Toronto Math Forum

For part c) we do essentially the same thing, calculating
\begin{equation}
\hat{f}(\mathbf{k})=a^2 \hat{g}(\mathbf{k})- i\hat{h}'(k)
\end{equation}

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:

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

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

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 $\mathbf{x}=(r,\theta)$.

\begin{equation}
\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}
\end{equation}
(since in cartesian coordinates $k_x=k$ and $k_y=0$, we can simplify the dot product)


Doing a change of coordinates we get,
\begin{equation}
\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
\end{equation}

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

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

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:
\begin{equation}
||Fu(\mathbf{x})||=||\hat{u}(\mathbf{x})||=||u(\mathbf{x})||
\end{equation}

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)
\begin{equation}
(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})
\end{equation}

c)
This is obvious from b.
\begin{equation}
(F^4f)(\mathbf{x})=(F^2F^2f)(\mathbf{x})=(F^2f)(-\mathbf{x})=f(\mathbf{x})
\end{equation}

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

Then
\begin{equation}
\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}
\end{equation}

Expanding the complex exponential:
\begin{equation}
\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}
\end{equation}

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

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:
\begin{equation}
\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}
\end{equation}

Adding (5) and (7) and dividing by 2 we get:
\begin{equation}
\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}
\end{equation}
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:
\begin{equation}
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}
\end{equation}

Expanding the complex exponential
\begin{equation}
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}
\end{equation}

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:

\begin{equation}
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}
\end{equation}

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

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

Since $L$ does not depend on x explicitly, we have
\begin{equation}
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
\end{equation}

Which simplifies to
\begin{equation}
H=-\frac{1}{c\left(u(x)\right)} \frac{1}{\sqrt{1+u'^2}}=C
\end{equation}

Letting $C=-1/A$,

\begin{equation}
-\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}
\end{equation}

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

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

We need to minimize the Lagrange functional. By subtracting the constraint multiplied by the Lagrange multiplier $\lambda$ from the original functional, we get:
\begin{equation}
\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
\end{equation}

The variation of $\Phi$ is:
\begin{equation}
\delta\Phi=\int_0^a \left(\frac{\partial L}{\partial u} - \frac{\partial}{\partial x}\frac{\partial L}{\partial u'}\right)\delta u dx
\end{equation}
(The boundary term is zero since $\delta u|_0 = \delta u|_a = 0$)

We need to solve the Euler-Lagrange equation.
\begin{equation}
\frac{\partial L}{\partial u} - \frac{\partial}{\partial x}\frac{\partial L}{\partial u'} = 0
\end{equation}
for
\begin{equation}
L = (\rho gu-\lambda)\sqrt{1 + u'^2}
\end{equation}

Calculating the two terms...
\begin{equation}
\frac{\partial L}{\partial u} = \rho g\sqrt{1+u'^2}
\end{equation}
\begin{equation}
\frac{\partial L}{\partial u'} = (\rho gu-\lambda) \frac{1}{\sqrt{1+u'^2}} u'
\end{equation}
\begin{equation}
\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}}}
\end{equation}

Plugging these results into $(3)$:
\begin{equation}
\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
\end{equation}

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

Dividing by $\rho g$,
\begin{equation}
(1+u'^2)=\left(u - \frac{\lambda}{\rho g}\right) u''
\end{equation}

Doing some rearranging:
\begin{equation}
\frac{1}{1+u'^2} u''=\frac{1}{u - \frac{\lambda}{\rho g}}
\end{equation}

Multiplying both sides by u'
\begin{equation}
\frac{1}{1+u'^2} u' u''=\frac{1}{u - \frac{\lambda}{\rho g}} u'
\end{equation}

Integrating both sides with respect to $x$
\begin{equation}
\frac{1}{2} \ln{\left(1+u'^2\right)} = \ln{\left(u-\frac{\lambda}{\rho g}\right)} + C
\end{equation}

Exponentiating:
\begin{equation}
1+u'^2=A^2\left(u-\frac{\lambda}{\rho g}\right)^2
\end{equation}
By separation of variables
\begin{equation}
\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}}
\end{equation}

We get
\begin{equation}
\cosh(Ax+B)=A\left(u-\frac{\lambda}{\rho g}\right)
\end{equation}

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

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

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

\begin{equation}
\Delta{u} = 0 \\
\Delta{v} \geq 0
\end{equation}

Define $g=v-u$. Then:
\begin{equation}
\Delta g \geq 0\\
g|_\Sigma = 0
\end{equation}

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 $v\geq u$

For (b),

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

\begin{equation}
\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)
\end{equation}

So

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


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