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### Messages - Andrew Hardy

Pages: [1] 2
1
##### Web Bonus Problems / Re: Exam Week
« on: April 14, 2018, 02:12:26 PM »
Writing the change of coordinates Sheng applied to the boundary conditions I found,  $w|_{\pi=0}=0$ and $w|_{r=1}=2\sin^2\theta$ I can apply separation of variables to the function $w = P(\theta)R(r)$ which I can solve for this half disk as
$$w = \sum A r^n \sin(n\theta)$$ and therefore
$$A = \frac{4}{\pi}\int_{0}^{\pi} \sin^2(\theta) \sin(n\theta) d \theta =\frac{ (4 (2 \cos(π n) - 2))}{(π (n^3 - 4 n)) }$$
and this is $\frac{-8}{(π (n^3 - 4 n)) }$ when n is odd. So I have
$$w = \sum_{n \text{ odd}} \frac{-8}{(π (n^3 - 4 n)) } r^n \sin(n\theta)$$
and therefore I can write $$u = w + v = \sum_{n \text{ odd}} \frac{-8}{(π (n^3 - 4 n)) } r^n \sin(n\theta) + r^2 \cos^2(\theta) - r^2 sin^2(\theta)$$ which can be written most concisely as
$$u = r^2 \cos(2 \theta) + \sum_{n \text{ odd}} \frac{-8}{(π (n^3 - 4 n)) } r^n \sin(n\theta)$$

2
##### Final Exam / Re: FE-P6
« on: April 14, 2018, 01:46:20 PM »
If this is still open for karma, I will go into more detail missing from the solution?

3
##### Final Exam / Re: FE-P5
« on: April 13, 2018, 04:03:03 PM »
Begin via separation of variables.
$$X''+\lambda X = 0$$
$$Y''-\lambda Y=0$$
with this the solution to the  new Dirichlet Boundary conditions gives  $\lambda = n^2$  $n = 1,2,...$
so
$$X= A_n \sin(2nx)$$ where we don't have any constants that might be forgotten...
Now the solution to the Y ODE are exponential, where we discard the infinite terms so our general solution is of the form
$$u(x,y)=\sum_{n=1}^\infty B_ne^{-2ny}\sin(2nx)$$ Then the Robin Boundary Condition gives us
$$- \sum _{n=1}^\infty B_n (2n+1)\sin(2nx)=1.$$ This allows us to solve the traditional method that
$$B_n = -\int_{0}^{\frac{\pi}{2}} (2n+1)^{-1}\sin(2nx) \text{dx}$$ Which has the nice solution that
$$= B_n ((2n(2n+1))^{-1}(\cos(n\pi) - 1)$$
$$B_n = -(n(2n+1))^{-1}$$ for n odd

and therefore we have what would have been a much nicer solution of
$$u(x,y)=-\sum_{n \text{ odd}}^\infty (n(2n+1))^{-1} e^{-2ny}\sin(2nx)$$

4
##### Web Bonus Problems / Re: Exam Week
« on: April 13, 2018, 09:08:11 AM »
To calculate $w|_{x^2+y^2=1}$ I make a substitution that if  $$u|_{x^2+y^2 =1}=1$$ then  $$v|_{x^2+y^2 =1}= 1 - y^2 - y^2 = 1-2y^2$$
$$v|_{x^2+y^2 =1}= x^2 - (1-x^2) = 2x^2 - 1$$
so then  $$w|_{x^2+y^2=1}=2y^2 = 2- 2x^2$$

Not sure if I'm on the right track, I'll return to this later today.

5
##### Web Bonus Problems / Re: Exam Week
« on: April 12, 2018, 10:17:00 PM »
Only considering (1) and (2),
My candidate is
$$v = x^2 -y^2$$
which is harmonic, (has zero laplacian) and $$v(x,0) = x^2$$
By homogeneous (2) do you mean?
$$w|_{y=0}=0$$
How do I modify (3) ?

6
##### Final Exam / Re: FE-P6
« on: April 12, 2018, 03:06:16 PM »
That's what I had originally before JX confused me.

7
##### Final Exam / Re: FE-P6
« on: April 11, 2018, 07:07:54 PM »
The solution is spherically symmetric because the question has that symmetry.
Then using the hint I reduce the Laplacian to $$(ru)" = v"$$ then do separation of variables on v and I get the solution to the eigenvalue problem of v is that  $R = \sin(r)$ because the eigenvalue must be one and the cos term has a singularity at r = 0
So now my general solution is that $$u = \frac{\sin(r)}{r} (A\cos(t) + Bsin(t))$$ To satisfy the boundary conditions, A = 0 and
$$u = \frac{\sin(r)}{r} \sin(t) \text{ for } r< \theta$$ B = 1.
For the other boundary condition, the only conceivable option is that u is defined as the trivial solution outside of this domain, B = 0

8
##### Final Exam / Re: FE-P7
« on: April 11, 2018, 06:18:47 PM »
We begin by transforming $$\hat{u} = \hat{u}_{xx} -\eta^2 \hat{u}_{\eta\eta}$$
and without having a singularity we only have one term. Then we know from the hint
$$\hat{h}=-i\eta e^{-|\eta|}$$ and so
$$\hat{u} = -i\frac{\eta}{|\eta|} e^{-|\eta|(x+1)}$$ of which we can inverse Fourier transform by the properties in the appendix

$$u = \frac{-2y}{y^2+(x+1)^2}$$

I missed the $\frac{1}{|\eta|}$ term on the final and hope Dr. Ivrii is merciful.

9
##### Final Exam / Re: FE-P4
« on: April 11, 2018, 05:55:02 PM »
a) separation of variables gives us that $$P'' + \lambda P = 0$$
$$r^2R''+ rR' -\lambda R = 0$$
The boundary conditions translate to Dirichlet Boundary Conditions $P(0) = P(\theta) = 0$
b) With the eigenvalue problem of  $$P \lambda = (n)^2$$ gives $$P = \sin(n\theta)$$
c) and then solving the ODE for R we know $$R = Ar^n + Br^{-n}$$
d) Now solving for coefficients $$u = \sin(n\theta)(Ar^n + Br^{-n} )$$
Now we know $A = -B$  because  of (2),(3) and so
$$A_n =\frac{2}{\pi} \int_{0}^{\pi} \frac{1}{2^{-n}-2^{n}} \sin(n\theta)$$ and via computation
becomes $$u =\sum_{n \text{ odd}} \frac{4}{n\pi(2^n-2^{-n})} \sin(n\theta) ( r^n + r^{-n})$$

10
##### Final Exam / Re: FE-P2
« on: April 11, 2018, 05:23:54 PM »
Very similar to Term Test 1. Here instead though,  apply even continuation and then because k = 1/2

$$u=\frac{1}{\sqrt{2t\pi}}\int_0^{\infty} \exp(-\frac{(y+x)^2}{2t}-y) \,dx +\frac{1}{\sqrt{t\pi}}\int_0^{\infty} \exp(-\frac{(y-x)^2}{2t}-y)\,dx\\$$
and then completing the square
$$=\frac{\exp(x+t)}{\sqrt{t\pi}}\int_0^{\infty} \exp(-\frac{(y+x+t)^2}{2t})\,dx + \frac{\exp(-x+t)}{\sqrt{t\pi}}\int_0^{\infty} \exp(-\frac{(y-x+t)^2}{2t})\,dx\\$$
and then via change of variables
$$=\frac{\exp(x+t/2)}{\sqrt{\pi}}\int_{\frac{x+t/2}{\sqrt{2t}}}^{\infty} e^{-z^2} \,dz + \frac{\exp(-x+t/2)}{\sqrt{\pi}}\int_{\frac{-x+t}{\sqrt{2t}}}^{\infty} e^{-z^2} \,dz\\$$
and in conclusion
$$=\frac{\exp(x+t/2)}{2}(1-\text{erf}(\frac{x+t}{\sqrt{2t}})) + \frac{\exp(-x+t/2)}{2}(1-\text{erf}(\frac{-x+t}{\sqrt{2t}}))$$

corrected

11
##### Web Bonus Problems / Re: Week 13 -- BP3, 4, 5, 6
« on: April 09, 2018, 09:37:19 AM »
d) *from Gel'fand*
Equations 5 and 6 come from the natural log of  the generalized expression
$$(x+iy)^{\nu} = \exp[\nu \log(x+iy) + iArg(x+iy)] \tag{A}$$  which is just the representation on the complex plane
Then we can find
$$(x+i0)^{-\nu} = x^{-\nu} - \frac{i\pi(-1)^{\nu-1}}{(\nu -1)!}\delta^ {(nu -1)}(x) \tag{B}$$
$$(x-i0)^{-\nu} = x^{-\nu} + \frac{i\pi(-1)^{\nu-1}}{(\nu -1)!}\delta^ {(nu -1)}(x) \tag{C}$$
and by combining these and setting $\nu =1$ we arrive at
$$(x-i0)^{-1} + (x+i0)^{-1}=2 x^{-1} \tag{D}$$
$$(x-i0)^{-1}- (x+i0)^{-1}=2\pi i \delta(x) \tag{E}$$

12
##### Web Bonus Problems / Re: Week 13 -- BP3, 4, 5, 6
« on: April 09, 2018, 09:21:53 AM »
Dr. Ivrii? Are Part a and b correct now? In Gel'fand's "Normalized Functions" he has what seems the same argument, but rather does the integrals explicitly. I can go that in depth if I haven't earned the points yet. I'm stilling looking for c and d.

13
##### Web Bonus Problems / Re: Week 13 -- BP3, 4, 5, 6
« on: April 07, 2018, 04:10:16 PM »
a) Consider
$$(x\pm i0)^{\nu}= f'$$
Now via definitions $$(f',\theta) = - (f,\theta')$$ and then f is defined as
$$\frac{1}{\nu+1}(x \pm i0)^{\nu+1} = f$$ This  exists for  $\Re\nu >-1$. Therefore the limit exists for this condition.
b) Extending the proof from part a, define $$(x\pm i0)^{\nu}= f^{(n)}$$ then with the same rules of distribution derivatives I can write $$[(x\pm \varepsilon i)^{\nu}]^{(n)}= (\nu ! )(x\pm \varepsilon i)^{\nu-n}$$ which is then defined for corresponding $\Re\nu \neq -n$ where  $n = 2, 3,\ldots$

14
##### Term Test 2 / Re: TT2--P3N
« on: April 07, 2018, 10:17:33 AM »
Just sloppy Latex'ing as I copied them. Thanks

15
##### Term Test 2 / Re: TT2--P3N
« on: April 06, 2018, 03:07:29 PM »
Above is incomplete.
we begin with separation of variables $U = X(x)Y(y)$
The equation simplifies to
$$\frac{X"}{X} + \frac{Y"}{Y} = -\lambda$$

We know that these fractions must remain constant and so we have corresponding $-\lambda_x + -\lambda_y = -\lambda$
We now have  Not enough letters? V.I. I'm sorry I don't follow
$$X" + \lambda_xX = 0$$
$$Y" + \lambda_yY = 0$$
Furthermore boundary conditions state that  $$X(0) = X'(a)$$ and $$Y(0) = Y'(b)$$ These are mixed boundary conditions (Dirichlet and Nuemann). They dictate the eigenvalues and corresponding eigenfunctions for the ODEs.
\begin{align*}
&\lambda_x =( \frac{\pi(2n+1)}{2a})^2 , &&X_n = \sin(\frac{\pi(2n+1)}{2a}) && n = 0,1,2,...\\
&\lambda_y =( \frac{\pi(2m+1)}{2b})^2  &&X_n = \sin(\frac{\pi(2m+1)}{2b}) &&m = 0,1,2,...
\end{align*}
We can and must then conclude that
$$\lambda =\pi^2( (\frac{(2n+1)}{2a})^2+ (\frac{(2m+1)}{2b})^2)$$
ERROR above. V.I. corrected
and our eigenfunctions are of the formula
$$U_{n,m}(x,y) = \sin(\frac{\pi(2n+1)}{2a})\sin(\frac{\pi(2m+1)}{2b})$$

This is the complete answer. No need to go on a tangent.

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