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Questions tagged [integral-equations]

This tag is about questions regarding the integral equations. An integral equation is an equation in which the unknown function appears under the integral sign. There is no universal method for solving integral equations. Solution methods and even the existence of a solution depend on the particular form of the integral equation.

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13 votes
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I tried to come up with an integral equation for fun, and made this creature:$\def\d{\mathrm d}$ $$ f(x)-\int_x^{2x}f(t)\,\d t=0, $$ so I followed these steps: $$ \begin{aligned} &f(x)+\int_x^{2x}...
A FFMAX's user avatar
  • 131
1 vote
2 answers
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I have found a Volterra integral equation of the first kind of the following type: $$y(r)=\int_0^r\,K(t,r)\,f(t,r)\,\mathrm dt,$$ in which my unknown function is $f(t,r)$. I have tried to find ...
jesusvaleo's user avatar
0 votes
1 answer
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Given a specific integral equation \begin{equation} f(x) = 2- \frac{1}{\pi}\lim_{C\rightarrow \infty} \int^C_{-C}\frac{f(y)}{(x-y)^2+1} d y, \end{equation} I want to show that $$\lim_{C\rightarrow ...
Geigercounter's user avatar
1 vote
0 answers
107 views

Fix a discrete-time system with input sequence $(x_t)_{t\ge 0}\subset\mathbb{R}^d$, output $(y_t)_{t\ge 0}\subset\mathbb{R}^p$, and $L$ internal levels. For each level $\ell\in\{1,\dots,L\}$ choose a ...
Deborah Roselle's user avatar
0 votes
0 answers
51 views

We are looking for a continuous function defined for all inner points of a unit $3D$ ball (sphere) such that its integral over nonempty intersection of unit ball with any plane is $1$. This question ...
Vladimir_U's user avatar
3 votes
1 answer
130 views

If we assume that our function depends only on distance from origin we can come up with differential equation for new function of one variable (distance from origin), I was doing it once but could not ...
Vladimir_U's user avatar
1 vote
0 answers
97 views

I am trying to solve the following Abel type integral equation, in which $f\left( s\right)$ is not a symmetric function: $$\int_{-r}^{r}\frac{f\left( s\right) }{\sqrt{r^{2}-s^{2}}} \ ds=g\left( r\...
jesusvaleo's user avatar
0 votes
0 answers
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I am considering the following type of equation $$ A(t) = F(t) + \int_0^t \mu(t,s) A(s) \, {\rm d} s, \qquad 0\le t\le T, $$ where $F : [0,T] \to M_n, \mu: \Delta \to M_n$ are given continuous ...
Kelvin's user avatar
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1 vote
0 answers
91 views

I have found the following integral equation: $$\int_{a}^{a+r}\frac{f\left( s\right) ds}{\sqrt{s-c}\sqrt{a+r-s}}=-\frac {\mu\pi}{2R}\left( a+c-r\right) \text{, }a>0 \text{, }r>0 \text{ and }...
jesusvaleo's user avatar
5 votes
1 answer
116 views

I'm studying the book Integral Equations by M. Rahman. To solve the integral equation $$ u(x) = f(x) + \lambda \int_0^x e^{x - t} u(t) \, dt, $$ the resolvent kernel method gives the solution: $$ u(x) ...
Faoler's user avatar
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2 votes
0 answers
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Consider the Hilbert-Schmidt operator $K$ defined on $L^2([0,T])$: \begin{align} Kf(\tau) = \int_0^T \left(\mathrm{e}^{-\frac{|\tau-s|}{a}}-\mathrm{e}^{\frac{\tau + s - 2T_0}{a}}\right)f(s)\,\mathrm{d}...
amrit 's user avatar
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4 votes
1 answer
166 views

The Question I was just watching this video that proves the result $$ \int_0^1 x^{-x} \mathrm{d} x = \sum_{k=1}^{\infty} k^{-k} $$ (So we already know one possibility: $f(x) = x^{-x}$) But I was ...
abluegiraffe's user avatar
6 votes
2 answers
226 views

Suppose that $f$ is continuous function on $(0,1]$ and satisfies $$ f(x) + \int_0^x \frac{f(y)}{\sqrt{x-y}} dy = 0, \qquad x \in (0,1]. $$ Does it follow that $f(x) = 0$ for $x \in (0,1]$? The ...
xen's user avatar
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6 votes
1 answer
119 views

Physics told us that when a conductor reaches the electrostatic state all charges reside on its surface, and the electric potential is constant within the conductor Viewing this statement from a ...
Lee's user avatar
  • 12k
0 votes
1 answer
77 views

Consider the following definition of non-local Laplacian operator, $$\Delta^K u(x) = \int_{\Omega} dy\,K(x,y)\left(u(y)-u(x)\right),\tag{1}$$ where $K(x,y)=K(y,x)$ is measurable and "good" ...
Artem Alexandrov's user avatar

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