Hello! Could you tell me about how to take the next numerical calculation in mathematica? (perhaps there are special packages).(adsbygoogle = window.adsbygoogle || []).push({});

I have an expression (in reality slightly more complex):

## V=x^2 + \int_a^b x \sqrt{x^2-m^2} \left(\text Log \left(e^{-\left(\beta \left(\sqrt{\left(\sqrt{l^2-m^2}+U\right)^2+(m+x)^2+N}+u\right)\right)}+1\right)\right) \, dl ##

where ##x## is function of ##l##; ##m##, ##N## are constants; ##\beta##, ##u##, ##U## are parameters.Code ( (Unknown Language)):

V=x^2+\!\(

\*SubsuperscriptBox[\(\[Integral]\), \(a\), \(b\)]\(l

\*SqrtBox[\(

\*SuperscriptBox[\(l\), \(2\)] -

\*SuperscriptBox[\(m\), \(2\)]\)] \(Log(1 +

\*SuperscriptBox[\(e\), \(-\(\[Beta](u +

\*SqrtBox[\(

\*SuperscriptBox[\((

\*SqrtBox[\(

\*SuperscriptBox[\(l\), \(2\)] -

\*SuperscriptBox[\(m\), \(2\)]\)] + U)\), \(2\)] +

\*SuperscriptBox[\((m + x)\), \(2\)] + N\)])\)\)])\) \[DifferentialD]l\)\)

I need to find the dependence ##U## on ##u## and ##\beta## (in order to draw graph) from an equation:

##\frac {\partial V} {\partial x}=0##

(##x## will be needed to set a constant after differentiation; In reality,there is not the derivative, but a variation)

If I have an integral (without parameters) rather than equation, I would try to do the following ones:

1) to define the a region of integration (due to graphical representation of function)

2) to tabulate integrand

3) to calculate the integral that is to get a number.

Nevertheless I have the equation, which probably requires other method. I would appreciate Mathematica literature on this subject, or help.

I do not know how actual it is to calculate integral equation, but the integral can be led to another kind:

## x \sqrt{x^2-m^2} \left(\text {Log} \left(e^{-\left(\beta \left(\sqrt{\left(\sqrt{l^2-m^2}+U\right)^2+(m+x)^2+N}+u\right)\right)}+1\right)\right) \to ##

## \left(x^2-m^2\right)^{3/2} \frac{\text{$\cosh(\beta $u)} +\exp \left(-\beta \sqrt{\left(\sqrt{l^2-m^2}+U\right)^2+(m+x+y)^2+(q+z)^2}\right)}{\text{$\cosh (\beta $u)} -\cosh \left(\beta \sqrt{\left(\sqrt{l^2-m^2}+U\right)^2+(m+x+y)^2+(q+z)^2}\right)} ##

Perhaps it was necessary to start with something simpler. LetCode ( (Unknown Language)):

(x^2-m^2)^(3/2) (cosh(\[Beta]u)+exp(-\[Beta] Sqrt[(Sqrt[l^2-m^2]+U)^2+(m+x+y)^2+(z+q)^2]))/(cosh(\[Beta]u)-cosh(\[Beta] Sqrt[(Sqrt[l^2-m^2]+U)^2+(m+x+y)^2+(z+q)^2]))

##V=x^2+\int_{a}^{b} x u U \beta l dl##

##\frac {\partial V} {\partial x}=0##

##x## is assumed a constant after differentiating.

##\int_a^b \beta l u U \, dl+2 x=0##

##U=\frac{4 x}{\beta u \left(a^2+b^2\right)}##

In all I get the dependency ##U##, on ##u## and ##\beta##

I need to do the same if the integral is not taken analytically.

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Mathematica Numerical solution of integral equation with parameters

Tags:

Have something to add?

Draft saved
Draft deleted

Loading...

Similar Threads - Numerical solution integral | Date |
---|---|

Matlab Parameter fitting with a numerical solution | Feb 9, 2018 |

Working with numerical solutions in mathematica | Apr 8, 2015 |

Problem with numerical solution to Sch eq in Mathematica | Jan 12, 2012 |

Integration of Numerical Solution Mathematica | Jul 20, 2011 |

Numerical solution in Mathematica | Sep 22, 2010 |

**Physics Forums - The Fusion of Science and Community**