1. ## 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

2. 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.

3. 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

4. At first glance your problem seems difficult but notice that

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

That helps tremendously!

5. 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?

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!!

6. 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?

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 $u = F(I(x,y))$.

7. 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. 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?

8. Originally Posted by RoyalFlush
I am really having trouble with this one, and I have a mid-term that I am worried about. 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?