I'm asking this question on behalf of another person who is not as familiar as me with computers. He has the following problem to solve and unfortunately Mathcad can't solve such type of equations.

He seeks help to find the function $u = u(x) = ?$ for

$$ \frac{d^2u(x)}{dx^2} =\left[U(x) + \alpha^2\right]u(x) $$

where $$U(x) = \frac{2}{x}\left(\exp\left(-\frac{x}{\lambda}\right)-\int_0^x 4\pi\left(u(z)\right)^2\,dz\right)$$

The parameters $\lambda$ and $\alpha$ are real.

Is another tool (i.e. Mathematica) able to solve it ? Do you have another suggestion ?

  • 1
    $\begingroup$ well you may use $v(x)=\int u(x)^2 dx$ to get a not easy 'third-order non linear ODE' using $u=\sqrt{v'}$ and derivatives (I didn't reverify this...). $\endgroup$ – Raymond Manzoni Jun 16 '12 at 9:30

A short answer to the question in the title is:

Isolate the integral term and differentiate.

In the case at hand, one could write $\displaystyle U'(x)=-\frac1xU(x)+\frac2x\left(-\frac1\lambda\mathrm e^{-x/\lambda}-4\pi u(x)^2\right)$ and $\displaystyle U(x)=\frac{u''(x)}{u(x)}-\alpha^2$ to deduce the third-order differential equation with no integral term $$ u'''(x)=\frac{u''(x)}{u(x)}u'(x)-\frac1x\left(\frac{u''(x)}{u(x)}-\alpha^2\right)u(x)+\frac2x\left(-\frac1\lambda\mathrm e^{-x/\lambda}-4\pi u^2(x)\right)u(x). $$ Or, one could transform this into a system of two equations of lower degree, namely, $$ xu''(x)=v(x)u(x),\qquad v'(x)=-8\pi u^2(x)+\alpha^2-\frac2\lambda\mathrm e^{-x/\lambda}. $$

  • $\begingroup$ Here is the answer of the person who asked me for help “I thank you very much for indicating an elegant way to solve the equation, by rewriting it in another form. Without the integral. I can now go on with Mathcard.” $\endgroup$ – chmike Jun 16 '12 at 16:33

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