This question is from my Exterior Differential System course. I have not seen a question that concerned a system so I do not know anyway of attacking it as of yet. Any help appreciated. Thanks !
The first line of attack on any pde is separation of variables. This system is unusual in that it is nonlinear, and yet yields separable solutions. Dym's equation is the only other nonlinear pde I know of that does that. If I were you, I would carry out a separation of variables procedure on this system and see what you get.
Question: what is the characteristic space of a system of pde's?
You might try looking up the term in your textbook's index (or, if you don't have a required text, ask your professor where to find a definition of that term).
As for the solution, like I said, go for a separable solution. That is, assume that
$\displaystyle u(x,y)=X(x)Y(y)$.
Plug this into your system, and you should be able to get all the x's to one side, and the y's to the other, in both original equations. From that, you can deduce that each of the sides is constant. Use that to convert into 4 ODE's.
From post #8, we will solve
$\displaystyle v_x + \frac{1}{3}v_y^3 = 0$.
To this we need a BC. From
$\displaystyle u(x,0)=x$, then $\displaystyle u_x(x,0) = 1$ so $\displaystyle v(x,0)=1$.
We now introduce an Ampere contact transformation
$\displaystyle x=X,\;\;\;y = V_Y,\;\;\;v = Y V_Y - V.$
With this, the derivatives transform as
$\displaystyle v_x = - V_X,\;\;\; v_y = Y.$
So now your problem is
$\displaystyle V_X = \frac{1}{3} Y^3$
which is easily integrated
$\displaystyle V = \frac{1}{3} X Y^3 + F(Y).$
To determine $\displaystyle F(Y)$ we need the BC. In our new coordinates it will be
$\displaystyle X = G(Y) $ some function (it really doesn't matter - we only need $\displaystyle F(Y)$ )
and on this boundary
$\displaystyle y = V_Y = 0\, \text{ and}\; v = Y V_Y - V = 1$
or
$\displaystyle V = -1,\;\;\;V_Y = 0$.
Imposing these two on our solution gives
$\displaystyle \frac{1}{3} Y^3 G(Y) + F(Y) = -1,$
$\displaystyle Y^2 G(Y) + F'(Y) = 0$
which eliminating $\displaystyle G(Y)$ and solving the resulting ODE for $\displaystyle F(Y) $ gives
$\displaystyle F(Y) = \frac{1}{3}c_1Y^3 -1$. (nb. the one third is to clean things up later)
Thus,
$\displaystyle V = \frac{1}{3} X Y^3 + \frac{1}{3}c_1Y^3 -1,$
or
$\displaystyle V = \frac{1}{3}(X+c_1)Y^3 - 1$.
This in turn gives
$\displaystyle x = X,\;\;\;y = V_Y = (X+c_1)Y^2,\;\;\;v = Y V_Y-V = \frac{2}{3}(X+c_1) Y^3+1$
Eliminating X and Y gives
$\displaystyle v = \dfrac{2}{3} \dfrac{y^{2/3}}{(x+c_1)^{1/2}} + 1$
Since $\displaystyle u_x = v$ then we integrate giving
$\displaystyle u = \frac{4}{3} (x+c_1)^{1/2}y^{3/2} + x + H(y)$
Finally, substitute into both original PDEs and you'll find that you'll need $\displaystyle H''(y) = 0$. Also, to get $\displaystyle u(x,0)=x$, you'll also need $\displaystyle H(0) = 0$. With this $\displaystyle H(y) = c_2 y$ and your final solution is
$\displaystyle u = \frac{4}{3} (x+c_1)^{1/2}y^{3/2} + x + c_2 y.$
Whew! It was long but you do have an exact solution to your nonlinear system of PDEs.