6z^2)-6c+20c(1-z^2)\\
&=12z^2-6a-12bz^2-6c+20c-20cz^2\\
&=(12-12b-20c)z^2-6a-6c=0
\end{align}
Then we have
\begin{align}
12-12b-20c=0\\
-6a-6c=0
\end{align}
Then $a = -c$, $b=1-\frac{5}{3}c$, and c is arbitrary.
5
Web Bonus Problems / Re: Web Bonus Problem 4
« on: March 11, 2015, 09:58:31 PM »
Now hint:
a) consider $u_s (x,t)= s^l u(s x, s^k t)$ and prove that if $u$ satisfies original problem then $u_s$ satisfies it for all $s>0$ iff $k=1$, $l=0$.
b) So, $u_s(x,t)= u(sx, st)$ satisfies it and we are interested in self-similar solution $u(x,t)=u(sx,st)$ for all $s>0$. Plugging $s=t^{-1}$ we arrive to $u(x,t)=v (xt^{-1})$ (with $v(y)=u(y, 1)$.
c) Plugging $u(x,t)=v (xt^{-1})$ into original equation we have an ODE. Which?
d) Find continuous solution of this ODE such that $v(y)=-1$ as $y<-1$ and $v(y)=1$ as $y>1$ (Think why).
e) Plug into $u$
a) $s^k = s$ because one derivative with respect to t is also one derivative with respect x. Thus $k = 1$;
By total energy rule, $\int_{-\infty}^{\infty} u_s dx = s^l\int _{-\infty}^{\infty}u dx$ Thus $s^l = 1\implies l=0$
b) $u_s= u(sx,st)$. Let $s= t^{-1}$ we have $u_s= u(t^{-1}x, 1) = v(t^{-1}x)$
c) plug in $v(t^{-1}x)$
\begin{align}
ut&=-t^{-2}xv'(t^{-1}x)\\
ux&=v'(t^{-1}x)t^{-1}\\
-t^{-2}xv'(t^{-1}x)&=v(t^{-1}x)(v'(t^{-1}x)t^{-1}\\
-t^{-1}xv'(t^{-1}x)&=v(t^{-1}x)(v'(t^{-1}x)\\
\end{align}
Let $t^{-1}x=\xi$ we get
\begin{equation}
-\xi v'(\xi)=v(\xi)v'(\xi)
\label{A}
\end{equation}
$-\xi = v(\xi)$ is the ODE?
6
Web Bonus Problems / Re: Web Bonus Problem 4
« on: March 07, 2015, 07:57:40 PM »No, this does not fly. To construct a solution you need to fill by characteristics the whole half-plane $t>0$. So far you covered only $x<-t$ where $u=-1$ and $x>t$ where $u=1$ leaving sector $-t<x<t$ empty.Sorry professor I quoted my post accidentally, could you help me delete the previous post? -t<x<t sector has two values which is definitely not right but I don't know what do you mean apply the self-similar solution, can you give me more hints please?
You need to apply the method of self-similar solutions to find continuous $u(x,t)$ there.
7
Web Bonus Problems / Re: Web Bonus Problem 4
« on: March 06, 2015, 12:13:48 AM »
Characteristic equation:
\begin{align}
\frac{dt}{1} &=\ \frac{dx}{-u} = \frac{du}{0}\\
\frac{du}{dt}&=\ 0 \implies u=f(C)\\
x &=\ C - tu\\
C &=\ x+tu
\end{align}
Impose boundary condition: $t=0$ and $x = C$,
\begin{align}
u &=\ f(C) =f(x+tu)\\
u|_{t=0}&=f(x) =\left\{\begin{aligned}
-1& &x\le 0,\\
1& &x> 0
\end{aligned}\right.
\end{align}
then,
\end{align}
\begin{align}
\frac{dt}{1} &=\ \frac{dx}{-u} = \frac{du}{0}\\
\frac{du}{dt}&=\ 0 \implies u=f(C)\\
x &=\ C - tu\\
C &=\ x+tu
\end{align}
Impose boundary condition: $t=0$ and $x = C$,
\begin{align}
u &=\ f(C) =f(x+tu)\\
u|_{t=0}&=f(x) =\left\{\begin{aligned}
-1& &x\le 0,\\
1& &x> 0
\end{aligned}\right.
\end{align}
then,
\end{align}
8
HA6 / Re: HA6 problem 1
« on: March 05, 2015, 10:15:48 PM »
a.
\begin{align}
\hat{f}(\omega) &=\ \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\alpha |x|}e^{-i\omega x}dx\\
&=\ \frac{1}{2\pi}[\int_{-\infty}^{0} e^{\alpha x -i\omega x} +\int_{0}^{\infty} e^{-\alpha x -i\omega x}]dx\\
&= \frac{1}{2\pi}[\int_{-\infty}^{0} (\cos (\omega x)-i\sin (\omega x))e^{\alpha x } +\int_{0}^{\infty} (\cos (\omega x)-i\sin (\omega x))e^{-\alpha x}]dx\\
&= \frac{1}{2\pi}[\int_{0}^{\infty} (\cos (\omega x)+i\sin (\omega x))e^{-\alpha x } +\int_{0}^{\infty} (\cos (\omega x)-i\sin (\omega x))e^{-\alpha x}]dx\\
&= \frac{1}{2\pi}[\int_{0}^{\infty} (\cos (\omega x)+i\sin (\omega x)+\cos (\omega x)-i\sin (\omega x))e^{-\alpha x}]dx\\
&= \frac{1}{2\pi}[\int_{0}^{\infty} (\cos (\omega x)+\cos (\omega x))e^{-\alpha x}]dx\\
&= \frac{2}{2\pi}\int_{0}^{\infty}\cos(\omega x)e^{-\alpha x} dx\\
&= \frac{\alpha}{\pi (\alpha ^2 +\omega ^2)}
\end{align}
b.
1)
Let $ f(\omega) = e^{-\alpha |x|}$, and let
\begin{align}
g(x) &=\ e^{-\alpha |x|}\cos(\beta x)\\
&=\ e^{-\alpha |x|}(\frac{1}{2}e^{i\beta x}+\frac{1}{2}e^{-i\beta x})\: \mbox{ (By Trig identity)}\\
&=\ \frac{1}{2}e^{-\alpha |x|}e^{i\beta x}+\frac{1}{2}e^{-\alpha |x|}e^{-i\beta x}
\end{align}
Which satisfies the form $\hat{g}(\omega) = \frac{1}{2}\hat{f}(\omega - \beta) + \frac{1}{2}\hat{f}(\omega + \beta)$.
From part a we have found $\hat{f}\frac{\alpha}{\pi (\alpha ^2 +\omega ^2)}$
Thus\begin{align}
\hat{g}(\omega) = \frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega - \beta) ^2})+\frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega + \beta) ^2)})
\end{align}
2)
Let $ f(\omega) = e^{-\alpha |x|}$, and let
\begin{align}
g(x) &=\ e^{-\alpha |x|}\sin(\beta x)\\
&=\ e^{-\alpha |x|}(\frac{1}{2i}e^{i\beta x}-\frac{1}{2i}e^{-i\beta x})\: \mbox{ (By Trig identity)}\\
&=\ \frac{1}{2i}e^{-\alpha |x|}e^{i\beta x}-\frac{1}{2i}e^{-\alpha |x|}e^{-i\beta x}
\end{align}
Which satisfies the form $\hat{g}(\omega) = \frac{1}{2i}\hat{f}(\omega - \beta) - \frac{1}{2i}\hat{f}(\omega + \beta)$.
From part a we have found $\hat{f}\frac{\alpha}{\pi (\alpha ^2 +\omega ^2)}$
Thus\begin{align}
\hat{g}(\omega) = \frac{1}{2i}(\frac{\alpha}{\pi (\alpha ^2 +(\omega - \beta) ^2)})-\frac{1}{2i}(\frac{\alpha}{\pi (\alpha ^2 +(\omega + \beta) ^2)})
\end{align}
c.
Let $ f(\omega) = e^{-\alpha |x|}$, and let
$g(x) = xe^{-\alpha |x|}$
By the theorem $ \hat{g}(\omega) = i\hat{f'}(\omega)$, We have
\begin{align}
\hat{g}(\omega) &=\ i \frac{d\hat{f}(\omega)}{d\omega}\\
&=\ i \frac{d}{d\omega }(\frac{\alpha }{\pi (\alpha ^2 +\omega ^2)})\\
&=\ i \frac{\alpha}{\pi}\frac{-2\omega}{(\alpha ^2 +\omega ^2)^2}\\
\end{align}
d.
Similar to part c),
Let $ f(\omega) = e^{-\alpha |x|}cos(\beta x)$, and let $g(x) = xe^{-\alpha |x|}cos(\beta x)$
By the theorem $ \hat{g}(\omega) = i\hat{f'}(\omega)$, where
$\hat{f}(\omega) = \frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega - \beta) ^2})+\frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega + \beta) ^2)})$
We have
\begin{align}
\hat{g}(\omega) &=\ i \frac{d\hat{f}(\omega)}{d\omega}\\
&=\ i \frac{d}{d\omega }\hat{f}(\omega)\\
& =\ i \frac{d}{d\omega }(\frac{\alpha}{\pi (\alpha ^2 +(\omega - \beta) ^2})+\frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega + \beta) ^2)})\\
&=\ i\frac{\alpha}{\pi}[(\frac{-(\omega - \beta)}{(\alpha^2 + (\omega - \beta)^2)^2})+\frac{-(\omega + \beta)}{(\alpha^2 + (\omega + \beta)^2)^2}]\\
&=\ \frac{-\alpha i}{\pi}[\frac{\omega - \beta}{(\alpha^2 + (\omega - \beta)^2)^2}+\frac{\omega + \beta}{(\alpha^2 + (\omega + \beta)^2)^2}]
\end{align}
Similar to above, we can get $g(x) = xe^{-\alpha |x|}sin(\beta x)$ by applying part b.2 formula and use the method we used in part c, then we get:
$\hat{g}(\omega) = \frac{-\alpha }{\pi}[\frac{\omega - \beta}{(\alpha^2 + (\omega - \beta)^2)^2}-\frac{\omega + \beta}{(\alpha^2 + (\omega + \beta)^2)^2}]$
\begin{align}
\hat{f}(\omega) &=\ \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\alpha |x|}e^{-i\omega x}dx\\
&=\ \frac{1}{2\pi}[\int_{-\infty}^{0} e^{\alpha x -i\omega x} +\int_{0}^{\infty} e^{-\alpha x -i\omega x}]dx\\
&= \frac{1}{2\pi}[\int_{-\infty}^{0} (\cos (\omega x)-i\sin (\omega x))e^{\alpha x } +\int_{0}^{\infty} (\cos (\omega x)-i\sin (\omega x))e^{-\alpha x}]dx\\
&= \frac{1}{2\pi}[\int_{0}^{\infty} (\cos (\omega x)+i\sin (\omega x))e^{-\alpha x } +\int_{0}^{\infty} (\cos (\omega x)-i\sin (\omega x))e^{-\alpha x}]dx\\
&= \frac{1}{2\pi}[\int_{0}^{\infty} (\cos (\omega x)+i\sin (\omega x)+\cos (\omega x)-i\sin (\omega x))e^{-\alpha x}]dx\\
&= \frac{1}{2\pi}[\int_{0}^{\infty} (\cos (\omega x)+\cos (\omega x))e^{-\alpha x}]dx\\
&= \frac{2}{2\pi}\int_{0}^{\infty}\cos(\omega x)e^{-\alpha x} dx\\
&= \frac{\alpha}{\pi (\alpha ^2 +\omega ^2)}
\end{align}
b.
1)
Let $ f(\omega) = e^{-\alpha |x|}$, and let
\begin{align}
g(x) &=\ e^{-\alpha |x|}\cos(\beta x)\\
&=\ e^{-\alpha |x|}(\frac{1}{2}e^{i\beta x}+\frac{1}{2}e^{-i\beta x})\: \mbox{ (By Trig identity)}\\
&=\ \frac{1}{2}e^{-\alpha |x|}e^{i\beta x}+\frac{1}{2}e^{-\alpha |x|}e^{-i\beta x}
\end{align}
Which satisfies the form $\hat{g}(\omega) = \frac{1}{2}\hat{f}(\omega - \beta) + \frac{1}{2}\hat{f}(\omega + \beta)$.
From part a we have found $\hat{f}\frac{\alpha}{\pi (\alpha ^2 +\omega ^2)}$
Thus\begin{align}
\hat{g}(\omega) = \frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega - \beta) ^2})+\frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega + \beta) ^2)})
\end{align}
2)
Let $ f(\omega) = e^{-\alpha |x|}$, and let
\begin{align}
g(x) &=\ e^{-\alpha |x|}\sin(\beta x)\\
&=\ e^{-\alpha |x|}(\frac{1}{2i}e^{i\beta x}-\frac{1}{2i}e^{-i\beta x})\: \mbox{ (By Trig identity)}\\
&=\ \frac{1}{2i}e^{-\alpha |x|}e^{i\beta x}-\frac{1}{2i}e^{-\alpha |x|}e^{-i\beta x}
\end{align}
Which satisfies the form $\hat{g}(\omega) = \frac{1}{2i}\hat{f}(\omega - \beta) - \frac{1}{2i}\hat{f}(\omega + \beta)$.
From part a we have found $\hat{f}\frac{\alpha}{\pi (\alpha ^2 +\omega ^2)}$
Thus\begin{align}
\hat{g}(\omega) = \frac{1}{2i}(\frac{\alpha}{\pi (\alpha ^2 +(\omega - \beta) ^2)})-\frac{1}{2i}(\frac{\alpha}{\pi (\alpha ^2 +(\omega + \beta) ^2)})
\end{align}
c.
Let $ f(\omega) = e^{-\alpha |x|}$, and let
$g(x) = xe^{-\alpha |x|}$
By the theorem $ \hat{g}(\omega) = i\hat{f'}(\omega)$, We have
\begin{align}
\hat{g}(\omega) &=\ i \frac{d\hat{f}(\omega)}{d\omega}\\
&=\ i \frac{d}{d\omega }(\frac{\alpha }{\pi (\alpha ^2 +\omega ^2)})\\
&=\ i \frac{\alpha}{\pi}\frac{-2\omega}{(\alpha ^2 +\omega ^2)^2}\\
\end{align}
d.
Similar to part c),
Let $ f(\omega) = e^{-\alpha |x|}cos(\beta x)$, and let $g(x) = xe^{-\alpha |x|}cos(\beta x)$
By the theorem $ \hat{g}(\omega) = i\hat{f'}(\omega)$, where
$\hat{f}(\omega) = \frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega - \beta) ^2})+\frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega + \beta) ^2)})$
We have
\begin{align}
\hat{g}(\omega) &=\ i \frac{d\hat{f}(\omega)}{d\omega}\\
&=\ i \frac{d}{d\omega }\hat{f}(\omega)\\
& =\ i \frac{d}{d\omega }(\frac{\alpha}{\pi (\alpha ^2 +(\omega - \beta) ^2})+\frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega + \beta) ^2)})\\
&=\ i\frac{\alpha}{\pi}[(\frac{-(\omega - \beta)}{(\alpha^2 + (\omega - \beta)^2)^2})+\frac{-(\omega + \beta)}{(\alpha^2 + (\omega + \beta)^2)^2}]\\
&=\ \frac{-\alpha i}{\pi}[\frac{\omega - \beta}{(\alpha^2 + (\omega - \beta)^2)^2}+\frac{\omega + \beta}{(\alpha^2 + (\omega + \beta)^2)^2}]
\end{align}
Similar to above, we can get $g(x) = xe^{-\alpha |x|}sin(\beta x)$ by applying part b.2 formula and use the method we used in part c, then we get:
$\hat{g}(\omega) = \frac{-\alpha }{\pi}[\frac{\omega - \beta}{(\alpha^2 + (\omega - \beta)^2)^2}-\frac{\omega + \beta}{(\alpha^2 + (\omega + \beta)^2)^2}]$
9
Test 1 / Re: TT1 Problem 2
« on: February 12, 2015, 10:23:59 PM »
I don't know how to attach a pdf properly such that we can see its preview without download it.
You cannot. So scan to png or jpg. V.I.
You cannot. So scan to png or jpg. V.I.
11
HA3 / Re: HA3 problem 4
« on: February 05, 2015, 09:09:54 PM »
a.$M(T)= $max $u(x,t)$, where $0\leq x\leq l, 0\leq t\leq T]$
We claim that M(T) is an non decreasing function of T.
Let $T_1 \leq T_2$ and the rectangle $R_1 = {0\leq x\leq l,0\leq t \leq T_1} $ and $R_2 = {0\leq x\leq l,0\leq t \leq T_2} $.
The bottom and lateral sides of $R_1$ contained in $R_2$. By the uniqueness property, the diffusion equation of u on $R_2$ is an extension of u of $R_1$ and thus$M(T1)\leq M(T2)$
b.$m(T)=$ min $u(x,t)$, where $0\leq x\leq l, 0\leq t\leq T]$
We claim that m(T) is an non increasing function of T.
Let $T_1 \leq T_2$ and the rectangle $R_1 = {0\leq x\leq l,0\leq t \leq T_1} $ and $R_2 = {0\leq x\leq l,0\leq t \leq T_2} $.
The bottom and lateral sides of $R_1$ contained in $R_2$. By the uniqueness property, the diffusion equation of u on $R_2$ is an extension of u of $R_1$ and thus $m(T2)\leq m(T1)$
We claim that M(T) is an non decreasing function of T.
Let $T_1 \leq T_2$ and the rectangle $R_1 = {0\leq x\leq l,0\leq t \leq T_1} $ and $R_2 = {0\leq x\leq l,0\leq t \leq T_2} $.
The bottom and lateral sides of $R_1$ contained in $R_2$. By the uniqueness property, the diffusion equation of u on $R_2$ is an extension of u of $R_1$ and thus$M(T1)\leq M(T2)$
b.$m(T)=$ min $u(x,t)$, where $0\leq x\leq l, 0\leq t\leq T]$
We claim that m(T) is an non increasing function of T.
Let $T_1 \leq T_2$ and the rectangle $R_1 = {0\leq x\leq l,0\leq t \leq T_1} $ and $R_2 = {0\leq x\leq l,0\leq t \leq T_2} $.
The bottom and lateral sides of $R_1$ contained in $R_2$. By the uniqueness property, the diffusion equation of u on $R_2$ is an extension of u of $R_1$ and thus $m(T2)\leq m(T1)$
12
HA2 / Re: HA2 problem 1
« on: January 29, 2015, 10:35:13 PM »
b. When $v=2$, we need to impose $u|_{x=2t}=0, t>0$
First when $x>3t$ then the domain satisfies both $\phi(x+3t)$ and $\psi(x-3t)$, thus $u(x, t)$ is same as part a)
Then when $2t<x<3t$, $x+3t>0$, but $x-3t<0$.
\begin{align}
\phi(x) = \frac{1}{2}\cos(x)+\frac{1}{6}\int_0^x \! \sin(y) dx.\\
\phi(x) = \frac{1}{2}\cos(x) - \frac{1}{6}(\cos(x)-1)\\
\phi(x+3t) = \frac{1}{3}\cos(x+3t) + \frac{1}{6}\\
\end{align}
We need to impose the extra condition such that
\begin{align}
u = \phi(5t) + \psi(-t) = 0\\
\phi(5t) = -\psi(-t)\\
\psi(t) = \phi(5t)\\
\psi(x-3t) = \frac{1}{3}\cos(5x-15t) + \frac{1}{6}\\
\end{align}
Then we get $u(x, y) = \frac{1}{3}(\cos(x+3t) +\cos(5x-15t)) + \frac{1}{3}$ for $2t<x<3t$.
c. When $v=-4$, we need to impose $u|_{x=-4t}=ux|_{x=-4t}=0, t>0$
Firstly, when $x>3t$ then the domain satisfies both $\phi(x+3t)$ and $\psi(x-3t)$, thus $u(x, t)$ is same as part a)
Then when $-4t<x<-3t$, $x+3t<0$ and $x-3t<0$.
\begin{align}
\phi(5t) + \psi(-t) = 0\\
\phi'(5t) + \psi'(-t) = 0\\
\end{align}
Then we get $\phi'(5t)=0$ and $\psi(-t) = 0 \implies u(x,t) = 0$ for $-4t<x<-3t$.
Lastly, -3t<x<3t, the situation is similar to part b) $u(x, y) = \frac{1}{3}(\cos(x+3t) +\cos(5x-15t)) + \frac{1}{3}$ .
d. When $v=-3$, we need to impose $u|_{x=-3t}=0, t>0$
First, when $x>3t$ then the domain satisfies both $\phi(x+3t)$ and $\psi(x-3t)$, thus $u(x, t)$ is same as part a)
Then when $ -3t<x<3t$, $x+3t>0$ and $x-3t<0$.
Then $\phi(x+3t) = \frac{1}{3}\cos(x+3t) + \frac{1}{6}$
We need to find $\psi(x)$ by imposing the condition:
\begin{align}
\phi(0) + \psi(-6t) = 0\\
0 + \psi(-6t) = 0\\
\psi(x-3t) = 0
\end{align}
Then we get $u(x,t) = \phi(x+3t) = \frac{1}{3}\cos(x+3t) + \frac{1}{6}$ for $ -3t<x<3t$.
First when $x>3t$ then the domain satisfies both $\phi(x+3t)$ and $\psi(x-3t)$, thus $u(x, t)$ is same as part a)
Then when $2t<x<3t$, $x+3t>0$, but $x-3t<0$.
\begin{align}
\phi(x) = \frac{1}{2}\cos(x)+\frac{1}{6}\int_0^x \! \sin(y) dx.\\
\phi(x) = \frac{1}{2}\cos(x) - \frac{1}{6}(\cos(x)-1)\\
\phi(x+3t) = \frac{1}{3}\cos(x+3t) + \frac{1}{6}\\
\end{align}
We need to impose the extra condition such that
\begin{align}
u = \phi(5t) + \psi(-t) = 0\\
\phi(5t) = -\psi(-t)\\
\psi(t) = \phi(5t)\\
\psi(x-3t) = \frac{1}{3}\cos(5x-15t) + \frac{1}{6}\\
\end{align}
Then we get $u(x, y) = \frac{1}{3}(\cos(x+3t) +\cos(5x-15t)) + \frac{1}{3}$ for $2t<x<3t$.
c. When $v=-4$, we need to impose $u|_{x=-4t}=ux|_{x=-4t}=0, t>0$
Firstly, when $x>3t$ then the domain satisfies both $\phi(x+3t)$ and $\psi(x-3t)$, thus $u(x, t)$ is same as part a)
Then when $-4t<x<-3t$, $x+3t<0$ and $x-3t<0$.
\begin{align}
\phi(5t) + \psi(-t) = 0\\
\phi'(5t) + \psi'(-t) = 0\\
\end{align}
Then we get $\phi'(5t)=0$ and $\psi(-t) = 0 \implies u(x,t) = 0$ for $-4t<x<-3t$.
Lastly, -3t<x<3t, the situation is similar to part b) $u(x, y) = \frac{1}{3}(\cos(x+3t) +\cos(5x-15t)) + \frac{1}{3}$ .
d. When $v=-3$, we need to impose $u|_{x=-3t}=0, t>0$
First, when $x>3t$ then the domain satisfies both $\phi(x+3t)$ and $\psi(x-3t)$, thus $u(x, t)$ is same as part a)
Then when $ -3t<x<3t$, $x+3t>0$ and $x-3t<0$.
Then $\phi(x+3t) = \frac{1}{3}\cos(x+3t) + \frac{1}{6}$
We need to find $\psi(x)$ by imposing the condition:
\begin{align}
\phi(0) + \psi(-6t) = 0\\
0 + \psi(-6t) = 0\\
\psi(x-3t) = 0
\end{align}
Then we get $u(x,t) = \phi(x+3t) = \frac{1}{3}\cos(x+3t) + \frac{1}{6}$ for $ -3t<x<3t$.
13
Web Bonus Problems / Re: Web Bonus Problem 1
« on: January 25, 2015, 08:43:56 AM »
The general solution for $u$ is $ u = f(x_0) = f(x-ut)$, now consider the first set of initial data.
When t = 0,
Case1. If $x < -a$, then we have $u = 1$
Case2. If $x > a$, then we have $u = -1$
Case3. If $-a\le x \le a$, then we have $u =- \frac{x}{a}$.
\begin{equation}
u = -\frac{x-ut}{a}\rightarrow u = \frac{x}{t-a}
\end{equation}
Then the solution looks like
\begin{align}
u(x,t)=&\left\{\begin{aligned}
1& && x<-a+t,\\
\frac{x}{t-a}& && -a+t\le x \le a-t,\\
-1& && x>a-t;
\end{aligned}\right.
\\
\end{align}
Thus as t increase, -a+t and a-t runs toward each other, the the solution graph is squeezing. The solution is define for $0<t<a$.
When t = 0,
Case1. If $x < -a$, then we have $u = 1$
Case2. If $x > a$, then we have $u = -1$
Case3. If $-a\le x \le a$, then we have $u =- \frac{x}{a}$.
\begin{equation}
u = -\frac{x-ut}{a}\rightarrow u = \frac{x}{t-a}
\end{equation}
Then the solution looks like
\begin{align}
u(x,t)=&\left\{\begin{aligned}
1& && x<-a+t,\\
\frac{x}{t-a}& && -a+t\le x \le a-t,\\
-1& && x>a-t;
\end{aligned}\right.
\\
\end{align}
Thus as t increase, -a+t and a-t runs toward each other, the the solution graph is squeezing. The solution is define for $0<t<a$.
14
Web Bonus Problems / Re: Web Bonus Problem 1
« on: January 24, 2015, 12:42:21 AM »
I interpret this question in the following:
This is a quasilinear equation with coefficient u.
As we solved in HA1, the general solution for $u$ is $ u = f(x_0) = f(x-ut)$, now consider the first set of initial data.
Case1. If $x_0 < -a$, then we have $u = -1$ and also $x_0 = x+t < -a$.
Case2. If $x_0 > a$, then we have $u = 1$ and also $x_0 = x-t > a$.
Case3. If $-a\le x \le a$, then we have $u = \frac{x}{a}$.
\begin{equation}
x_0 = x - \frac{x}{a}t\rightarrow u = \frac{x}{a+t}
\end{equation}
Then the solution looks like
\begin{align}
u(x,t)=&\left\{\begin{aligned}
-1& && x<-a-t,\\
\frac{x}{a+t}& && -a-t\le x \le a+t,\\
1& && x>a+t;
\end{aligned}\right.
\\
\end{align}
This is a quasilinear equation with coefficient u.
As we solved in HA1, the general solution for $u$ is $ u = f(x_0) = f(x-ut)$, now consider the first set of initial data.
Case1. If $x_0 < -a$, then we have $u = -1$ and also $x_0 = x+t < -a$.
Case2. If $x_0 > a$, then we have $u = 1$ and also $x_0 = x-t > a$.
Case3. If $-a\le x \le a$, then we have $u = \frac{x}{a}$.
\begin{equation}
x_0 = x - \frac{x}{a}t\rightarrow u = \frac{x}{a+t}
\end{equation}
Then the solution looks like
\begin{align}
u(x,t)=&\left\{\begin{aligned}
-1& && x<-a-t,\\
\frac{x}{a+t}& && -a-t\le x \le a+t,\\
1& && x>a+t;
\end{aligned}\right.
\\
\end{align}
15
HA1 / Re: HA1 problem 4
« on: January 22, 2015, 11:59:45 PM »
c.
Suppose in general case, $yu x - 4xu y = f(x, y)$
Use polar angles, define
\begin{equation}
x=r\cos \theta \rightarrow x {\theta} = -r \sin\theta\\
y = 2r\sin \theta \rightarrow y{\theta} = 2r\cos\theta\\
yu x - 4xu y = -2u\theta.
\end{equation}
Since trajectories are closed (elliptic shape) $\theta \in (0, 2\pi)$ periodically.
Let $-2u \theta = g(\theta, r)$ then $u(\theta, r) = \int_0^{2\pi}\! g(\theta, r) d\theta$ but g needs to be 0 over the period, i.e $\int_0^{2\pi}\! g(\theta, r) d\theta = 0$.
In part a) \begin{equation}
y = -2u\theta\\
r\cos\theta = u
\end{equation} and integral over this period is 0. Thus solution for part a) is valid.
In part b) \begin{equation}
x^2 = -2u\theta\\
-\frac{1}{2}r^2 \frac{1}{2}(\cos 2\theta +1) = u\theta\\
u = -\frac{1}{8}r^2\sin 2\theta - \frac{1}{4}r^2\theta = u\\
\end{equation} the integral over period is not zero. Thus the periodic trajectories would not work.
Suppose in general case, $yu x - 4xu y = f(x, y)$
Use polar angles, define
\begin{equation}
x=r\cos \theta \rightarrow x {\theta} = -r \sin\theta\\
y = 2r\sin \theta \rightarrow y{\theta} = 2r\cos\theta\\
yu x - 4xu y = -2u\theta.
\end{equation}
Since trajectories are closed (elliptic shape) $\theta \in (0, 2\pi)$ periodically.
Let $-2u \theta = g(\theta, r)$ then $u(\theta, r) = \int_0^{2\pi}\! g(\theta, r) d\theta$ but g needs to be 0 over the period, i.e $\int_0^{2\pi}\! g(\theta, r) d\theta = 0$.
In part a) \begin{equation}
y = -2u\theta\\
r\cos\theta = u
\end{equation} and integral over this period is 0. Thus solution for part a) is valid.
In part b) \begin{equation}
x^2 = -2u\theta\\
-\frac{1}{2}r^2 \frac{1}{2}(\cos 2\theta +1) = u\theta\\
u = -\frac{1}{8}r^2\sin 2\theta - \frac{1}{4}r^2\theta = u\\
\end{equation} the integral over period is not zero. Thus the periodic trajectories would not work.
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