# Quasi Linear PDE's

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• Nov 24th 2010, 01:04 PM
bugatti79
Quasi 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 PM
Jester
You're not doing anything wrong - just not going far enough.

(1) You have

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

If you write it as

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

(2) You have

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

Then

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

or

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

This also integrates.
• Nov 25th 2010, 09:08 AM
bugatti79
Quote:

Originally Posted by Danny
You're not doing anything wrong - just not going far enough.

(1) You have

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

If you write it as

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

(2) You have

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

Then

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

or

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

This also integrates.

Hi Danny,

1) I am not sure how to convert this expression into Bernoulli's. Have I got the correct form as attached?

2) I derived an implicit solution given the initial conditions. Not sure if it is right?.....

Thanks
• Nov 25th 2010, 09:21 AM
Jester
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 AM
bugatti79
Quote:

Originally Posted by Danny
In part (1), multiply by 2x and let X = x^2 (it's linear now)

In (2) - looks good to me :-)

Lovely, I will inform you on how I get on in a few days.

Thanks
• Nov 29th 2010, 07:11 AM
bugatti79
Quote:

Originally Posted by bugatti79
Lovely, I will inform you on how I get on in a few days.

Thanks

Getting stuck on this one. How does it become linear? I multiplied accros by 2x on both sides and let X=x^2.
Its bernoulli in x but arent we solving for u? See attached my attempt. THanks
• Nov 29th 2010, 02:58 PM
Jester
Multiplying by $2x$ gives

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

and letting $X = x^2$ gives

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

or

$\dfrac{dX}{du} + \dfrac{X}{u} = - 3B^2u-3u^2$. This is now linear!
• Jan 17th 2011, 09:41 AM
bugatti79
Quote:

Originally Posted by Danny
Multiplying by $2x$ gives

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

and letting $X = x^2$ gives

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

or

$\dfrac{dX}{du} + \dfrac{X}{u} = - 3B^2u-3u^2$. This is now linear!

I have brought this further based on Danny's tip. See attached. I am stuck now getting the particular solution for the given IC's
• Jan 17th 2011, 10:17 AM
Jester
First replace your $B$ with $B = \dfrac{y}{u}$ so your solution is

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

or

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

Are you sure about the IC b/c with $u(x,2x) = 0$, the only solution is $u \equiv 0$.
• Jan 17th 2011, 11:18 AM
bugatti79
Quote:

Originally Posted by Danny
First replace your $B$ with $B = \dfrac{y}{u}$ so your solution is

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

or

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

Are you sure about the IC b/c with $u(x,2x) = 0$, the only solution is $u \equiv 0$.

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

or

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

ie, you had u/y

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

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

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

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

$x_s=x^2 + 3y^2 + 3u^3$
$y_s =-2xy$
$u_s = -2xu$

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

$y u_s - u y_s = 0$ which integrates gives $\dfrac{u}{y} = a(r)$. Imposing the IC gives $a(r) = 0$ giving $u = 0$.
• Jan 17th 2011, 12:54 PM
bugatti79
Quote:

Originally Posted by Danny
$\dfrac{u}{y} = a(r)$. Imposing the IC gives $a(r) = 0$ giving $u = 0$.

I dont understand this last bit. How doyou impose the IC?

Thanks
• Jan 17th 2011, 02:35 PM
Jester
Quote:

Originally Posted by bugatti79
I dont understand this last bit. How doyou impose the IC?

Thanks

If $\dfrac{u}{y} = a(r)$. Imposing the IC $x = r, y = 2r, u = 0$ gives $\dfrac{0}{2r} = a(r)$ gives $a(r) = 0$ giving $u = 0$.
• Jan 17th 2011, 11:45 PM
bugatti79
Quote:

Originally Posted by Danny
If $\dfrac{u}{y} = a(r)$. Imposing the IC $x = r, y = 2r, u = 0$ gives $\dfrac{0}{2r} = a(r)$ gives $a(r) = 0$ giving $u = 0$.

Ok, that makes sense.

However, refering to $f(\lambda) = \bar{f}\left(\dfrac{1}{\lambda}\right)$

Ie you say it doesnt matter whether the LHS is y/u or u/y but if you have y/u and u=o then the LHS goes to infinity...
• Jan 18th 2011, 04:13 AM
Jester
For the general solution, it doesn't matter but for the IC $u(x,2x) = 0$, then it would matter.