Dgdrxrt

I have the following second order linear ordinary differential equation with variable coefficients:

$ c fracd^2 overlinevarphid E^2 + aE^k fracd overlinevarphid E + (akE^k-1 +s)overlinevarphi = varphi(E,0)$, where $overlinevarphi = overlinevarphi(E,s) = int_0^infty varphi(E,x)e^-sxdx$.

Before I actually show how I tried to solve this, it is perhaps good if I provide some background. This equation is obtained by taking the Laplace transform in the x variable of the following second order partial differential equation:

$c cdot fracpartial^2 varphi(E,x)partial E^2 + aE^k cdot fracpartial varphi(E,x)partial E + akE^k-1 cdot varphi + fracpartial varphipartial x = 0 $ where $E in (infty,+infty) , xin(0,infty)$ .

Coming back to the first equation, in order to solve this I have attempted to reduce this to an equation with constant coefficients, by following this document: http://www.mecheng.iisc.ernet.in/~sonti/ME261_variable_coeff_2nd_order.pdf.

However, when calculating whether the equation is exact, via the $P''-Q'+R$ formula, the result is the $s $ constant. Thus, I have concluded that the equation cannot be reduced to an equation with constant coefficients.

I have also tried using the variation of parameters on the equation. This was mostly by following the method given here: http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx

However, that requires knowledge of the fundamental solutions of the equation. The fundamental solutions are obtained by solving the following equation:

$ c fracd^2 overlinevarphid E^2 + aE^k fracd overlinevarphid E + (akE^k-1 +s)overlinevarphi = 0$

I attempted thereafter to solve the following polynomial:

$cr^2+aE^kr+(akE^k-1+s)=0$

The constants in the equation are: $c = 0.1/2$, $a = e^148$ and $k = -1.4$. When I plotted the delta from the well known abc formula it always turned out positive. So I have concluded that the roots of the equation are real and thus the fundamental solutions are

$overlinevarphi_1 = e^Er_1(s)$ and $overlinevarphi_2 = e^Er_2(s)$, with $r_1$ and $r_2$ the roots of the characteristic polynomial.

Thus, the variation of parameters method states that the solution to this equation is:

$overlinevarphi(E,s) = -overlinevarphi_1 int_0^infty fracoverlinevarphi_2varphi(E,0)W(overlinevarphi_1,overlinevarphi_2) + overlinevarphi_2 int_0^infty fracoverlinevarphi_1varphi(E,0)W(overlinevarphi_1,overlinevarphi_2)$

Using this the solution to the original PDE can be obtained via:

$varphi(E,x) = frac12pi iint_gamma-iinfty^gamma+iinfty overlinevarphi(E,s)e^sxds$

So, my questions are:

1. Does the first equation in the post have an analytical solution?


2. Is my treatment correct?


3. Do you have suggestions for improvements/ alternative methods?

Thank you in advance

Hint:

For $cdfracd^2overlinevarphidE^2+aE^kdfracdoverlinevarphidE+(akE^k-1+s)overlinevarphi=0$ ,

Let $overlinevarphi=e^mEoverline U$ ,

Then $dfracdoverlinevarphidE=e^mEdfracdoverline UdE+me^mEoverline U$

$dfracd^2overlinevarphidE^2=e^mEdfracd^2overline UdE^2+me^mEdfracdoverline UdE+me^mEdfracdoverline UdE+m^2e^mEoverline U=e^mEdfracd^2overline UdE^2+2me^mEdfracdoverline UdE+m^2e^mEoverline U$

$therefore cleft(e^mEdfracd^2overline UdE^2+2me^mEdfracdoverline UdE+m^2e^mEoverline Uright)+aE^kleft(e^mEdfracdoverline UdE+me^mEoverline Uright)+(akE^k-1+s)e^mEoverline U=0$

$cleft(dfracd^2overline UdE^2+2mdfracdoverline UdE+m^2overline Uright)+aE^kleft(dfracdoverline UdE+moverline Uright)+(akE^k-1+s)overline U=0$

$cdfracd^2overline UdE^2+(aE^k+2cm)dfracdoverline UdE+(amE^k+akE^k-1+cm^2+s)overline U=0$

Take $cm^2+s=0$ , i.e. $m=pm isqrtdfracsc$ , the ODE becomes

$cdfracd^2overline UdE^2+(aE^kpm2isqrtcs)dfracdoverline UdE+left(pm iasqrtdfracscE^k+akE^k-1right)overline U=0$