New answers tagged

1

Intuitively, the delta is on the line $t=\tau$. If the cell $[k,k-1]\times[j,j-1]$ doesn't contain that line, the integral is $0$. So now consider the integral $$\int_{j-1}^j \int_{k-1}^k \delta(t-\tau)dtd\tau = \int_{j-1}^j H(t-\tau)\Bigr|_{k-1}^k d\tau$$ $$ = \int_{k-1}^k H(k-\tau) - H(k-1-\tau) d\tau = \int_{k-j}^{k-j+1} H(\tau')-H(\tau'-1)d\tau'$$ ...


0

$$ x(-t) \ast \delta(-t-t_0) = x(-t) \ast \delta(t+t_0) = x(-t) \ast \delta(t-(-t_0)) \\ = x(-(t-(-t_0))) = x(-(t+t_0)). $$ First equality: $\delta$ is even. Second equality: small rewrite. Third equality: $f(t) \ast \delta(t-t_0) = f(t-t_0)$. Fourth equality: obvious rewrite.


3

Based on Wikipedia, the Dirac delta can be loosely thought of as a function on the real line that is defined by $$ \delta (t)=\begin{cases}+\infty ,&t=0\\0,&t\neq 0\end{cases}$$ and which satisfies $$ \int _{-\infty }^{\infty }\delta (t)\,dt=1.$$ This is merely a heuristic characterization. The Dirac delta is not a function in the traditional ...


1

Recall that the Green's function $G(x,s)$ for this BVP solves $$y''-2y'+y = \delta(x-s), \ \ \ y(-1)=y(1)=0.$$ Since this equation is linear, we can add solutions and all that will change is the right-hand side. If we choose $y(x) = G(x,1/2)+G(x,-1/2)$, then $y$ solves the equation with $g(x) = \delta(x-1/2)+\delta(x+1/2)$


0

When we define the distributional derivative, the form is like \begin{align*} \left<\partial^{\alpha}\varphi,f\right>=(-1)^{|\alpha|}\left<\varphi,\partial^{\alpha}f\right> \end{align*} for good functions $f$, we define in this way such that the integration by parts is allowed.


1

Technically yes. The first term is just zero; in fact this always happens when $f$ is a test function and $d$ is a distribution*, so that distributional integration by parts reads $\int f d' = -\int f' d$ (in shorthand). The second term is just evaluated using the definition of $\delta$. * Here I mean either "$f$ is compactly supported smooth and $d$ is a ...


1

You should integrate by parts twice ti shift the derivatives off the delta and on to the function f(p, p'). Doing this, and assuming the boundary terms cannot contribute (they should not saturate the delta function) will leave $$(-1)^2\int dp dp' f''(p, p') \delta(p-p'). $$ Can you finish from here?


0

If we regularize OP's expressions as a smooth function in $C^{\infty}(\mathbb{R}^4)$, in the sense of generalized functions$^1$ $$ \frac{1}{p^2+\varepsilon}~\rightarrow~ {\rm P.V.}\frac{1}{p^2} \quad\text{for}\quad\varepsilon\to 0^+, \tag{2i}$$ then the derivatives $\partial_{\mu} ~:=~\frac{\partial}{\partial p_{\mu}}$ are well-defined: $$ \partial_{\mu}\...


1

In the classical sense, the integrals do not exist for any $n$, at the value $x=0$. For $n\geq 0$, this is because $a^n$ does not decay at infinity. For $n<0$, the singularity at $0$ is too poor. If we instead consider the tempered distributions associated with $a^n$, that is the functional $T_{a^n}\in \mathcal{S}'(\mathbb{R})$ given by $T_{a^n} (\phi) = \...


1

The best way to think of the dirac delta function is: $$\delta_n(x)=\begin{cases} 0 & x<0 \\ n & 0<x<\frac1n \\ 0 & x>\frac1n \end{cases}$$ $$\delta(x)=\lim_{n\to\infty}\delta_n(x)$$ Now try and integrate $\delta_n(x)$ first then take the limit: $$\int_{-\infty}^\infty\delta_n(x)dx=\int_{-\infty}^0\delta_n(x)dx+\int_0^{\frac1n}\...


1

Formally, the dirac delta is not a function in the same sense as the heaviside function. Rather, it is a distribution.


0

The linear map $\delta_{t}$ can act on the functions, as long as the function is defined at $t$, and the act is simply $\left<\delta_{t},f\right>=f(t)$. The problem is that, whether this linear map is continuous with respect to certain topology. For Schwartz functions $f$, not necessarily have compact support, $\delta_{t}$ is still continuous on the ...


0

Note that here $\delta_x$ is a measure, i.e. a function taking a set (belonging to some $\sigma$-algebra) and giving a nonnegative value. In this case, for any $A \subseteq \mathbb{R}$, $$ \delta_x(A) := \begin{cases} 1 & \text{if } x\in A \\ 0 & \text{if } x\not\in A \end{cases} $$ I won't go into the definitions of integrals of functions w.r.t. a ...


0

You can completely avoid going over the Dirac-delta terms. Your equation has the form $x+ax'=c(F'+bF)$, $x(t)=0=F(t)$ for $t<0$. Now apply the integrating factor $e^{at}$ and compress the right side by a similar factor $$ (e^{at}x(t))'=ce^{(a-b)t}(e^{bt}F(t))' $$ and apply the product rule/partial integration $$ e^{at}x(t)=ce^{at}F(t)-c(a-b)\int_0^te^{as}...


1

Here's an idea for a proof that seems to be semi-rigorous and within the limitations of the class material (I would give full points to it at least :P). Using complex analysis arguments it is easy to prove that $$\theta_{\epsilon}(x)=e^{-\epsilon x}\theta(x)~,~x\neq0$$ If we define $\theta'(x)\equiv\lim_{\epsilon\to 0}\theta'_{\epsilon}(x)$ then it is ...


-2

Notice that the first derivative of $|x|$ is the sign function. This is why the second derivative is the Delta Dirac function.


0

It is $\frac{3}{2}A_1'(t)$. \begin{align*} &\int_0^L \sum_{i=1}^3 \delta(x-x_i) \cos \left( \frac{2\pi x}{L} \right) \left( A_1'(t) \cos \left( \frac{2\pi x}{L} \right) + B_1'(t) \sin \left( \frac{2\pi x}{L} \right)\right) \,\mathrm{d}x \\ &= \sum_{i=1}^3 \int_0^L \delta(x-x_i) \cos \left( \frac{2\pi x}{L} \right) \left( A_1'(t) \cos \left( \frac{2\...


2

I suggest reading about distributions (generalised functions), e.g. https://en.wikipedia.org/wiki/Distribution_(mathematics) . The point is, Dirac's $\delta$-function is not a function at all, at least not in the usual sense. What you have given as a "definition" is merely an intuition for it. What $\delta$-function is is this: it is a linear functional ...


1

I'm assuming you are using the physicist definition of the Dirac delta distribution, where: $$\int_{-\infty}^{\infty} f(x)\delta(x)dx = f(0)$$ If this is the case, then the properties follow from: 1) $\int_{-\infty}^{\infty} f(x) \delta(-x)dx = -\int_{\infty}^{-\infty} f(-y)\delta(y)dy =\int_{-\infty}^{\infty}f(y)\delta(y)dy$ 2) $\int_{-\infty}^{\infty}f(...


0

Continuity of $y$ at $x=p$ implies $$(D_1 e^{ap} + D_2 e^{-ap}) - (C_1 e^{ap} + C_2 e^{-ap}) = 0.$$ A unit step of $y'$ at $x=p$ implies $$(D_1 a e^{ap} + D_2 (-a) e^{-ap}) - (C_1 a e^{ap} + C_2 (-a) e^{-ap}) = 1.$$ Using these two conditions you can eliminate two of the constants $C_1, C_2, D_1, D_2.$ But you probaly want another pair of boundary ...


0

By substitution and composition,$$P(k,\,t)=\int_{k_0(t)}^{k_s(t)}\delta[k-k_s(t)]\frac{\partial s}{\partial k}dk=-\left(\frac{\partial s}{\partial k}\right)_{k=k_s(t)},$$where the overall $-$ sign is due to the order of the integral's limits. But of course, we can rewrite this as$$-\left(\frac{\partial k_s(t)}{\partial s}\right)^{-1}_{s=s(k,\,t)},$$as ...


0

We have that \begin{align*} 1+2\sum_{n=1}^\infty \cos 2n\pi x=\sum_{k=-\infty}^\infty \exp(2k\pi i x), \end{align*} and the right side is the Dirichlet kernel which means that \begin{align*} &\int_{-\infty}^\infty \sum_{k=-\infty}^\infty \exp(2k\pi i x) \phi(x)dx\\ &=\lim_{N\to\infty}\int_{-\infty}^\infty 2\pi D_N(2\pi x)\phi(x)dx\\ &=\lim_{N\...


0

I would integrate it and show that both sides are equal. The Dirac Delta is defined by its action under the integral, so it should be enough. But I am physicists, so maybe it is more subtle.


0

I'm going to assume that $0 < \xi < b$. Let $\epsilon>0$, then take your differential equation and integrate it like so: $$\int_{\xi-\epsilon}^{\xi+\epsilon} A''(y) - \lambda^2 A(y) dy = \int_{\xi-\epsilon}^{\xi+\epsilon} \delta(y-\xi)dy$$ $$\implies A'\left(\xi+\epsilon\right) - A'\left(\xi-\epsilon\right) -\lambda^2\int_{\xi-\epsilon}^{\xi+\...


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