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Final Exam / Re: FE-P6
« on: April 12, 2018, 01:17:19 PM »
I have the same answer as Andrew, except without the constant B, and I have a $\sin(t)$ term instead, due to the fact that $u(r,0)=0$
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Final Exam / Re: FE-P5
« on: April 11, 2018, 08:11:27 PM »
I shall expand upon this post later with the full list of steps I took to cancel out all my eigenfunctions and eigenvalues. For the moment, my final answer was u=-1, which is bounded and satisfies all boundary conditions (trivially, $u_x$ and $u_y$ are 0 as needed).
Update (apologies for the comparatively short response, I'm not feeling great at the moment): so we first write $u=XY$ so that the problem becomes $\frac{Y''}{Y}+\frac{X''}{X}=0$. Condition (2) implies periodicity in x, so we have associated problems $X''+\lambda X = 0$, and $Y''-\lambda Y=0$. Then $X=A_n \cos(\omega_n x) + B_n \sin(\omega_n x)$, but we can set $B_n=0$ immediately from (2). Additionally, $X'(0)=X'(\frac{\pi}{2}=0) $ so $\omega_n=2n, \lambda_n=4n^2$. Then $Y = A_n e^{-2ny} + B_n e^{2ny} = A_n e^{-2ny} + 0$ , because u is bounded on $y>0$
So $u=A_0 + \sum_{n=1}^\infty e^{-2ny} \cos (2nx) \rightarrow (u_y-u)(x,0)=-A_0 - \sum_{n=1}^\infty (2n+1) A_n \cos (2nx) = 1 \rightarrow A_n(2n+1) = \frac{4}{\pi}\int_0^\frac{\pi}{2}\cos(2nx)$
In fact, this integral is 0 for all n. Therefore, we are just left with the condition $-A_0=1 \rightarrow u=-1$
Update (apologies for the comparatively short response, I'm not feeling great at the moment): so we first write $u=XY$ so that the problem becomes $\frac{Y''}{Y}+\frac{X''}{X}=0$. Condition (2) implies periodicity in x, so we have associated problems $X''+\lambda X = 0$, and $Y''-\lambda Y=0$. Then $X=A_n \cos(\omega_n x) + B_n \sin(\omega_n x)$, but we can set $B_n=0$ immediately from (2). Additionally, $X'(0)=X'(\frac{\pi}{2}=0) $ so $\omega_n=2n, \lambda_n=4n^2$. Then $Y = A_n e^{-2ny} + B_n e^{2ny} = A_n e^{-2ny} + 0$ , because u is bounded on $y>0$
So $u=A_0 + \sum_{n=1}^\infty e^{-2ny} \cos (2nx) \rightarrow (u_y-u)(x,0)=-A_0 - \sum_{n=1}^\infty (2n+1) A_n \cos (2nx) = 1 \rightarrow A_n(2n+1) = \frac{4}{\pi}\int_0^\frac{\pi}{2}\cos(2nx)$
In fact, this integral is 0 for all n. Therefore, we are just left with the condition $-A_0=1 \rightarrow u=-1$
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Final Exam / Re: FE-P7
« on: April 11, 2018, 06:33:56 PM »
By taking a partial fourier transform w.r.t. y, our Laplacian becomes $\hat{u}_{xx}-k^2\hat{u}=0$ which resolves into $\hat{u}=A(k)e^{|k|x}+B(k)e^{-|k|x}$. We can eliminate the first term, because it is unbounded on domain $x>0$, so $\hat{u}(x)=B(k)e^{-|k|x}$. Then $\hat{u}_x=|k|e^{-|k|x}B(k)$.
Before this can be solved, we must take the fourier transform of h(y). Using the hint that $h(y)=-g'(y), \hat{g}=e^{-|k|} \rightarrow \hat{h}=-ike^{-|k|}$.
Now: $\hat{u}_x(0,k)=|k|B(k)=-ike^{-|k|} \rightarrow B(k)=-i sgn(k) e^{-|k|} \rightarrow \hat{u}(x,k)=-i sgn(k)e^{-|k|(x+1)}$.
Let us take an inverse fourier transform back to (x,y) coordinates:
$u(x,y)=-i\int_{-\infty}^{\infty} sgn(k)e^{-|k|(x+1)}e^{iky}dk = -i \left ( \int_0^\infty e^{(iy-(x+1))k}dk - \int_{-\infty}^0 e^{(iy+(x+1))k}dk \right ) = -i \left ( \frac{1}{iy-(x+1)} + \frac{1}{iy+(x+1)} \right ) = \frac{2y}{y^2+(x+1)^2}$
Before this can be solved, we must take the fourier transform of h(y). Using the hint that $h(y)=-g'(y), \hat{g}=e^{-|k|} \rightarrow \hat{h}=-ike^{-|k|}$.
Now: $\hat{u}_x(0,k)=|k|B(k)=-ike^{-|k|} \rightarrow B(k)=-i sgn(k) e^{-|k|} \rightarrow \hat{u}(x,k)=-i sgn(k)e^{-|k|(x+1)}$.
Let us take an inverse fourier transform back to (x,y) coordinates:
$u(x,y)=-i\int_{-\infty}^{\infty} sgn(k)e^{-|k|(x+1)}e^{iky}dk = -i \left ( \int_0^\infty e^{(iy-(x+1))k}dk - \int_{-\infty}^0 e^{(iy+(x+1))k}dk \right ) = -i \left ( \frac{1}{iy-(x+1)} + \frac{1}{iy+(x+1)} \right ) = \frac{2y}{y^2+(x+1)^2}$
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Final Exam / Re: FE-P4
« on: April 11, 2018, 06:07:34 PM »
First, we convert the Laplacian to polar coordinates: $\Delta u =u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}$. Write $u=RP$, so that we can solve the problem of form $\frac{r^2R''+rR'}{R}+\frac{P''}{P}=0$
Now, notice that boundary condition (4) becomes $u(r,0)=u(r,\pi)=0$ in polar coordinates. This gives us a condition for periodicity in P, so that we can associate a negative eigenvalue $-\lambda_n=-\omega_n^2$ to it. Solving $P''+\omega_n^2 P=0$ gives us $P=A_n \cos (\omega_n \theta) + B_n \sin (\omega_n \theta)$. We can immediately eliminate the $A_n$ term, because of the boundary condition on $\theta=0$, and we can also conclude that $\omega_n \pi = n \pi$ means that our $\omega_n=n$, $\lambda = n^2$.
Now, we expect some equations of the form $r^m$ for the solution to R. Solving the Euler problem $r^2R+rR'=\lambda R$ gives $m(m-1)+m=n^2\rightarrow m = \pm n$. So $R=A_n r^n+B_n r^{-n}$. Neither term can be negated, as r does not tend to 0 or infinity. However, $A_0$ and $B_0$ can be ignored as they do not obey the periodic behaviour we expect.
This gives general form $u(r,\theta) = \sum_{n=1}^\infty(A_n r^n+B_n r^{-n})\sin (n \theta)$. Now, we consider boundary conditions (2) and (3) in the cases $r=2, r=2^{-1}$ respectively.
Then
$u(2,\theta) = \sum_{n=1}^\infty(A_n 2^n+B_n 2^{-n})\sin (n \theta) = 1$ (2)
$u(2^{-1},\theta) = \sum_{n=1}^\infty(A_n 2^{-n}+B_n 2^{n})\sin (n \theta) = -1$ (3)
If we add these two summations together, this implies that $B_n=-A_n$. Use this simplification to find the fourier coefficients in (2):
$u(2,\theta) = \sum_{n=1}^\infty A_n (2^n - 2^{-n})\sin (n \theta) = 1$
Substitute $C_n=A_n(2^n-2^{-n})$. Then $C_n=\frac{2}{\pi}\int_0^\pi sin (n \theta) d\theta = \frac{4}{n\pi}$ for n odd, $0$ for n even.
So $u(r,\theta) = \sum_{n=1}^\infty \frac{4}{n\pi(2^n-2^{-n})} (r^n - r^{-n})\sin (n \theta)$
Now, notice that boundary condition (4) becomes $u(r,0)=u(r,\pi)=0$ in polar coordinates. This gives us a condition for periodicity in P, so that we can associate a negative eigenvalue $-\lambda_n=-\omega_n^2$ to it. Solving $P''+\omega_n^2 P=0$ gives us $P=A_n \cos (\omega_n \theta) + B_n \sin (\omega_n \theta)$. We can immediately eliminate the $A_n$ term, because of the boundary condition on $\theta=0$, and we can also conclude that $\omega_n \pi = n \pi$ means that our $\omega_n=n$, $\lambda = n^2$.
Now, we expect some equations of the form $r^m$ for the solution to R. Solving the Euler problem $r^2R+rR'=\lambda R$ gives $m(m-1)+m=n^2\rightarrow m = \pm n$. So $R=A_n r^n+B_n r^{-n}$. Neither term can be negated, as r does not tend to 0 or infinity. However, $A_0$ and $B_0$ can be ignored as they do not obey the periodic behaviour we expect.
This gives general form $u(r,\theta) = \sum_{n=1}^\infty(A_n r^n+B_n r^{-n})\sin (n \theta)$. Now, we consider boundary conditions (2) and (3) in the cases $r=2, r=2^{-1}$ respectively.
Then
$u(2,\theta) = \sum_{n=1}^\infty(A_n 2^n+B_n 2^{-n})\sin (n \theta) = 1$ (2)
$u(2^{-1},\theta) = \sum_{n=1}^\infty(A_n 2^{-n}+B_n 2^{n})\sin (n \theta) = -1$ (3)
If we add these two summations together, this implies that $B_n=-A_n$. Use this simplification to find the fourier coefficients in (2):
$u(2,\theta) = \sum_{n=1}^\infty A_n (2^n - 2^{-n})\sin (n \theta) = 1$
Substitute $C_n=A_n(2^n-2^{-n})$. Then $C_n=\frac{2}{\pi}\int_0^\pi sin (n \theta) d\theta = \frac{4}{n\pi}$ for n odd, $0$ for n even.
So $u(r,\theta) = \sum_{n=1}^\infty \frac{4}{n\pi(2^n-2^{-n})} (r^n - r^{-n})\sin (n \theta)$
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Final Exam / Re: FE-P3
« on: April 11, 2018, 05:40:16 PM »
Write: $u = X(x)T(t)$ so that the Laplacian becomes $T''X - X''T + 4XT = 0 $, giving us $ \frac{T''}{T}+4-\frac{X''}{X}=0$, giving us eigenvalue problems in X and T.
Boundary condition (2) divided by T gives $X(0) = X(\pi) = 0 $ which implies a periodicity in X, so therefore we can associate a negative eigenvalue to it. So $X''+\lambda_n X = 0$ and $T''+(\lambda_n+4) T=0$. Write $\lambda_n=\omega^2$.
In general, we have $X=A_n \cos (\omega_n x) + B_n \sin (\omega_n x)$. From the boundary condition used above, we can determine that $A_n=0, \omega_n = n$.
Similarly, $T$ has the solution $T=A_n \cos(\sqrt{n^2+4}t) + B_n \sin(\sqrt{n^2+4}t)$, where we can set $A_n = 0$ from boundary condition (3).
Then $u(x,t)=\sum_{n=1}^\infty A_n \sin(\sqrt{n^2+4}t) \sin (n x)$.
Now to apply initial conditions: $u_t(x,0)=\sum_{n=0}^\infty A_n \sqrt{n^2+4} \sin (\omega_n x)=x^2-\pi x$ and substitute: $C_n=A_n \sqrt{n^2+4}$.
Therefore $C_n=\frac{2}{\pi}\int_0^\pi (x^2-\pi x) \sin(nx) dx=\frac{8}{\pi n^3}$ for odd n, $0$ for even n
So $u(x,t)=u(x,t)=\sum_{n=1}^\infty \frac{8}{\pi n^3 \sqrt{n^2+4}} \sin(\sqrt{n^2+4}t) \sin (n x)$ for all n odd
Unfortunately on the final itself, I made a couple minor mistakes such as using 1 instead of 4, and using $\lambda_n$ instead of $\sqrt{\lambda_n}$
Boundary condition (2) divided by T gives $X(0) = X(\pi) = 0 $ which implies a periodicity in X, so therefore we can associate a negative eigenvalue to it. So $X''+\lambda_n X = 0$ and $T''+(\lambda_n+4) T=0$. Write $\lambda_n=\omega^2$.
In general, we have $X=A_n \cos (\omega_n x) + B_n \sin (\omega_n x)$. From the boundary condition used above, we can determine that $A_n=0, \omega_n = n$.
Similarly, $T$ has the solution $T=A_n \cos(\sqrt{n^2+4}t) + B_n \sin(\sqrt{n^2+4}t)$, where we can set $A_n = 0$ from boundary condition (3).
Then $u(x,t)=\sum_{n=1}^\infty A_n \sin(\sqrt{n^2+4}t) \sin (n x)$.
Now to apply initial conditions: $u_t(x,0)=\sum_{n=0}^\infty A_n \sqrt{n^2+4} \sin (\omega_n x)=x^2-\pi x$ and substitute: $C_n=A_n \sqrt{n^2+4}$.
Therefore $C_n=\frac{2}{\pi}\int_0^\pi (x^2-\pi x) \sin(nx) dx=\frac{8}{\pi n^3}$ for odd n, $0$ for even n
So $u(x,t)=u(x,t)=\sum_{n=1}^\infty \frac{8}{\pi n^3 \sqrt{n^2+4}} \sin(\sqrt{n^2+4}t) \sin (n x)$ for all n odd
Unfortunately on the final itself, I made a couple minor mistakes such as using 1 instead of 4, and using $\lambda_n$ instead of $\sqrt{\lambda_n}$
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Quiz-5 / Re: Quiz 5, T5102
« on: March 07, 2018, 06:51:46 PM »
I went with a more brute force method, wherein I think I made a minor mistake (I did not have |k| at the end like we should have gotten) so it would be great to find out where.
I immediately plugged in the function $f(x)=\frac{1}{(x^2+a^2)}$ into the integral $F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{e^{-ikx}}{(x-ia)(x+ia)}$ (where I rewrote the denominator to make use of residue theorem ($\int f = 2\pi i Res (f;-ia)$)), and integrated over the contour:
Giving me $F(k)=\frac{\sqrt{\pi}}{\sqrt{2}a}e^{-ak}$
For the 2nd part, I simply applied the property that $g(x)=x*f(x)\rightarrow G(k)=ikF(x)$
I immediately plugged in the function $f(x)=\frac{1}{(x^2+a^2)}$ into the integral $F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{e^{-ikx}}{(x-ia)(x+ia)}$ (where I rewrote the denominator to make use of residue theorem ($\int f = 2\pi i Res (f;-ia)$)), and integrated over the contour:
Giving me $F(k)=\frac{\sqrt{\pi}}{\sqrt{2}a}e^{-ak}$
For the 2nd part, I simply applied the property that $g(x)=x*f(x)\rightarrow G(k)=ikF(x)$
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Quiz-3 / Re: Q3-T0101
« on: February 11, 2018, 06:19:59 PM »
I took a crack at the first part, but I'm not entirely confident in my steps and I was in the other section.
As always in the wave eqn, we have $u=\varphi(x+ct)+\psi$
I first solved in the region $x>ct$, where we simply have $\varphi(x)+\psi(x)=\phi(x)$ and $\varphi'(x)-\psi'(x)=\phi'(x)$, giving us $\varphi(x)=\phi(x)$ and $\psi(x)=0$.
For the region $0<x<ct$:
$\varphi'(ct)+\psi'(-ct)+\alpha\varphi(ct)+\alpha\psi(-ct)=0$
Let $t=-\frac{x}{c}$, so $\varphi'(-x)+\psi'(x)+\alpha\varphi(-x)+\alpha\psi(x)=0$
Then $\psi'(x)+ \alpha\psi(x) =-(\varphi'(-x)+\alpha\varphi(-x))$
We can see then, that $(e^{\alpha x}\psi(x))'=(e^{\alpha x}\varphi(x))'$
Integrating tells us that $\psi(x)=e^{-\alpha x}\int e^{\alpha x}\varphi(x)dx$
So $u(x,t)=\varphi(x+ct)$ for $x>ct$
And $u(x,t)=\varphi(x+ct)+e^{-\alpha x}\int_0^{ct-x} e^{\alpha x'}\varphi(x')dx'$ for $0<x<ct$
In particular, I'm not sure where the $-2\alpha\phi(-s)$ outside of the e^ derivatives in Jingxuan's solutions come from.
As always in the wave eqn, we have $u=\varphi(x+ct)+\psi$
I first solved in the region $x>ct$, where we simply have $\varphi(x)+\psi(x)=\phi(x)$ and $\varphi'(x)-\psi'(x)=\phi'(x)$, giving us $\varphi(x)=\phi(x)$ and $\psi(x)=0$.
For the region $0<x<ct$:
$\varphi'(ct)+\psi'(-ct)+\alpha\varphi(ct)+\alpha\psi(-ct)=0$
Let $t=-\frac{x}{c}$, so $\varphi'(-x)+\psi'(x)+\alpha\varphi(-x)+\alpha\psi(x)=0$
Then $\psi'(x)+ \alpha\psi(x) =-(\varphi'(-x)+\alpha\varphi(-x))$
We can see then, that $(e^{\alpha x}\psi(x))'=(e^{\alpha x}\varphi(x))'$
Integrating tells us that $\psi(x)=e^{-\alpha x}\int e^{\alpha x}\varphi(x)dx$
So $u(x,t)=\varphi(x+ct)$ for $x>ct$
And $u(x,t)=\varphi(x+ct)+e^{-\alpha x}\int_0^{ct-x} e^{\alpha x'}\varphi(x')dx'$ for $0<x<ct$
In particular, I'm not sure where the $-2\alpha\phi(-s)$ outside of the e^ derivatives in Jingxuan's solutions come from.
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