Hi All,

I have 2 problems that I am stuck on (see attached), I dont know how to go any further. Any tips or advice on what Im doing wrong?

Thanks

Printable View

- Nov 24th 2010, 01:04 PMbugatti79Quasi Linear PDE's
Hi All,

I have 2 problems that I am stuck on (see attached), I dont know how to go any further. Any tips or advice on what Im doing wrong?

Thanks - Nov 24th 2010, 01:59 PMJester
You're not doing anything wrong - just not going far enough.

(1) You have

$\displaystyle \dfrac{du}{dx} = \dfrac{-2xu}{x^2 + 3B^2u^2+3u^3}$

If you write it as

$\displaystyle \dfrac{dx}{du} = \dfrac{x^2 + 3B^2u^2+3u^3}{-2xu}$. It's Bernoulli in $\displaystyle x$.

(2) You have

$\displaystyle \dfrac{d(x+y)}{y-x} = \dfrac{du}{x-y}$ which integrates giving $\displaystyle x+y+u = c_1$.

Then

$\displaystyle \dfrac{d(x-y)}{x+ y - 2u} = \dfrac{du}{x-y}$

or

$\displaystyle (x-y)d(x-y) = (c_1 - 3u)du $.

This also integrates. - Nov 25th 2010, 09:08 AMbugatti79
- Nov 25th 2010, 09:21 AMJester
In part (1), multiply by 2x and let X = x^2 (it's linear now)

In (2) - looks good to me :-) - Nov 25th 2010, 09:33 AMbugatti79
- Nov 29th 2010, 07:11 AMbugatti79
- Nov 29th 2010, 02:58 PMJester
Multiplying by $\displaystyle 2x$ gives

$\displaystyle \dfrac{2x dx}{du} = \dfrac{x^2 + 3B^2u^2+3u^3}{-u}$

and letting $\displaystyle X = x^2$ gives

$\displaystyle \dfrac{dX}{du} = \dfrac{X + 3B^2u^2+3u^3}{-u}$

or

$\displaystyle \dfrac{dX}{du} + \dfrac{X}{u} = - 3B^2u-3u^2$. This is now linear! - Jan 17th 2011, 09:41 AMbugatti79
- Jan 17th 2011, 10:17 AMJester
First replace your $\displaystyle B$ with $\displaystyle B = \dfrac{y}{u}$ so your solution is

$\displaystyle \dfrac{u}{y} = f\left(x^2u+y^2u + \frac{3}{4}u^4\right)$

or

$\displaystyle x^2u+y^2u + \frac{3}{4}u^4\right = f\left(\dfrac{u}{y}\right)$.

Are you sure about the IC b/c with $\displaystyle u(x,2x) = 0$, the only solution is $\displaystyle u \equiv 0$. - Jan 17th 2011, 11:18 AMbugatti79
You are right Danny...once again! The IC was for another similar question and this question is to find general solution only.

However shouldnt the GS be

$\displaystyle \dfrac{y}{u} = f\left(x^2u+y^2u + \frac{3}{4}u^4\right)$

or

$\displaystyle x^2u+y^2u + \frac{3}{4}u^4\right = f\left(\dfrac{y}{u}\right)$

ie, you had u/y

Assuming I am correct, how do you know that the only soltion is $\displaystyle u \equiv 0$ - Jan 17th 2011, 12:21 PMJester
As for your first question, it doesn't matter. Define

$\displaystyle f(\lambda) = \bar{f}\left(\dfrac{1}{\lambda}\right)$ then drop the bar.

For your second question, let go back when we introduced new coordiantes such that

$\displaystyle u_s = u_x x_s + u_y y_s$ gave your PDE by choosing

$\displaystyle x_s=x^2 + 3y^2 + 3u^3$

$\displaystyle y_s =-2xy$

$\displaystyle u_s = -2xu$

with the IC as $\displaystyle x = r, y = 2r, u = 0$ when $\displaystyle s = 0.$ From the second two we find

$\displaystyle y u_s - u y_s = 0$ which integrates gives $\displaystyle \dfrac{u}{y} = a(r)$. Imposing the IC gives $\displaystyle a(r) = 0$ giving $\displaystyle u = 0$. - Jan 17th 2011, 12:54 PMbugatti79
- Jan 17th 2011, 02:35 PMJester
- Jan 17th 2011, 11:45 PMbugatti79
- Jan 18th 2011, 04:13 AMJester
For the general solution, it doesn't matter but for the IC $\displaystyle u(x,2x) = 0$, then it would matter.