1. ## Systems of PDE help please

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 !

2. 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?

3. I don't know about anybody else but I can't see this attachment.

4. Originally Posted by Danny
I don't know about anybody else but I can't see this attachment.
Hm...I can see it.

Here's the OP's question (for those who can't see it)

5. I'm not sure what the characteristic space is for a system of PDE's. All I know is that if it is indeed one dimensional, then there exist a solution (which may not be unique). Thise course is a killer =s

6. 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

$u(x,y)=X(x)Y(y)$.

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.

7. OK I asked the professor, what I'm meant to do is rewrite the equations as a system of first order PDE's. Problem now is what change of variables will I use?

8. If you set $u_x = v$ and $u_y = w$ then you have

$w_x -v_y = 0$

$v_x + \frac{1}{3}v_y^3 = 0$

$w_x w_y = 1$.

Notice that they also de-couple.

9. @ the op. Are you interested in actually solving this problem with an exact solution. If so, I believe that I can do this for you.

10. Yes that is my main goal given the initial data. It would be great if you showed me how ! Thanks for the help !

11. From post #8, we will solve

$v_x + \frac{1}{3}v_y^3 = 0$.

To this we need a BC. From

$u(x,0)=x$, then $u_x(x,0) = 1$ so $v(x,0)=1$.

We now introduce an Ampere contact transformation

$x=X,\;\;\;y = V_Y,\;\;\;v = Y V_Y - V.$

With this, the derivatives transform as

$v_x = - V_X,\;\;\; v_y = Y.$

$V_X = \frac{1}{3} Y^3$

which is easily integrated

$V = \frac{1}{3} X Y^3 + F(Y).$

To determine $F(Y)$ we need the BC. In our new coordinates it will be

$X = G(Y)$ some function (it really doesn't matter - we only need $F(Y)$ )

and on this boundary

$y = V_Y = 0\, \text{ and}\; v = Y V_Y - V = 1$

or

$V = -1,\;\;\;V_Y = 0$.

Imposing these two on our solution gives

$\frac{1}{3} Y^3 G(Y) + F(Y) = -1,$
$Y^2 G(Y) + F'(Y) = 0$

which eliminating $G(Y)$ and solving the resulting ODE for $F(Y)$ gives

$F(Y) = \frac{1}{3}c_1Y^3 -1$. (nb. the one third is to clean things up later)

Thus,

$V = \frac{1}{3} X Y^3 + \frac{1}{3}c_1Y^3 -1,$

or

$V = \frac{1}{3}(X+c_1)Y^3 - 1$.

This in turn gives

$x = X,\;\;\;y = V_Y = (X+c_1)Y^2,\;\;\;v = Y V_Y-V = \frac{2}{3}(X+c_1) Y^3+1$

Eliminating X and Y gives

$v = \dfrac{2}{3} \dfrac{y^{2/3}}{(x+c_1)^{1/2}} + 1$

Since $u_x = v$ then we integrate giving

$u = \frac{4}{3} (x+c_1)^{1/2}y^{3/2} + x + H(y)$

Finally, substitute into both original PDEs and you'll find that you'll need $H''(y) = 0$. Also, to get $u(x,0)=x$, you'll also need $H(0) = 0$. With this $H(y) = c_2 y$ and your final solution is

$u = \frac{4}{3} (x+c_1)^{1/2}y^{3/2} + x + c_2 y.$

Whew! It was long but you do have an exact solution to your nonlinear system of PDEs.

12. THanks alot Danny !