1. ## Pde

Hi, just working on some problems but I'm getting stuck.

1)In trying to express the Laplacian in parabolic cylindrical coordinates, I can't get u = a function of x and y or v = a function of x and y...so I'm stuck on how to keep on going. (Here the parabolic cylindrical coordinates are given as (u,v,z) and rectangular coordinates as (x,y,z)

2)Find a transformation of the form: u(x,t) = v(x,t)e^(qx-wt) where a and w have to be determined, that would bring or transform the following equation for u(x,t): u_t-au_xx-bu_x-cu = 0 where a b and c are non zero constants to a diffusion partial differential equation.

Thanks!

2. Originally Posted by angels_symphony
Hi, just working on some problems but I'm getting stuck.

1)In trying to express the Laplacian in parabolic cylindrical coordinates, I can't get u = a function of x and y or v = a function of x and y...so I'm stuck on how to keep on going. (Here the parabolic cylindrical coordinates are given as (u,v,z) and rectangular coordinates as (x,y,z)

Mr F says: Try reading this and see how things go .....

2)Find a transformation of the form: u(x,t) = v(x,t)e^(qx-wt) where q and w have to be determined, that would bring or transform the following equation for u(x,t): u_t-au_xx-bu_x-cu = 0 where a b and c are non zero constants to a diffusion partial differential equation. Mr F edit in red.

Thanks!

First note that the diffusion p.d.e. is $\frac{\partial \phi}{\partial t} = k \frac{\partial^2 \phi}{\partial x^2}$ where k is a constant (the thermal diffusivity).

Now note that if $\displaystyle u(x, t) = v(x, t) e^{qx - wt}$ then (using the product rule):

1. $\displaystyle \frac{\partial u}{\partial t} = \frac{\partial v}{\partial t} e^{qx - wt} - w v e^{qx - wt} = \left( \frac{\partial v}{\partial t} - wv \right) e^{qx - wt}$.

2. $\displaystyle \frac{\partial u}{\partial x} = \frac{\partial v}{\partial x} e^{qx - wt} + q v e^{qx - wt} = \left( \frac{\partial v}{\partial x} + q v \right) e^{qx - wt}$.

3. $\displaystyle \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}\left( \frac{\partial u}{\partial x} \right) = \left( \frac{\partial^2 v}{\partial x^2} + q \frac{\partial v}{\partial x} \right) e^{qx - wt} + q \left( \frac{\partial v}{\partial x} + q v \right) e^{qx - wt}$

$\displaystyle = \left( \frac{\partial^2 v}{\partial x^2} + 2q \frac{\partial v}{\partial x} + q^2 v\right) e^{qx - wt}$.

Substitute all that lot into $\displaystyle \frac{\partial u}{\partial t} - a \frac{\partial^2 u}{\partial x^2} - b \frac{\partial u}{\partial x} - cu = 0$ and divide through by the common (non-zero) factor of $\displaystyle e^{qx - wt}$:

$\displaystyle \left( \frac{\partial v}{\partial t} - wv \right) - a \left( \frac{\partial^2 v}{\partial x^2} + 2q \frac{\partial v}{\partial x} + q^2 v\right) - b \left( \frac{\partial v}{\partial x} + q v \right) - cv = 0$

$\displaystyle \Rightarrow \frac{\partial v}{\partial t} = a \frac{\partial^2 v}{\partial x^2} + \frac{\partial v}{\partial x} (2aq + b) + v(c + aq^2 + bq + w)$.

Comparing this with the diffusion equation, you clearly require the coefficients of $\displaystyle \frac{\partial v}{\partial x}$ and v to equal zero:

$2aq + b = 0 \Rightarrow q = -\frac{b}{2a}$ .... (1)

$c + aq^2 + bq + w = 0$ .... (2)

Substitute (1) into (2) and solve for w:

$\displaystyle w = -c - aq^2 - bq = -c + \frac{b^2}{4a} = \frac{b^2 - 4ac}{4a}$.

There are probably numerous typos in the above - work through it carefully, double check everything .....

3. Originally Posted by angels_symphony
1)In trying to express the Laplacian in parabolic cylindrical coordinates, I can't get u = a function of x and y or v = a function of x and y...so I'm stuck on how to keep on going. (Here the parabolic cylindrical coordinates are given as (u,v,z) and rectangular coordinates as (x,y,z)
$x(r,\theta)=r\cos\theta \mbox{ and }y(r,\theta) = r\sin \theta$

Then, (confirm this by chain rule),
$\frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} = \frac{\partial ^ 2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{ \partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}$