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4

The identity $$ \int_{-\infty}^{+\infty}dt\ f(t) \delta(t) = f(0)\tag{*} $$ is meaningless without context. Also this notation is a convenient abuse of notation, and not a standard (Riemann or Lebesgue) integral. Let's say you are considering $\delta:\mathcal{S}(\mathbb{R})\to\mathbb{R}$ as a tempered distribution on the Schwartz class $\mathcal{S}(\...


2

My guess is that the author makes a hypothesis that the solution is a distribution which can be represented by a $L^2_{loc}$ function in the sense that there exists a function $g\in L^2_{loc}$ such that for any test function $\phi$ $$\langle u,\phi\rangle=\int_{(0,T)\times \Bbb R^d}g(t,x)\phi(t,x)\,\mathrm dt\,\mathrm dx$$


2

Your computation is correct. Someone dropped the minus sign when stating the exercise. Remembering the calculus formula $(1/x)=-1/x^2$ would help prevent that. The calculus formula doesn't prove the result for distributions, but it must be consistent with it in the sense that differentiation away from singularities is the same as classical derivative.


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As has been mentioned in some comments, this is the definition of distributional derivative. In general, if $\alpha \in \mathbb{N}^n$ is a multi-index, the distributional derivative of order $\alpha$, of a distribution $T \in \mathcal{D}'(\Omega)$ is defined by placing $(\star)$ $\displaystyle \langle D^\alpha T, \varphi \rangle =(-1)^{|\alpha|} \langle T, ...


1

For every point $x\in V_p\cap V_q$, you can use bump functions $\varphi_n$ (with integral $1$) supported in $(1/n)$-neighborhood of $x$ to conclude that $$u_p(x) = \lim_{n\to\infty} \langle u_p, \varphi_n\rangle = \lim_{n\to\infty} \langle v_p, \varphi_n\rangle = v_p(x)$$ Put another way, $\varphi_n$ converge to the Dirac delta at $x$ in the sense of ...


1

One can consider the distributional derivative of any locally integrable function, which includes all continuous functions. This doesn't mean there's much to say about them other than "they exist". am I correct in understanding that the distributional derivative of the brownian motion is a measure No, it is not a measure. The distributional ...


1

Ok let me see if I can't answer your question from a mathematical perspective in a way that might be helpful for an engineer. The first thing you need to understand is that $\varphi_{n}(x)=\frac{1}{\sqrt{2\pi}}e^{ikx} , \, k \in \mathbb{Z}$ forms an orthonormal sequence in $L^{2}([-\pi,\pi]))$. Now, anytime you take the inner product of two orthonormal ...


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I assume that the OP actually wants to solve the following ODE $$y'' + y = \delta + \delta'$$ rather than $y'' + y' = \delta + \delta'$. Assuming zero initial conditions and taking the Laplace transform of both sides, $$s^2 Y (s) + Y (s) = 1 + s$$ and, thus, $$Y (s) = \dfrac{s+1}{s^2+1} = \dfrac{s}{s^2+1} + \dfrac{1}{s^2+1}$$ Taking the inverse Laplace ...


1

$$y''(t)+y'(t)=\delta(t)+\delta'(t)$$ $\delta(t)$ is the Dirac function. First integration : $$y'+y=u(t)+\delta(t)+c_1$$ $u(t)$ is the Heaviside function. Seconf integration : The solution of the homogeneous ODE $y'+y=0$ is $y=c\:e^{-t}$ Let $y(t)=f(t)e^{-t}$ $(f'-f)e^{-t}+fe^{-t}=u(t)+\delta(t)+c_1$ $f'=(u(t)+\delta(t)+c_1)e^t$ $f=(e^t-1)u(t)+u(t)+...


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Here are some ideas to get you started: If you were using the space $W^{1,p}(\newcommand{\R}{\mathbb R} \R^+)$ you would define the reflection $\bar u(x) = u(-x)$ for $x < 0$ and $\bar u(x) = u(x)$ for $x > 0$. This gives you continuity of $\bar u$ at $x = 0$ from which the absolute continuity of $\bar u$ follows. Suppose instead you are using the ...


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The first question follows from the fact that $u_m \rightarrow \overline{u}$ in $W^{1,p}(\mathbb{R}^n)$ implies $||u_m-u_l||_{L^{p^{\ast}}(\mathbb{R}^n)} \leq C ||Du_m-Du_l||_{L^p(\mathbb{R}^n)} \leq C[||Du_m-D\overline{u}||_{L^p(\mathbb{R}^n)} + ||D\overline{u}-Du_l||_{L^p(\mathbb{R}^n)}] \rightarrow 0$ as $m,j \rightarrow \infty$, and $\lbrace u_m \...


1

I worked this out referring to what you have already done and what Ian suggested. Also refer to https://proofwiki.org/wiki/Scaling_Property_of_Dirac_Delta_Function . To prove $\delta(kx)=\frac{1}{|k|} \delta(x)$ , instead prove ${|k|}\delta(kx)= \delta(x)$, where k is a nonzero real constant. Use the definition of $\delta(x)$ consiting of the first ...



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