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### Messages - Jeremy Li 2

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1
##### Web Bonus = Nov / Re: Web Bonus Problem to Week 8 (#2)
« on: December 02, 2015, 12:10:30 PM »
d)
We calculate

\hat{f}(\mathbf{k})=\int_0^a e^{-\alpha r} e^{-ikr} rdr=\frac{1-e^{-(\alpha+ik)a} (1+\alpha a +ika)}{(\alpha+ik)^2}

2
##### Web Bonus = Nov / Re: Web Bonus Problem to Week 8 (#2)
« on: December 02, 2015, 09:04:46 AM »
For part c) we do essentially the same thing, calculating

\hat{f}(\mathbf{k})=a^2 \hat{g}(\mathbf{k})- i\hat{h}'(k)

Where $\hat{g}(\mathbf{k})$ is equation 8 from post above, and $\hat{h}(k)$ is equation 10 from above. Like the last question, a lot of stuff cancels and we get the answer:

\hat{f}(\mathbf{k})=\frac{-k^2a^2-6+(6+6ika-2k^2a^2) e^{-ika}}{k^4}

3
##### Web Bonus = Nov / Re: Web Bonus Problem to Week 8 (#2)
« on: December 02, 2015, 06:48:15 AM »
Note that $k$ not boldfaced means the magnitude of $\mathbf{k}$.

Since $f(\mathbf{x})$ depends only on $|\mathbf{x}|$, $f(\mathbf{x})$ is rotationally symmetric and so is $\hat{f}(\mathbf{k})$. Then we might as well rotate the vector $\mathbf{k}$ onto an arbitrary axis. So

\hat{f}(\mathbf{k})=\hat{f}(0,0,...,k)

For concreteness, I'll do this in 2D, with polar coordinates. Then, since we're free to choose the direction of $\mathbf{k}$, we let $\mathbf{k}=(k,0)$ in polar coordinates, and $\mathbf{x}=(r,\theta)$.

Then

\hat{f}(\mathbf{k})=\frac{1}{2\pi}\int_0^\infty \int_0^{2\pi} f(\mathbf{x})e^{-ikr}rdrd\theta=\frac{1}{2\pi}\int_0^\infty \int_0^{2\pi} f(r)e^{-ikr}rdrd\theta

Since $f$ depends on $r$ only. The equation falls into a 1d fourier transform.

We get the forumla

\hat{f}(\mathbf{k})=\int_0^{\infty} f(r)e^{-ikr}rdr

a) It's a straightforward calculation:

\hat{f}(\mathbf{k})=\int_0^a r e^{-ikr}dr=\frac{-1+e^{-i k a} (1+i k a)}{k^2}

b) We first calculate the fourier transform of $a$ alone, and then $|\mathbf{x}|$ alone and subtract the difference. The fourier transform of $a$ is just $a$ times eq 8. The fourier transform of $|\mathbf{x}|$ can be calculated using 5.2.3a, via the formula $\hat{f}(\mathbf{k})=i\hat{g}'(k)$ where $\hat{g}(k)$ is the fourier transform from part a, eq 8.

\int_0^a ar e^{-ikr}dr=\frac{-a +a e^{-i k a} (1+i k a)}{k^2}

\int_0^a r^2 e^{-ikr}dr=
\frac{2i + e^{-ika} (2ka + i k^2 a^2 -2 i)}{k^3}

Taking the difference, and simplifying (a lot of terms cancel)

\hat{f}(\mathbf{k})=\frac{-ka-2i +e^{-ika} (2 i-ka)}{k^3}

c) and d) to follow...

4
##### Web Bonus = Nov / Re: Web Bonus Problem to Week 8 (#2)
« on: December 01, 2015, 10:59:13 AM »
I'm really not sure if I'm doing this correctly but...

Since $f(\mathbf{x})$ depends only on $|\mathbf{x}|$, $f(\mathbf{x})$ is rotationally symmetric and so is $\hat{f}(\mathbf{k})$. Then we might as well rotate the vector $\mathbf{k}$ onto an arbitrary axis. So

\hat{f}(\mathbf{k})=\hat{f}(0,0,...,k)

For concreteness, lets do this in 2D, with polar coordinates. Then, since we're free to choose the direction of $\mathbf{k}$, we let $\mathbf{k}=(k,0)$ in polar coordinates, and $\mathbf{x}=(r,\theta)$.

\hat{f}(\mathbf{k})=\frac{1}{2\pi}\iint_{\mathbb{R}^n} f(\mathbf{x}) e^{-i\mathbf{k}\cdot\mathbf{x}} d^2 \mathbf{x}=\frac{1}{2\pi}\iint_{\mathbb{R}^n} f(\mathbf{x}) e^{-ikx} d^2 \mathbf{x}

(since in cartesian coordinates $k_x=k$ and $k_y=0$, we can simplify the dot product)

Doing a change of coordinates we get,

\hat{f}(\mathbf{k})=\frac{1}{2\pi}\int_0^\infty \int_0^{2\pi} f(\mathbf{x}) e^{-ikr\cos\theta} r dr d\theta

Is this correct? This formula is very difficult to apply even for the simplest problems like (a). I'm guessing this is because I calculated $\mathbf{k}\cdot\mathbf{x}$ as the cartesian dot product, which in polar coordinates is $rk\cos(\theta-\phi)$. This style of fourier transform decomposes the circularly-symmetric function into complex plane waves. I tried this on a computer and got a Bessel function.

If you meant to do a multi-dimensional fourier transform such that the function is decomposed into complex radial waves and complex tangential waves, the FT is much simpler. This isn't the usual definition of a dot product though.

\hat{f}(\mathbf{k})=\frac{1}{2\pi}\int_0^\infty \int_0^{2\pi} f(\mathbf{x}) e^{-ikr} r dr d\theta

I will assume that you want the latter and I will try doing it that way.

5
##### Web Bonus = Nov / Re: Web Bonus Problem to Week 8 (#1)
« on: December 01, 2015, 07:21:31 AM »
PART B
The transformation $Q$ is orthogonal. That means $Q^T=Q^{-1}$.

Then

\det(Q)\det(Q)=\det(Q)\det(Q^T)=\det(QQ^T)=\det(QQ^{-1})=\det(I)=1

This means that $|\det(Q)|=1$. By 5.2.A Eq(1):

f(\mathbf{x})=f(Q\mathbf{x}) \iff \hat{f}(\mathbf{k})=\hat{f}(Q^T\mathbf{k})

It's evident that if $f$ is rotationally invariant (left side is true), than the right side holds for any $Q^T$ orthogonal, and so $\hat{f}$ is rotationally invariant. It's evident that if $\hat{f}$ is rotationally invariant (right side is true), than the left side holds for any $Q$ orthogonal, and so $f$ is rotationally invariant.

6
##### Web Bonus = Nov / Re: Web Bonus Problem to Week 8 (#1)
« on: December 01, 2015, 06:52:43 AM »
PART A
In case the boldfacing isn't clear below, $\mathbf{k}$ and $\mathbf{x}$ are vectors.

a)
Let $\kappa$ be $\sqrt{2\pi}$. Then, applying the Plancherel theorem to each dimension:

||Fu(\mathbf{x})||=||\hat{u}(\mathbf{x})||=||u(\mathbf{x})||

Then $F^*F=I$.

This is an infinite dimensional operator, but right-multiplying the previous equation by $F^{-1}$ we get $F^*=F^{-1}$. Therefore, $FF^*=FF^{-1}=I$

b)

(F^2 f)(\mathbf{x}) = F(F(f(\mathbf{x})))=F(\hat{f}(\mathbf{k}))
=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) e^{-i\mathbf{k}\cdot\mathbf{x}}d^n\mathbf{k}
=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) e^{i\mathbf{k}\cdot(-\mathbf{x})}d^n\mathbf{k}
=f(-\mathbf{x})

c)
This is obvious from b.

(F^4f)(\mathbf{x})=(F^2F^2f)(\mathbf{x})=(F^2f)(-\mathbf{x})=f(\mathbf{x})

d)
Firstly, assume $f(\mathbf{x})$ is even,

Then

\hat{f}(\mathbf{k})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) e^{-i \mathbf{k} \cdot \mathbf{x}} d^n\mathbf{x}

Expanding the complex exponential:

\hat{f}(\mathbf{k})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) \cos(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x} - \frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) i\sin(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x}

Since $f(\mathbf{x})$ is even, we can repeat the calculation of the Fourier Transform with $f(-\mathbf{x})$ instead.

\hat{f}(\mathbf{k})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{-x}) \cos(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x} - \frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{-x}) i\sin(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x}

We can do the substitution $\mathbf{-x}=\mathbf{x}$. We get a negative factor in the differential when we do this, but the limits of integration are also reversed on each coordinate, so the negative factor cancels. Therefore:

\hat{f}(\mathbf{k})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) \cos(-\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x} - \frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) i\sin(-\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x}\\
=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) \cos(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x} + \frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) i\sin(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x}

Adding (5) and (7) and dividing by 2 we get:

\hat{f}(\mathbf{k})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) \cos(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x}

which is real if $f(\mathbf{x})$ is real.

Now assume that $\hat{f}(\mathbf{k})$ is real. We want to prove that $f(\mathbf{x})$ is even. The IFT is:

f(\mathbf{x})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x}} d^n\mathbf{k}

Expanding the complex exponential

f(\mathbf{x})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) \cos(\mathbf{k} \cdot \mathbf{x}) d^n\mathbf{k} + \frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) i\sin(\mathbf{k} \cdot \mathbf{x}) d^n\mathbf{k}

If $\hat{f}(\mathbf{k})$ is real, than the integrand of the term on the right is purely imaginary, whereas the term on the left is purely real. With the given assumption that $f(\mathbf{x})$ is real, the second term must necessarily be zero. Then:

f(\mathbf{x})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) \cos(\mathbf{k} \cdot \mathbf{x}) d^n\mathbf{k}

Therefore $f(\mathbf{x})$ is even.

The case of odd $f(\mathbf{x})$ can be done in the same way.

7
##### Web Bonus = Oct / Re: Web bonus problem : Week 3 (#4)
« on: November 30, 2015, 05:49:35 PM »
I didn't really make it that far on this problem, but I thought I ought to post as far as I got in case anyone has any ideas.

Plugging $u(x,t)=\phi(x-vt)$ into the given PDEs:

v^2\phi''-\phi''+\phi-2\phi^3=0\\
v^2\phi''-\phi''-\phi+2\phi^3=0

And so we get

(v^2-1)\phi''=2\phi^3-\phi\\
(v^2-1)\phi''=\phi-2\phi^3

This ODE looks very difficult - any ideas?

8
##### HA10 / Re: HA10-P3
« on: November 30, 2015, 04:31:16 PM »
Since $L$ does not depend on x explicitly, we have

H=u'L_{u'}=\frac{1}{c\left(u(x)\right)} \left(\frac{u'^2}{\sqrt{1+u'^2}} - \sqrt{1+u'^2}\right) = C

Which simplifies to

H=-\frac{1}{c\left(u(x)\right)} \frac{1}{\sqrt{1+u'^2}}=C

Letting $C=-1/A$,

-\frac{1}{c\left(u(x)\right)} \frac{1}{\sqrt{1+u'^2}}=-\frac{1}{A}\\
\frac{A^2}{c^2\left(u(x)\right)}=u'^2+1\\
\frac{du}{dx}=\sqrt{\frac{A^2}{c^2\left(u(x)\right)}-1}

We get

x=\int\frac{1}{\sqrt{\frac{A^2}{c^2\left(u(x)\right)}-1}}du

I don't think there's much more to be done without a formula for $c(y)$.

9
##### HA10 / Re: HA10-P3
« on: November 30, 2015, 12:55:39 AM »
Part A

In this problem, the Lagrangian is

L=\frac{\sqrt{1+u'^2}}{c(x,u(x))}

We need to calculate the Euler-Lagrange equation:

\frac{\partial L}{\partial u} - \frac{\partial}{\partial x} \frac{\partial L}{\partial u'} = 0

Calculating each term

\frac{\partial L}{\partial u} = -\frac{\sqrt{1+u'^2}}{c^2\left(x,u(x)\right)} \frac{\partial c}{\partial y}

\frac{\partial L}{\partial u'} = \frac{1}{c\left(x,u(x)\right)} \frac{u'}{\sqrt{1+u'^2}}\\
\frac{\partial}{\partial x} \frac{\partial L}{\partial u'} =
-\frac{1}{c^2\left(x,u(x)\right)} \frac{u'}{\sqrt{1+u'^2}} \left(\frac{\partial c}{\partial x}+\frac{\partial c}{\partial y}u'\right)
+\frac{1}{c\left(x,u(x)\right)}\frac{1}{\left(1+u'^2\right)^{3/2}} u''

Plugging into $(2)$:

-\frac{\sqrt{1+u'^2}}{c^2\left(x,u(x)\right)} \frac{\partial c}{\partial y}
+\frac{1}{c^2\left(x,u(x)\right)} \frac{u'}{\sqrt{1+u'^2}} \left(\frac{\partial c}{\partial x}+\frac{\partial c}{\partial y}u'\right)
-\frac{1}{c\left(x,u(x)\right)}\frac{1}{\left(1+u'^2\right)^{3/2}}u''
=0

We can simplify a little bit

\frac{1}{c\left(x,u(x)\right)^2\sqrt{1+u'^2}}\left(u'\frac{\partial c}{\partial x} - \frac{\partial c}{\partial y}\right)
-\frac{1}{c\left(x,u(x)\right)}\frac{1}{\left(1+u'^2\right)^{3/2}} u''
=0

\frac{1}{\sqrt{1+u'^2}}\left(u'\frac{\partial c}{\partial x} - \frac{\partial c}{\partial y}\right)
=\frac{c\left(x,u(x)\right)}{\left(1+u'^2\right)^{3/2}} u''

Will type up B later, Or if someone else wants to volunteer below. Since in part (b) the Lagrangian doesn't depend on $x$ explicitly, we can calculate the Hamiltonian ($H=u'L_{u'}-L$) and set it to a constant.

10
##### HA10 / Re: HA10-P1
« on: November 29, 2015, 01:38:21 AM »
We need to minimize the Lagrange functional. By subtracting the constraint multiplied by the Lagrange multiplier $\lambda$ from the original functional, we get:

\Phi \lbrack u \rbrack = \int_0^a(\rho gu\sqrt{1 + u'^2}-\lambda \sqrt{1+u'^2}) dx\\
= \int_0^a(\rho gu-\lambda)\sqrt{1 + u'^2} dx

The variation of $\Phi$ is:

\delta\Phi=\int_0^a \left(\frac{\partial L}{\partial u} - \frac{\partial}{\partial x}\frac{\partial L}{\partial u'}\right)\delta u dx

(The boundary term is zero since $\delta u|_0 = \delta u|_a = 0$)

We need to solve the Euler-Lagrange equation.

\frac{\partial L}{\partial u} - \frac{\partial}{\partial x}\frac{\partial L}{\partial u'} = 0

for

L = (\rho gu-\lambda)\sqrt{1 + u'^2}

Calculating the two terms...

\frac{\partial L}{\partial u} = \rho g\sqrt{1+u'^2}

\frac{\partial L}{\partial u'} = (\rho gu-\lambda) \frac{1}{\sqrt{1+u'^2}} u'

\frac{\partial}{\partial x}\frac{\partial L}{\partial u'} = \frac{\rho gu'^2}{\sqrt{1+u'^2}} + \frac{(\rho gu-\lambda)u''}{\sqrt{1+u'^2}} - \frac{(\rho gu-\lambda) u'^2 u''}{(1+u'^2)^{\frac{3}{2}}}

Plugging these results into $(3)$:

\rho g \sqrt{1+u'^2} - \frac{\rho gu'^2}{\sqrt{1+u'^2}} - (\rho gu-\lambda)u''  \left( \frac{1}{\sqrt{1+u'^2}}-\frac{u'^2}{(1+u'^2)^{\frac{3}{2}}} \right) = 0

Multiplying every term by $(1+u'^2)^{\frac{3}{2}}$:

\rho g(1+u'^2)^2 - \rho gu'^2 (1+u'^2) - (\rho gu - \lambda) u'' ((1+u'^2)-u'^2) = 0 \\
\rho g(1+u'^2)((1+u'^2)-u'^2) - (\rho gu - \lambda) u'' ((1+u'^2)-u'^2) = 0 \\
\rho g(1+u'^2)=(\rho gu - \lambda) u''

Dividing by $\rho g$,

(1+u'^2)=\left(u - \frac{\lambda}{\rho g}\right) u''

Doing some rearranging:

\frac{1}{1+u'^2} u''=\frac{1}{u - \frac{\lambda}{\rho g}}

Multiplying both sides by u'

\frac{1}{1+u'^2} u' u''=\frac{1}{u - \frac{\lambda}{\rho g}} u'

Integrating both sides with respect to $x$

\frac{1}{2} \ln{\left(1+u'^2\right)} = \ln{\left(u-\frac{\lambda}{\rho g}\right)} + C

Exponentiating:

1+u'^2=A^2\left(u-\frac{\lambda}{\rho g}\right)^2

By separation of variables

\frac{du}{dx}=\sqrt{A^2\left(u-\frac{\lambda}{\rho g}\right)^2-1}\\
dx=\frac{du}{\sqrt{A^2\left(u-\frac{\lambda}{\rho g}\right)^2-1}}

We get

\cosh(Ax+B)=A\left(u-\frac{\lambda}{\rho g}\right)

u=\frac{1}{A}\cosh(Ax+B)+\frac{\lambda}{\rho g}

What remains is to apply the boundary conditions. I'll figure the rest out tomorrow.

11
##### HA9 / Re: HA9-P4
« on: November 23, 2015, 01:52:09 AM »
This is my solution to 1a).

Proof of the maximum principle for subharmonic functions

Let $u$ be subharmonic.

\Delta u \geq 0

To prove that

\max_{\Omega}u=\max_{\Sigma}u

Assume, on the contrary, that the maximum is not on the boundary. Then the maximum must lie in the interior of $\Omega$. Call this point $y$. Then we draw a ball around it $B(y,r)$ such that it remains inside the domain $\Omega$. $u(y)$ is the global maximum of $\Omega$.

Now we assume the result of 2b, that $u(y)$ does not exceed the average of $u$ on this ball. But $u(y)$ is the global maximum, and thus the maximum on this ball. The only way these can hold simultaneously is if $u = u(y)$ on this entire ball. For if on $B(r,y)$, $u\neq u(y)$, then $u \leq u(y)$, and $u(y)$ would certainly exceed the average of $u$ on $B(r,y)$. Thus we conclude $u = u(y)$ on $B(r,y) \in \Omega$.

Now within this ball we take a new point $y'$. Then $g(y')$ is also the global maximum on $\Omega$. We repeat this over and over again and discover that on a connected $\Omega$, $g$ is identically equal to $g(y)$ including on the boundaries. Therefore, the $g$ is constant on all of $\Omega$, and so the maximum is again on the boundary, contradicting our original assumption.

Proof of the maximum principle for subharmonic functions

This proof is very similar.

12
##### HA9 / Re: HA9-P4
« on: November 23, 2015, 01:33:30 AM »
Hi again, this is my solution to part b. (Oops, I didn't realize P4 had 3 parts, so I did this problem before proving the maximum principle for subharmonic functions).

Let's prove that $u \geq v$. $u$ is harmonic and $v$ is subharmonic, so

\Delta{u} = 0 \\
\Delta{v} \geq 0

Define $g=v-u$. Then:

\Delta g \geq 0\\
g|_\Sigma = 0

We only need to prove that $g\leq0$ to obtain our desired result.

Let's assume, on the other hand that $g>0$. Then since $g$ is zero on the boundary, $g$ must have a maximum that is on the interior of $\Omega$.

Let $y$ denote the point on $\Omega$ where $g$ is maximum. Then we draw a ball around it $B(y,r)$ such that it remains inside the domain $\Omega$.

Now we assume the result of 2b, that $g(y)$ does not exceed the average of $g$ on this ball. But $g(y)$ is the global maximum, and also the maximum on this ball. The only way these can hold simultaneously is if $g = g(y)$ on this entire ball. If on $B(r,y)$, $g\neq g(y)$, then $g \leq g(y)$, and $g(y)$ would certainly exceed the average of $g$ on $B(r,y)$. Thus we conclude $g = g(y)$ on $B(r,y) \in \Omega$.

Now within this ball we take a new point $y'$. Then $g(y')$ is also the global maximum on $\Omega$. We repeat this over and over again and discover that on a connected $\Omega$, $g$ is identically equal to $g(y)$ and so $g>0$ on the boundary. This contradicts $g|_\Sigma = 0$. Therefore our original assumption cannot be correct, and so $g\leq0$. This proves that $u\geq v$.

Doing the exact same thing with $f=u-v$ proves that $v\geq u$

13
##### HA9 / Re: HA9-P2
« on: November 23, 2015, 12:57:32 AM »
(c):

u(y) is at least the mean value of u over the sphere S(y,r) bounding the ball $B(y,r)$ in which $\Delta u=0$:

u(y) \geq \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u dS

u(y) is at least the mean value of u over the ball $B(y,r)$ in which $\Delta u=0$:

u(y) \geq \frac{1}{\omega_n r^{n}} \int_{B(y,r)} u dV

14
##### HA9 / Re: HA9-P2
« on: November 23, 2015, 12:52:27 AM »
For (b),

Knowing that $u(y) \leq \frac{1}{\sigma_n\rho^{n-1}}\int_{S(y,\rho)} u(x) dS$

\int_{B(y,r)} u(x)dV = \int_0^r(\int_{S(y,\rho)}u(x)dS)d\rho \geq \int_0^r(\sigma_n \rho^{n-1} u(y))d\rho = u(y) (\sigma_n \int_0^r \rho^{n-1} d\rho)

So

u(y)\leq\frac{1}{\sigma_n \int_0^r \rho^{n-1} d\rho} \int_{B(y,r)} u(x)dV

Where $(\sigma_n \int_0^r \rho^{n-1} d\rho)$ is the volume of the ball.

15
##### HA9 / Re: HA9-P2
« on: November 23, 2015, 12:30:29 AM »
Hi everyone. I want to learn LaTeX, so I'm trying to solve (a) problem (using 7.2.7, in a different way that the guy above me). I haven't figured it out fully, and I'm not even sure if this is right. Can someone help?

We begin with equation 7.2.7.

u(y)=\int_{\Omega}G(x,y) \Delta u(x) dV + \int_{\Sigma}-u(x) \frac{\partial G}{\partial \nu_x}(x,y) dS + \int_{\Sigma}G(x,y) \frac{\partial u}{\partial \nu}(x)dS

Let's look at the second term. Let $y$ be at the centre of the sphere, $\Omega$ is a ball, and $\Sigma$ is it's bounding sphere. We can pull $\frac{\partial G}{\partial v_x}$ out of the integral, since when $y$ is at the centre of the sphere, $\frac{\partial G}{\partial \nu_x}(x,y)$ is constant over $\Sigma$. The second term becomes:

\frac{1}{\sigma_n r^{n-1}} \int_{\Sigma}u(x)dS

This is straight out of the textbook. This term represents the average of $u(x,y)$ over the sphere.

All that's left is to prove the remaining terms are negative (or zero) when evaluated at $y$. I'm not sure how to solve this problem for $n<3$, but it is easy for $n\geq3$, because when $n\geq3$, $G(x,y)$ is always negative (see 7.2).

Let's look at the first term. Since $G(x,y)<0$ and $\Delta u(x)\geq0$ on $\Omega$, the integrand of the first term is always non-positive, so the integral is non-positive.

\int_{\Omega}G(x,y) \Delta u(x) dV \leq 0

The third term is easy too. $G(x,y)$ depends on $|x-y|$ only. Thus, if $y$ is at the centre of the sphere, $G(x,y)$ is constant over the sphere $\Sigma$. Pulling it out of the integral, we get

G(x,y) \int_{\Sigma} \frac{\partial u}{\partial \nu} (x) dS

By the divergence theorem, this term is

G(x,y) \int_{\Omega} \Delta u(x) dS

Since $\Delta u(x)$ is non-negative on the domain $\Omega$, so is the integral. Since $G(x,y)$ is always negative at $y$ (at least for $n\geq3$), this term is negative (or zero).

\int_{\Sigma}G(x,y) \frac{\partial u}{\partial \nu}(x)dS \leq 0

Any idea how to approach this problem for lower $n$? If 7.2.7 is true for $n=2$, it seems possible to create a function $u$ such both the first and third term are positive.

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