exercise09.tex

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% % \def\hidesolutions{}
\input{header}
\begin{document}
\exsheet{9}{14}{November} % parameters are the number of the session and the day




\begin{exercise}
    Show that 
    \begin{align}
        \int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}. 
    \end{align}
    \textit{Hint: square the integral and convert it to polar/radial coordinates.}
\end{exercise}
\begin{solution}
    Following the hint, we find
    \begin{align}
        \left(\int_{-\infty}^\infty e^{-x^2} dx\right)^2 &= \int_{-\infty}^\infty e^{-x^2} dx \int_{-\infty}^\infty e^{-y^2} dy = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2 + y^2)} dx dy\\
        &= \int_0^{2\pi} \int_0^\infty e^{-r^2} r dr d\theta = 2\pi \int_0^\infty e^{-r^2} r dr \\
        &= \pi \int_0^\infty e^{-u} du = \pi [-e^u]_{u=0}^{u=\infty} = \pi,
    \end{align}
    where we have used Fubini's Theorem and the substitution $u = r^2$. Taking the square root of both sides gives the desired result.
\end{solution}


\begin{exercise}
    We check whether the following are distributions or not.
    \begin{itemize}    
        \item 
        Show that $T$ is a distribution, where 
        \begin{align}\label{eq:ex2 dist1}
            T(\phi) = \int_{-1}^{1} \phi(x) \ dx
            .
        \end{align}
        \item 
        Show that $T$ is a distribution, where 
        \begin{align}\label{eq:ex2 dist2}
            T(\phi) = \int_{-\infty}^{\infty} \phi(x) \ dx
            .
        \end{align}
        \item 
        Show that $S$ is not a distribution, where 
        \begin{align}\label{eq:ex2 dist3}
            S(\phi) = \int_{0}^{1} |\phi(x)| \ dx 
            .
        \end{align}
    \end{itemize}
\end{exercise}
\begin{solution}
    The definition in \eqref{eq:ex2 dist1} is clearly linear in $\phi$ as the integral is linear
     and finite for any $\phi \in \mathcal{D}$ since continuous functions on compact sets are bounded. In particular, we have
     \begin{align}
            |T(\phi)| = \left| \int_{-1}^{1} \phi(x) \ dx \right| \leq \int_{-1}^{1} |\phi(x)| \ dx \leq \sup_{x \in [-1,1]} |\phi(x)| \int_{-1}^{1} dx = 2 \sup_{x \in [-1,1]} |\phi(x)|,
     \end{align}
     which proves the continuity condition.\\

    The definition in \eqref{eq:ex2 dist2} is also linear in $\phi$ for the same reason as before. To show that the integral is finite for any $\phi \in \mathcal{D}$, we use the fact that $\phi$ is compactly supported and hence bounded.
     In particular, there exists for any $\phi \in \mathcal{D}$ a compact interval $[a, b] \subset \mathbb{R}$ such that $\mathrm{supp}(\phi) \subset [a, b]$. Then we have
    \begin{align}
        |T(\phi)| = \left| \int_{-\infty}^{\infty} \phi(x) \ dx \right| \leq \int_{a}^{b} |\phi(x)| \ dx \leq \sup_{x \in [a,b]} |\phi(x)| \int_{a}^{b} dx = (b-a) \sup_{x \in [a,b]} |\phi(x)| < \infty,
    \end{align}
    which shows that the integral is finite for any $\phi \in \mathcal D$. Conversely, for any compact interval $[a, b] \in \mathbb{R}$ and $\phi \in \mathcal{D}$, such that $\mathrm{supp}(\phi) \subset [a, b]$, we have
    \begin{align}
        |T(\phi)| = \left| \int_{-\infty}^{\infty} \phi(x) \ dx \right| \leq \int_{a}^{b} |\phi(x)| \ dx \leq \sup_{x \in [a,b]} |\phi(x)| \int_{a}^{b} dx = (b-a) \sup_{x \in [a,b]} |\phi(x)|,
    \end{align}
    which shows the continuity condition.
    \\

    The definition in \eqref{eq:ex2 dist3} is not a distribution since it is not linear in $\phi$. 
    There are many different ways to see this. 
    For example, consider $\phi \in \mathcal{D}$ such that $\phi(x) \geq 0$ for all $x \in \mathbb{R}$. 
    We have seen such a function in the lecture. 
    Set $\psi = -\phi \in \mathcal D$. Then we have
    \begin{align}
        S(\phi + \psi) = \int_0^1 |\phi(x) + \psi(x)| \ dx = \int_0^1 0 \ dx = 0,
    \end{align}
    but
    \begin{align}
        S(\phi) + S(\psi) = \int_0^1 |\phi(x)| \ dx + \int_0^1 |\psi(x)| \ dx = 2 \int_0^1 |\phi(x)| \ dx > 0.
    \end{align}
\end{solution}


\begin{exercise}
    Suppose that $f : \mathbb R \rightarrow \mathbb R$ is an integrable function. 
    Whenever $a > 0$, show that 
    \begin{align}
        T( \phi ) = a \langle f, \phi(a \cdot) \rangle 
    \end{align}
    is a distribution. 
\end{exercise}
\begin{solution}
    With the substitution $u = ax$, hence $du = a dx$, we find 
    \begin{align}
        \int f(u) \phi(u/a) \frac 1 a \ d u
        = 
        \int f(ax) \phi(x) \ d x 
        .
    \end{align}
    Clearly, the function $g(x) := f(ax)$ is integrable. That means 
    \begin{align}
        T( \phi ) = \int g(x) \phi(x) \ dx,
    \end{align}
    which is a distribution. 
\end{solution}



\begin{exercise}
    Find the distributional derivative of the function 
    \begin{align}
        f(x) = |x|.
    \end{align}
\end{exercise}
\begin{solution}
    For any test function $\phi \in \mathcal{D}$, we have 
    \begin{align}
        \langle f', \phi \rangle 
        &= 
        - \langle f, \phi' \rangle 
        = 
        - \int_{-\infty}^{\infty} |x| \phi'(x) \ dx 
        \\&
        = 
        - \int_0^\infty x \phi'(x) \ dx + \int_{-\infty}^0 x\phi'(x) \ dx 
        \\
        &= 
        -[x \phi(x)]_0^\infty + \int_0^\infty \phi(x) \ dx + [x \phi(x)]_{-\infty}^0 - \int_{-\infty}^0 \phi(x) \ dx 
        \\
        &= 
        \int_0^\infty \phi(x) \ dx - \int_{-\infty}^0 \phi(x) \ dx 
        ,
    \end{align}
    where we have used that $\phi$ is compactly supported and hence the boundary terms vanish.
    We define the function $g$ by 
    \begin{align}
        g(x) = \begin{cases} -1 & \text{ if } x < 0, \\ 1 & \text{ if } x \geq 0. \end{cases}
    \end{align}
    Apparently,
    \begin{align}
        \langle f', \phi \rangle = \langle g , \phi \rangle. 
    \end{align}
    Note that we can also write this in terms of the Heaviside step function, 
    \begin{align}
        H(x) = \begin{cases}
                0 & \text{ if } x    < 0, 
                \\
                1 & \text{ if } x \geq 0.
               \end{cases}
    \end{align}
    One easily sees that $g(x) = 2 H(x) - 1$ is an alternative way of writing the distributional derivative. 
\end{solution}


\begin{exercise}
    Consider the function with period $2$ that satisfies 
    \begin{align}
        f(x) = \begin{cases}
                -x & \text{ if } -1 < x \leq 0 \\
                 x & \text{ if }  0 < x \leq 1 \\
               \end{cases}
    \end{align}
    \begin{itemize}
     \item Draw a plot of this function from $-4$ to $4$.
     \item Is this function differentiable? What is the first distributional derivative of this function?
     \item What is the second distributional derivative of this function?
    \end{itemize}
\end{exercise}
\begin{solution}
    \begin{figure}
    \centering
    \begin{tikzpicture}
        \begin{axis}[
            axis lines=middle,
            xlabel={$x$},
            ylabel={$f(x)$},
            xmin=-4, xmax=4,
            ymin=0, ymax=2,
            samples=400,
            domain=-4:4,
            xtick={-4,-3,...,4},
            ytick={-1, 0, 1},
            grid=major
        ]
        % Plot the piecewise periodic function
        \addplot[thick, blue, domain=-4:4] 
            {(mod(abs(x), 2)) * (mod(abs(x), 2) <= 1) + (2 - mod(abs(x), 2)) * (mod(abs(x), 2) > 1)};
        \end{axis}
    \end{tikzpicture}
    \caption{Plot of the periodic function $f$.}
    \label{fig:ex5}
    \end{figure}
    The solution is plotted in Figure \ref{fig:ex5}. 
    The function is not differentiable at $x \in \mathbb{Z}$ since the left and right derivatives do not agree:
    for example, at $x=0$ the derivative from the left equals $-1$ but the derivative from the right equals $1$.
    
    It remains to compute the distributional derivative. 
    If we look at the drawing we see that the function is continuous and differentiable over the intervals between integers. 
    Over the interval $(n,n+1)$, the derivative is either $1$ if $n$ is even or it is $-1$ if $n$ is odd. 
    Moreover, the function does not jump. 
    This gives us an idea what the distributional derivative should look like:
    \begin{align}
        f'(x) = \begin{cases}
                 1 & \text{ if } x \in (n,n+1), \text{ $n$ even, } 
                 \\
                -1 & \text{ if } x \in (n,n+1), \text{ $n$ odd. } 
               \end{cases}
    \end{align}
    Let us formalize this intuition. We use the definition of distributional derivative. 
    For any test function $\phi \in \mathcal{D}$, 
    we have by definition of the distributional derivative that
    \begin{align*}
        \langle f', \phi \rangle 
        &
        = 
        - \langle f, \phi' \rangle 
        \\&
        = 
        - \int_{-\infty}^{\infty} f(x) \phi'(x) \ dx 
        \\&
        = 
        - 
        \sum_{\substack{n \in \mathbb{Z}\\ n \text{ even}}} \int_{n}^{n+1} f(x) \phi'(x) \ dx
        -
        \sum_{\substack{n \in \mathbb{Z}\\ n \text{ odd} }} \int_{n}^{n+1} f(x) \phi'(x) \ dx
        \\&
        = 
        \sum_{\substack{n \in \mathbb{Z}\\ n \text{ even}}} \int_{n}^{n+1} \phi(x) \ dx - f(n+1)\phi(n+1) - f(n)\phi(n)
        \\&\qquad 
        -
        \sum_{\substack{n \in \mathbb{Z}\\ n \text{ odd} }} \int_{n}^{n+1} \phi(x) \ dx - f(n+1)\phi(n  ) - f(n)\phi(n)
        \\&
        = 
        \sum_{\substack{n \in \mathbb{Z}\\ n \text{ even}}} \int_{n}^{n+1} \phi(x) \ dx
        -
        \sum_{\substack{n \in \mathbb{Z}\\ n \text{ odd} }} \int_{n}^{n+1} \phi(x) \ dx
        \\&\qquad 
        +
        \sum_{\substack{n \in \mathbb{Z}\\ n \text{ even}}} - f(n+1)\phi(n+1) - f(n)\phi(n) 
        -
        \sum_{\substack{n \in \mathbb{Z}\\ n \text{ odd} }} - f(n+1)\phi(n  ) - f(n)\phi(n) 
        .
    \end{align*}
    We simplify this. The point evaluations cancel out each other.
    Then 
    \begin{align}
        \langle f', \phi \rangle 
        &
        = 
        \sum_{\substack{n \in \mathbb{Z} }} \int_{n}^{n+1} (-1)^{n} \phi(x) \ dx
        .
    \end{align}
    This is exactly the form of $f'$ as proposed above. It also corresponds to our observations in the lecture. 

    For the second distributional derivative, we again use the definition and simplify:
    \begin{align}
        \langle f'', \phi \rangle 
        &
        = 
        -
        \langle f', \phi' \rangle 
        \\&
        = 
        -
        \sum_{\substack{n \in \mathbb{Z} }} (-1)^{n} \int_{n}^{n+1} \phi'(x) \ dx
        \\&
        = 
        -
        \sum_{\substack{n \in \mathbb{Z} }} (-1)^{n} \left( \phi(n+1) - \phi(n) \right)
        \\&
        = 
        \sum_{\substack{n \in \mathbb{Z} }} (-1)^{n+1} \left( \phi(n+1) - \phi(n) \right)
        .
    \end{align}
    As we now carefully observe for term in the sum:
    when $n$ is odd, the summand is $\phi(n+1) - \phi(n)$, and when $n$ is even, then the summand is $- \phi(n+1) + \phi(n)$. 
    In other words, for every even integer $n$ we receive two positive Dirac deltas, and for every odd integer $n$ we receive two negative Dirac deltas. 
    Thus,
    \begin{align}
        \langle f'', \phi \rangle 
        =
        \sum_{\substack{n \in \mathbb{Z} }} (-1)^{n} 2 \delta_{n}
        .
    \end{align}
	This corresponds to the jumps of the distributional derivative $f'$.
	we jump (up)   by  $2$ at every even integer 
	and 
	we jump (down) by $-2$ at every odd integer.
\end{solution}



\end{document}