Given is the dimensionless system of temperature and concentration profile for a solid fuel pill of a spherical structure.

$\frac{\text{d}\text{e}\text{l}\text{Y}}{\text{d}\text{e}\text{l}}=\frac{1}{z^2}\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{z}}(z^2\frac{\text{d}\text{e}\text{l}\text{Y}}{\text{d}\text{e}\text{l}\text{z}})-{}^2$Yexp$((1-\frac{1}{}))$

$Le\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}}=\frac{1}{z^2}\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{z}}(z^2\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{z}})+{}^2$Yexp$((1-\frac{1}{}))$;

The initial and boundary conditions become:

$Y(0,z)=Y(,1)=1$;

$(0,z)=(,1)=1$;

$\frac{\text{d}\text{e}\text{l}\text{Y}}{\text{d}\text{e}\text{l}\text{z}}(,0)=\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{z}}(,0)=0$

The equations may be further simplified with the help of the following "trick". Introduce the new

variables $Y=\frac{y}{z},$ and $=\frac{}{z}.$ Show that the equations for the new variables become:

$\frac{\text{d}\text{e}\text{l}\text{y}}{\text{d}\text{e}\text{l}}=\frac{de{l}^{2}y}{del{z}^2}-{}^2$yexp$((1-\frac{z}{}))$

$Le\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}}=\frac{de{l}^2}{del{z}^2}+{}^2$yexp$((1-\frac{z}{}))$;

The initial and boundary conditions are $($why$?)$:

$y(0,z)=z$;$y(,1)=1$;

$(0,z)=z$;$(,1)=1$;

$z\frac{\text{d}\text{e}\text{l}\text{y}}{\text{d}\text{e}\text{l}\text{z}}(,0)-y(,0)=z\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{z}}(,0)-(,0)=0$

Can you help explain how the last boundary condition becomes?

z$($dely$)/($delz$)(\backslash$tau $,0)-$y$(\backslash$tau $,0)=0$

z$($del$\backslash$theta $)/($delz$)(\backslash$tau $,0)-\backslash$theta $(\backslash$tau $,0)=0$

From:

$($delY$)/($delz$)(\backslash$tau $,0)=0$

$($del$\backslash$Theta $)/($delz$)(\backslash$tau $,0)=0$