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Mathematica Numerical solution of integral equation with parameters

  1. Jan 13, 2017 #1
    Hello! Could you tell me about how to take the next numerical calculation in mathematica? (perhaps there are special packages).
    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 ##

    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\)\)

     
    where ##x## is function of ##l##; ##m##, ##N## are constants; ##\beta##, ##u##, ##U## are parameters.
    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)} ##

    Code ( (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]))
     
    Perhaps it was necessary to start with something simpler. Let

    ##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.
     
  2. jcsd
  3. Jan 15, 2017 #2
    Can you help me understand your simple example. For starters you arrive at the conclusion that U is a function of x. However, when you differentiate the integral term by x you ignore this dependence. Second you state the x is a function of l. If this is the case then you should not ignore this dependence when evaluating the integral.
     
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