# Partial Differential Equation

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

Derive the transformation $\displaystyle \xi = x + 2y, \eta = x -\frac{2}{3} y^3$ ,show that the equation reduces to $\displaystyle \omega_{\xi \eta} = 0$ where $\displaystyle \omega(\xi, \eta) = u(x(\xi, \eta), y(\xi,eta))$ , 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
$\displaystyle 4y^2u_{xx} + 2(1-y^2)u_{xy} - u_{yy}- \frac{2y}{1 + y^2}(2u_x - u_y) = 0$

Derive the transformation $\displaystyle \xi = x + 2y, \eta = x -\frac{2}{3} y^3$ ,show that the equation reduces to $\displaystyle \omega_{\xi \eta} = 0$ where $\displaystyle \omega(\xi, \eta) = u(x(\xi, \eta), y(\xi,eta))$ , 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 $\displaystyle A = 4y^2,\;\; B = 2(1-y^2),\;\; C = -1$ then

$\displaystyle B^2 - 4AC = 4(1-y^2)^2+ 16y^2 = 4(1+y^2)^2 > 0$ 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.

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

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

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

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

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

Substitute these into your PDE and group terms

$\displaystyle \left( 4y^2 r_x^2 +2(1-y^2)r_x r_y - r_y^2\right) u_{rr}$
$\displaystyle + \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}$$\displaystyle + \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 \displaystyle u_{rr} and \displaystyle u_{ss} to vanish. Thus, \displaystyle 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 \displaystyle \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 \displaystyle r and the other for \displaystyle s, i.e. \displaystyle 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 \displaystyle r = R(x + 2y),\;\;\; s = S(x - \frac{2}{3}y^3) which we choose the function \displaystyle R and \displaystyle S simple. So \displaystyle 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 \displaystyle A = 4y^2,\;\; B = 2(1-y^2),\;\; C = -1 then \displaystyle 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 \displaystyle \xi(x,y)=k be an integral of \displaystyle \frac{dy}{dx}=\frac{B+\sqrt{B^2-AC}}{A} (1) \displaystyle \eta(x,y)=K be an integral of \displaystyle \frac{dy}{dx}=\frac{B-\sqrt{B^2-AC}}{A} (2) I attempted to integrate (1) above I get \displaystyle \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 \displaystyle (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 \displaystyle (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 \displaystyle 2u_x-u_y=0 where \displaystyle u_x=u_rr_x+ u_ss_x but I dont know how you would use this method for \displaystyle 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 \displaystyle 2u_x-u_y=0 where \displaystyle u_x=u_rr_x+ u_ss_x but I dont know how you would use this method for \displaystyle 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 \displaystyle a(x,y)u_x + b(x,y)u_y = 0 the method of characteristics is to solve \displaystyle \dfrac{dx}{a(x,y)} = \dfrac{dy}{b(x,y)},\;\;\;du = 0 Solve the first (an ODE) and second gives \displaystyle I(x,y) = c_1,\;\;\; u = c_2 and solution of your PDE is \displaystyle c_2 = F(c_1) or \displaystyle 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 \displaystyle A = 4y^2,\;\; B = 2(1-y^2),\;\; C = -1 then \displaystyle 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. \displaystyle u_{rs} + lots = 0 (I'm using r and s instead of \displaystyle \xi and \displaystyle \eta. lots - lower order terms). We now introduce first and secord order transforms. If \displaystyle r = r(x,y), \;\;s = s(x,y) then \displaystyle u_x = u_r r_x + u_s s_x,\;(1a) \displaystyle u_y = u_r r_y + u_s s_y,\;(1b) \displaystyle 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) \displaystyle 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) \displaystyle 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 \displaystyle \left( 4y^2 r_x^2 +2(1-y^2)r_x r_y - r_y^2\right) u_{rr} \displaystyle + \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}$$\displaystyle + \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 $\displaystyle u_{rr}$ and $\displaystyle u_{ss}$ to vanish. Thus,

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

or

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

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

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

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

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

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

$\displaystyle r = x + 2y,\;\;\; s = x - \frac{2}{3}y^3$ (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 $\displaystyle \omega_{\xi \eta} = 0$ where $\displaystyle \omega(\xi, \eta) = u(x(\xi, \eta), y(\xi,eta))$

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 $\displaystyle \omega_{\xi \eta} = 0$ where $\displaystyle \omega(\xi, \eta) = u(x(\xi, \eta), y(\xi,eta))$

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?