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Chapter 12: Problem 3

Which of the following is a distribution? (a) \(T(\phi)=\sum_{n=1}^{m} \lambda_{n} \phi^{(n)}(0) \quad\left(\lambda_{n} \in \mathbb{R}\right)\) (b) \(T(\phi)=\sum_{n=1}^{m} \lambda_{n} \phi\left(x_{n}\right) \quad\left(\lambda_{n}, x_{n} \in \mathbb{R}\right)\). (c) \(T(\phi)=(\phi(0))^{2}\). (d) \(T(\phi)=\sup \phi\) (c) \(T(\phi)=\int_{-\infty}^{\infty}|\phi(x)| \mathrm{d} x\).

Short Answer

Expert verified
The distributions are (a) \(T(\phi)=\sum_{n=1}^{m} \lambda_{n} \phi^{(n)}(0) \quad(\lambda_{n}\in \mathbb{R})\) and (b) \(T(\phi)=\sum_{n=1}^{m} \lambda_{n} \phi\left(x_{n}\right) \quad(\lambda_{n}, x_{n} \in \mathbb{R})\)

Step by step solution

01

Checking Option (a)

Option (a) \(T(\phi)=\sum_{n=1}^{m} \lambda_{n} \phi^{(n)}(0) \quad(\lambda_{n} \in \mathbb{R})\) is a distribution because it is linear in \(\phi\) and also sends compactly supported smooth functions into continuous linear functionals.
02

Checking Option (b)

Option (b) \(T(\phi)=\sum_{n=1}^{m} \lambda_{n} \phi\left(x_{n}\right) \quad(\lambda_{n}, x_{n} \in \mathbb{R})\) is also a distribution for the same reasons as in Option (a). It is linear in \(\phi\) and sends compactly supported smooth functions into real numbers in a linear way.
03

Checking Option (c)

Option (c) \(T(\phi)=(\phi(0))^{2}\) is not a distribution because it is nonlinear which eventuates into failure to satisfy the required properties of the distributions.
04

Checking Option (d)

Option (d) \(T(\phi)=\sup \phi\) is not a distribution because it is not linear in \(\phi\). The supremum operation is a nonlinear operation, so this does not satisfy the linearity property that distributions must have.
05

Checking Option (e)

Option (e) \(T(\phi)=\int_{-\infty}^{\infty}|\phi(x)| \mathrm{d} x\) is not a distribution because the absolute value operation makes the function nonlinear in \(\phi\), so it does not satisfy the linearity property of distributions

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Most popular questions from this chapter

Find the Fourier transforms of the functions $$ f(x)= \begin{cases}1 & \text { if }-a \leq x \leq a \\ 0 & \text { otherwise }\end{cases} $$ and $$ g(x)= \begin{cases}1-\frac{\mid x}{2} & \text { if }-a \leq x \leq a \\ 0 & \text { otherwisc }\end{cases} $$

Show that the Fourier transform of the distribution $$ \delta_{0}+\delta_{d i}+\delta_{2 a}+\cdots+\delta_{(2 n-1)} $$ is a distribution with density $$ \frac{1}{\sqrt{2 \pi}} \frac{\sin (n a y)}{\sin \left(\frac{1}{2} a y\right)} \mathrm{e}^{-\left(n-\frac{1}{2}\right)+2 y} $$ Show that $$ \mathcal{F}^{-1}\left(f(y) e^{2 b y}\right)=\left(\mathcal{F}^{-1} f\right)(x+b) $$ Hcnce find the inverse Fourier transform of $$ g(y)=\frac{\text { sin } n a y}{\sin \left(\frac{1}{2} a y\right)} $$

Show that the Green ' function for the one-dimensional diffusion equation, $$ \left.\frac{\partial^{2} G(x, t)}{\partial x^{2}}-\frac{1}{\kappa} \frac{\partial G(x, t)}{\partial t}=\varepsilon\left(x-x^{\prime}\right)\right\\}\left(t-t^{\prime}\right) $$ is given by $$ G\left(x-x^{\prime}, t-t^{\prime}\right)=-\theta\left(t-t^{\prime}\right) \sqrt{\frac{K}{4 \pi\left(t-t^{\prime}\right)}} e^{-\left(x-r^{\prime}\right)^{2} / 4(t-h)} $$ and write out the corresponding solution of the inhomogencous equation $$ \frac{\partial^{2} \psi(x, t)}{\partial x^{2}}-\frac{1}{x^{\prime}} \frac{\partial \psi(x, t)}{\partial t}=F(x, t) $$ Do the same for the two- and thrce-dimensional diffusion equations $$ \nabla^{2} G(x, t)-\frac{1}{\kappa} \frac{\partial G(x, t)}{\partial t}=\delta^{n}\left(x-x^{\prime}\right) \delta\left(t-t^{\prime}\right) \quad(n=2,3) $$

Evaluate (a) \(\int_{-\infty}^{\infty} \mathrm{e}^{a t} \sin b t \delta^{(n)}(t) \mathrm{d} t \quad\) for \(n=0,1,2\). (b) \(\int_{-\infty}^{\infty}(\cos t+\sin t) \delta^{(n)}\left(t^{3}+t^{2}+t\right) d t \quad\) for \(n=0,1\).

Show the identities $$ \frac{d}{d x}(\delta(f(x)))=f^{\prime}(x) \delta^{\prime}(f(x)) $$ and $$ \delta(f(x))+f(x) \delta^{\prime}(f(x))=0 $$ Hence show that \(\phi(x, y)=\delta\left(x^{2}-y^{2}\right)\) is a solution of the partial differcntial equation $$ x \frac{\partial \phi}{\partial x}+y \frac{\partial \phi}{\partial y}+2 \phi(x, y)=0 $$
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