# Parabolic Linear 2nd Order PDE with Trig terms

• Mar 31st 2011, 11:15 AM
bugatti79
Parabolic Linear 2nd Order PDE with Trig terms
Hi Mathematicians!,

Make a suitable change of variables and hence transform to its canonical form. Heres goes...

$
Sin^{2}(x)u_{xx} -2ysin(x)u_{xy}+y^2u_{yy}=0
$

$
A= sin^{2}(x), B=-y sin(x), C=y^2
$

The discrimant B^2-AC is = $y^2 sin^{2}(x)-sin^{2}(x)(y^2)=0$ making it parabolic.

$
\displaystyle \implies \frac{dy}{dx}=\frac{B\pm\sqrt{B^2-AC}}{A}=\frac{-y}{sin(x)}
$

$
\implies ln y = ln(cot(x)+cosec(x))+C'
$
therefore $y = A(cot(x)+cosec(x))$where $A=e^{C'}$

Let $\displaystyle t(x,y)= \frac{y}{cot(x)+cosec(x)}$and $s(x,y)=x$

Check the determinant is not = 0 ie,

$
\displaystyle det\begin{bmatrix}\frac{\partial s}{\partial x} &\frac{\partial s}{\partial y}\\ \frac{\partial t}{\partial x} &\frac{\partial t}{\partial y}\end{bmatrix}=\frac{1}{cot(x)+cosec(x)}\neq 0
$

How am I doing so far?..
• Apr 1st 2011, 05:59 AM
Jester
Looking good - keep going :-)
• Apr 1st 2011, 03:21 PM
bugatti79
Quote:

Originally Posted by Danny
Looking good - keep going :-)

I had no idea what I was letting myself in for...
The derivatives are turning out to be a monster of a task! (Shake) Particualrly t_xx

$\displaystyle u_x=w_s s_x+w_t t_x=w_s+\overbrace{y \left[\frac{csc^2(x)+csc(x)cot(x)}{(cot(x)+csc(x))^2}\ri ght ]}^{A}w_t$

$u_{xx}=s_x(w_{ss}s_x+w_{ts}t_x)+t_x(w_{st}s_x+w_{t t}t_x)+w_tt_{xx}$

The $s_{xx}$ term is 0

$=w_{ss}+2Aw_{ts}+A^2w_t+w_t t_{xx}$

$\displaystyle u_y=w_t t_y=w_t\left [\frac{1}{cot(x)+csc(x)}\right ]$

$\displaystyle u_{yy}=t_y(w_{tt} t_y)= \frac{w_{tt}}{(cot(x)+csc(x))^2}$

$u_{xy}=w_{st}s_xt_y+w_{tt}t_xt_y+w_{tt}t_{xy}$

I have attached the derivation of t_xx

I have no idea where to start in order to simplify it. I am not sure if I will continue on with this unless I get some encouragement. I think its beyond the scope of a possible exam question! (Wondering) (Nerd)