Partial Differential Equation

• Aug 23rd 2010, 09:15 AM
RoyalFlush
Partial Differential Equation
Consider the equation
$4y^2u_{xx} + 2(1-y^2)u_{xy} - u_{yy}- \frac{2y}{1 + y^2}(2u_x - u_y) = 0$

Derive the transformation $\xi = x + 2y, \eta = x -\frac{2}{3} y^3$ ,show that the equation reduces to $\omega_{\xi \eta} = 0$ where $\omega(\xi, \eta) = u(x(\xi, \eta), y(\xi,eta))$ , and find the general solution.

I am a part-time maths student, so I havent done any calculus courses for about a year and a half (although I did a numerical analysis course last semester that looked at numerical methods for ODE's and PDE's) and I'm really rusty :(
This is my last semester, so I would really appreciate it if someone could help me understand this.
Thanks
• Aug 25th 2010, 04:32 AM
Jester
Quote:

Originally Posted by RoyalFlush
Consider the equation
$4y^2u_{xx} + 2(1-y^2)u_{xy} - u_{yy}- \frac{2y}{1 + y^2}(2u_x - u_y) = 0$

Derive the transformation $\xi = x + 2y, \eta = x -\frac{2}{3} y^3$ ,show that the equation reduces to $\omega_{\xi \eta} = 0$ where $\omega(\xi, \eta) = u(x(\xi, \eta), y(\xi,eta))$ , and find the general solution.

I am a part-time maths student, so I havent done any calculus courses for about a year and a half (although I did a numerical analysis course last semester that looked at numerical methods for ODE's and PDE's) and I'm really rusty :(
This is my last semester, so I would really appreciate it if someone could help me understand this.
Thanks

First off, let's determine the type.

If $A = 4y^2,\;\; B = 2(1-y^2),\;\; C = -1$ then

$B^2 - 4AC = 4(1-y^2)^2+ 16y^2 = 4(1+y^2)^2 > 0$ so it's hyperbolic. This means under a change of variable we should be able to transform you PDE to a standard form, i.e.

$u_{rs} + lots = 0$ (I'm using r and s instead of $\xi$ and $\eta$. lots - lower order terms). We now introduce first and secord order transforms. If $r = r(x,y), \;\;s = s(x,y)$ then

$u_x = u_r r_x + u_s s_x,\;(1a)$
$u_y = u_r r_y + u_s s_y,\;(1b)$

$u_{xx} = u_{rr} r_x^2 + 2 u_{rs} r_x s_x + u_{ss} s_x^2 +u_r r_{xx} + u_s s_{xx},\;(1c)$

$u_{xy} = u_{rr} r_x r_y + u_{rs}\left( r_x s_y + r_y s_x\right) + u_{ss} s_x s_y + u_r r_{xy} + u_s s_{xy},\;(1d)$

$u_{yy} = u_{rr} r_y^2 + 2 u_{rs} r_y s_y + u_{ss} s_y^2 +u_r r_{yy} + u_s s_{yy},\;(1e)$.

Substitute these into your PDE and group terms

$\left( 4y^2 r_x^2 +2(1-y^2)r_x r_y - r_y^2\right) u_{rr}$
$+ \left( 4y^2 r_x s_x +2(1-y^2)(r_x s_y + r_y s_x) - r_y s_y\right) u_{rs}$ $+ \left( 4y^2 s_x^2 +2(1-y^2)s_x s_y - s_y^2\right) u_{ss} + lots = 0$.

In order to hit the target we need the coefficient of $u_{rr}$ and $u_{ss}$ to vanish. Thus,

$4y^2 r_x^2 +2(1-y^2)r_x r_y - r_y^2 = 0,\;\;\;
4y^2 s_x^2 +2(1-y^2)s_x s_y - s_y^2 = 0$
,

or

$\left(2r_x - r_y\right)\left(2y^2 r_x + r_y\right) = 0,\;\;\;\left(2s_x - s_y\right)\left(2y^2 s_x + s_y\right) = 0$

They are the same so choose one for $r$ and the other for $s$, i.e.

$2r_x - r_y = 0,\;\;\;2y^2 s_x + s_y = 0$.

These are first oder PDEs, so by the method of characteristics, we fiind the solutions

$r = R(x + 2y),\;\;\; s = S(x - \frac{2}{3}y^3)$

which we choose the function $R$ and $S$ simple. So

$r = x + 2y,\;\;\; s = x - \frac{2}{3}y^3$ (as you said).

Now put these into your transformations (1) and put all of these into you PDE and simply. See how that goes.
• Aug 25th 2010, 01:30 PM
bugatti79
Quote:

Originally Posted by Danny
First off, let's determine the type.

If $A = 4y^2,\;\; B = 2(1-y^2),\;\; C = -1$ then

$B^2 - 4AC = 4(1-y^2)^2+ 16y^2 = 4(1+y^2)^2 > 0$ so it's hyperbolic. This means under a change of variable we should be able to transform you PDE to a standard form, i.e.

I am also self studying these types of DE. According to the book I follow, the transformations are obtained by letting
$\xi(x,y)=k$ be an integral of $\frac{dy}{dx}=\frac{B+\sqrt{B^2-AC}}{A}$ (1)

$\eta(x,y)=K$ be an integral of $\frac{dy}{dx}=\frac{B-\sqrt{B^2-AC}}{A}$ (2)

I attempted to integrate (1) above
I get $\frac{dy}{dx}=\frac{(1-y^2)+\sqrt{(1-y^2)^2+4y^2}}{4y^2}$
(Note that coefficients of the general form of the PDE for uxx, uxy and uyy are A, 2B and C repsectively.)

Is this integral solvable, suitable way to derive the transformations? Your method seems easier? It works for easier problems I solved, this is the hardest I have come across.

Thanks
• Aug 25th 2010, 02:49 PM
Jester
At first glance your problem seems difficult but notice that

$(1-y^2)^2 + 4y^2 = (1+y^2)^2$

That helps tremendously!
• Aug 26th 2010, 01:20 PM
bugatti79
Quote:

Originally Posted by Danny
At first glance your problem seems difficult but notice that

$(1-y^2)^2 + 4y^2 = (1+y^2)^2$

That helps tremendously!

Well spotted, thanks. So I can get the same transformations. I have a query regarding the last part of your method. How did you arrive at the following (particularly the captial R and S) using the method of characterictics?

http://www.mathhelpforum.com/math-he...c462f2d3fb.png
I have used the method of characteristics for different problems such as

$2u_x-u_y=0$

where $u_x=u_rr_x+ u_ss_x$ but I dont know how you would use this method for

$2r_x - r_y = 0,\;\;\;2y^2 s_x + s_y = 0$

Thanks
BTW, the substitution of the derived terms back into the PDE is extremely labourious!!
• Aug 27th 2010, 04:23 AM
Jester
Quote:

Originally Posted by bugatti79
Well spotted, thanks. So I can get the same transformations. I have a query regarding the last part of your method. How did you arrive at the following (particularly the captial R and S) using the method of characterictics?

http://www.mathhelpforum.com/math-he...c462f2d3fb.png
I have used the method of characteristics for different problems such as

$2u_x-u_y=0$

where $u_x=u_rr_x+ u_ss_x$ but I dont know how you would use this method for

$2r_x - r_y = 0,\;\;\;2y^2 s_x + s_y = 0$

Thanks
BTW, the substitution of the derived terms back into the PDE is extremely labourious!!

To solve a PDE of the form

$a(x,y)u_x + b(x,y)u_y = 0$

the method of characteristics is to solve

$\dfrac{dx}{a(x,y)} = \dfrac{dy}{b(x,y)},\;\;\;du = 0$

Solve the first (an ODE) and second gives

$I(x,y) = c_1,\;\;\; u = c_2$

and solution of your PDE is

$c_2 = F(c_1)$ or $u = F(I(x,y))$.
• Aug 29th 2010, 05:54 AM
RoyalFlush
Quote:

Originally Posted by Danny
First off, let's determine the type.

If $A = 4y^2,\;\; B = 2(1-y^2),\;\; C = -1$ then

$B^2 - 4AC = 4(1-y^2)^2+ 16y^2 = 4(1+y^2)^2 > 0$ so it's hyperbolic. This means under a change of variable we should be able to transform you PDE to a standard form, i.e.

$u_{rs} + lots = 0$ (I'm using r and s instead of $\xi$ and $\eta$. lots - lower order terms). We now introduce first and secord order transforms. If $r = r(x,y), \;\;s = s(x,y)$ then

$u_x = u_r r_x + u_s s_x,\;(1a)$
$u_y = u_r r_y + u_s s_y,\;(1b)$

$u_{xx} = u_{rr} r_x^2 + 2 u_{rs} r_x s_x + u_{ss} s_x^2 +u_r r_{xx} + u_s s_{xx},\;(1c)$

$u_{xy} = u_{rr} r_x r_y + u_{rs}\left( r_x s_y + r_y s_x\right) + u_{ss} s_x s_y + u_r r_{xy} + u_s s_{xy},\;(1d)$

$u_{yy} = u_{rr} r_y^2 + 2 u_{rs} r_y s_y + u_{ss} s_y^2 +u_r r_{yy} + u_s s_{yy},\;(1e)$.

Substitute these into your PDE and group terms

$\left( 4y^2 r_x^2 +2(1-y^2)r_x r_y - r_y^2\right) u_{rr}$
$+ \left( 4y^2 r_x s_x +2(1-y^2)(r_x s_y + r_y s_x) - r_y s_y\right) u_{rs}$ $+ \left( 4y^2 s_x^2 +2(1-y^2)s_x s_y - s_y^2\right) u_{ss} + lots = 0$.

In order to hit the target we need the coefficient of $u_{rr}$ and $u_{ss}$ to vanish. Thus,

$4y^2 r_x^2 +2(1-y^2)r_x r_y - r_y^2 = 0,\;\;\;
4y^2 s_x^2 +2(1-y^2)s_x s_y - s_y^2 = 0$
,

or

$\left(2r_x - r_y\right)\left(2y^2 r_x + r_y\right) = 0,\;\;\;\left(2s_x - s_y\right)\left(2y^2 s_x + s_y\right) = 0$

They are the same so choose one for $r$ and the other for $s$, i.e.

$2r_x - r_y = 0,\;\;\;2y^2 s_x + s_y = 0$.

These are first oder PDEs, so by the method of characteristics, we fiind the solutions

$r = R(x + 2y),\;\;\; s = S(x - \frac{2}{3}y^3)$

which we choose the function $R$ and $S$ simple. So

$r = x + 2y,\;\;\; s = x - \frac{2}{3}y^3$ (as you said).

Now put these into your transformations (1) and put all of these into you PDE and simply. See how that goes.

I am really having trouble with this one, and I have a mid-term that I am worried about. (Crying) I'm still having trouble getting out the general solution and showing that the equation reduces to $\omega_{\xi \eta} = 0$ where $\omega(\xi, \eta) = u(x(\xi, \eta), y(\xi,eta))$

My teacher said we will have a question very similar to this on the mid-term, and it will be the most difficult on the test. Where do I go from here?
• Aug 30th 2010, 01:34 AM
bugatti79
Quote:

Originally Posted by RoyalFlush
I am really having trouble with this one, and I have a mid-term that I am worried about. (Crying) I'm still having trouble getting out the general solution and showing that the equation reduces to $\omega_{\xi \eta} = 0$ where $\omega(\xi, \eta) = u(x(\xi, \eta), y(\xi,eta))$

My teacher said we will have a question very similar to this on the mid-term, and it will be the most difficult on the test. Where do I go from here?

Where exactly are you having trouble?