Oops! Smooth Operator

From the post "Stress When The Heat Is On,  Thermal Stress",

$\cfrac{\partial ({d}_{s})}{\partial x}=\cfrac{\partial (d_s)}{\partial T}.\cfrac{d T(x)}{d x}+\cfrac{\partial }{\partial T}\left\{\cfrac{\partial (d_s)}{\partial x}\right\}.T$

$\left\{1-T.\cfrac{\partial }{\partial T}\right\}\cfrac{\partial (d_s)}{\partial x}=\cfrac{\partial (d_s)}{\partial T}.\cfrac{d T(x)}{d x}$

$\cfrac{\partial (d_s)}{\partial x}=\left\{1-T.\cfrac{\partial }{\partial T}\right\}^{-1}\left\{\cfrac{\partial (d_s)}{\partial T}.\cfrac{d T(x)}{d x}\right\}$

where

$\left\{1-T.\cfrac{\partial }{\partial T}\right\}^{-1}$    is the inverse of the operator    $\left\{1-T.\cfrac{\partial }{\partial T}\right\}$

This turns out to be very difficult.

But consider this,

$\cfrac { \partial  }{ \partial  ln(T)  } \equiv\cfrac { \partial T }{ \partial  ln(T)  } \cfrac { \partial  }{ \partial T } \equiv T\cfrac { \partial  }{ \partial T }$


So,

$\left\{1-\cfrac{\partial }{\partial   ln(T)  }\right\}\cfrac{\partial (d_s)}{\partial x}=\cfrac{\partial (d_s)}{\partial T}.\cfrac{d T(x)}{d x}$

$\left\{1-\cfrac{\partial }{\partial ln(T) }\right\}\cfrac{\partial (d_s)}{\partial x}=\cfrac{\partial (d_s)}{\partial x}\cfrac{\partial x }{\partial T}.\cfrac{d T(x)}{d T}\cfrac{d T}{d x}$

that is to say,

$\left\{1-\cfrac{\partial }{\partial   ln(T) }\right\}\cfrac{\partial (d_s)}{\partial x}=\cfrac{\partial (d_s)}{\partial x}.\cfrac{d T(x)}{d T}$

$\left\{1-\cfrac{\partial }{\partial   ln(T)  }\right\}\cfrac{\partial (d_s)}{\partial x}=\cfrac{\partial (d_s)}{\partial x}$

which suggests  (just mathematical trickery),

$\cfrac{\partial }{\partial  ln(T) }\left\{\cfrac{\partial (d_s)}{\partial x}\right\}=0$

that    $\cfrac{\partial (d_s)}{\partial x}$    is independent of    ${  ln(T) }$    or

$\cfrac{\partial }{\partial x}\left\{\cfrac{\partial (d_s)}{\partial {  ln(T)  } }\right\}=0$

that      $\cfrac{\partial (d_s)}{\partial   ln(T)  }$    is independent of    $x$

One way this could happen is that    $d_s$    is a linear combination of  functions in   $x$    and    ${  ln(T)  }$.  This is to say,

$d_s=A.f(x)+B.h({  ln(T)  })$

where    $A$    and     $B$    are constants and    $f(x)$    is a function only of    $x$    and    $h({ln(T)})$    is a function only of  ${ ln(T) }$.

That space density    $d_s$    is the superposition of space distribution itself, expressed as    $A.f(x)$,    and the effect of thermal energy in the form of    $B.h({ ln(T) })$.

In the case of uniformly distributed space,

$f(x)=C$   where    $C$    is a constant

$d_s=d_n+B.h({ ln(T) })$

where    $A.C=d_n$    is the normal space density and,

$h({ ln(T) })|_{lim\,T\rightarrow0}=0$

such that when    $T=0$,    $d_s=d_n$

The points here are the    $ln(T)$    dependence of    $d_s$    and superposition    $d_s=A.f(x)+B.h({  ln(T)  })$.