Proof involving ##ω(ξ,n)=u(x,y)## - Partial differential equations

[chwala]

Homework Statement

kindly see attached.

Relevant Equations

Partial differential equations

I am going through this page again...just out of curiosity, how did they arrive at the given transforms?, ...i think i get it...very confusing...
in general,
$U_{xx} = ξ_{xx} =ξ_{x}ξ_{x}= ξ^2_{x}$ . Also we may have
$U_{xy} =ξ_{xy} =ξ_{x}ξ_{y}.$ the other transforms follow in a similar manner.

 

[Orodruin]

The chain rule for derivatives.

Homework Statement:: kindly see attached.
Relevant Equations:: Partial differential equations

i think i get it...very confusing...
in general,
Uxx=ξxx=ξxξx=ξx2 . Also we may have
Uxy=ξxy=ξxξy. the other transforms follow in a similar manner.

No, this is incorrect. You are missing the derivatives of w. For example:
$$
u_x = w_\xi \xi_x + w_\eta \eta_x
$$
and so on. There seems to be an implicit assumption that the transformation is linear or the higher derivatives would also contain higher derivatives of the new variables.

 

[chwala]

ok the way i understand it is,
$ω(ξ, n)=u(x,y)$
$ω(ξ, n)=ξ(x,y)+η(x,y)$
$u_x$=$\frac{∂ω}{∂ξ}⋅\frac{∂ξ}{∂x}$+$\frac{∂ω}{∂η}⋅\frac{∂η}{∂x}$
$u_x=ω_ξ⋅ξ_x+ω_η⋅η_x$

 

[Orodruin]

ok the way i understand it is,
$ω(ξ, n)=u(x,y)$
$ω(ξ, n)=ξ(x,y)+η(x,y)$
$u_x$=$\frac{∂ω}{∂ξ}⋅\frac{∂ξ}{∂x}$+$\frac{∂ω}{∂η}⋅\frac{∂η}{∂x}$
$u_x=ω_ξ⋅ξ_x+ω_η⋅η_x$

No, this is incorret. You do not know that w is a sum of xi and eta. It may be any function.

 

[chwala]

No, this is incorret. You do not know that w is a sum of xi and eta. It may be any function.

Lol...let me look at this again...
The correct approach should be, since $ξ =ξ (x,y)$ is a function of $x$ and $y$ and
$η= η(x,y)$ being also a function of $x$ and $y$, then using the transform $ω(ξ,n)=u(x,y)$, we shall have
$u_x$=$\frac{∂ω}{∂ξ}⋅\frac{∂ξ}{∂x}$+$\frac{∂ω}{∂η}⋅\frac{∂η}{∂x}$
$u_x=ω_ξ⋅ξ_x+ω_η⋅η_x$

 

[chwala]

Quite interesting how they came up with the transformations...its some bit of work. To go straight to the point we have,
$u_x=ξ_x ⋅ω_ξ+η_x⋅ω_η$ it follows that,
$u_{xx}=u_x ⋅u_x=(ξ_x ⋅ω_ξ+η_x⋅ω_η)(ξ_x ⋅ω_ξ+η_x⋅ω_η)=ξ^2_x⋅ω_{ξξ}+2ξ_x⋅ η_xω_{ξη}+η^2_x⋅ω_{ηη}$
$u_{xy}=u_x ⋅u_y=(ξ_x ⋅ω_ξ+η_x⋅ω_η)(ξ_y ⋅ω_ξ+η_y⋅ω_η)=ξ_xξ_y ω_{ξξ}+(ξ_x⋅ η_y+ξ_y⋅ η_x)ω_{ξη}+η_xη_yω_{ηη}$
$u_{yy}=u_y ⋅u_y=(ξ_y ⋅ω_ξ+η_y⋅ω_η)(ξ_y ⋅ω_ξ+η_y⋅ω_η)=ξ^2_y⋅ω_{ξξ}+2ξ_y⋅ η_yω_{ξη}+η^2_y⋅ω_{ηη}$

Now what they simply did in finding $α$ for e.g is by putting all coefficients of $ω_{ξξ}$ together, i.e on the pde of the form;

$au_{xx}+2bu_{xy}+cu_{yy}+...$

$α$= $[aξ^2_x+2bξ_xξ_y+cξ^2_y]ω_{ξξ}$

Coming to the coefficients of $ω_{ξη}$, we shall be getting $β$ i.e;
$β$=$[2aξ_x⋅ η_x+2b(ξ_x⋅ η_y+ξ_y⋅ η_x)+2cξ_y⋅ η_y]ω_{ξη}$
$β$=$[aξ_x⋅ η_x+b(ξ_x⋅ η_y+ξ_y⋅ η_x)+cξ_y⋅ η_y]ω_{ξη}$
...

the other transforms, $δ, ϒ$and $∈$ can be found similarly... i always try to use my own way of thinking ...and also utilizing your input...much appreciated guys...cheers @Orodruin thanks mate. Bingo!