# An interesting PDE perhaps?

1. Oct 4, 2005

### saltydog

In the HW section, someone proposed:

$$u^2\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=0;\quad u(x,0)=x$$

As per "Basic PDEs" by Bleecker and Csordas", treating this as:

$$F(x,y,u,p,q)=0\quad\text{with}\quad \frac{\partial u}{\partial x}=p\quad\text{and}\quad\frac{\partial u}{\partial y}=q$$

and solving for parametric functions as a function of two variables t and s such that:

$$\frac{dx(s,t)}{dt}=\frac{\partial F}{\partial p}$$

$$\frac{dy(s,t)}{dt}=\frac{\partial F}{\partial q}$$

$$\frac{du(s,t)}{dt}=p\frac{\partial F}{\partial p}+q\frac{\partial F}{\partial q}$$

Solving the ODEs subject to boundary conditions:

$$u(x,0)=u(f(s),g(s))=s$$

yields:

$$x(s,t)=s^2t$$

$$y(s,t)=t$$

$$u(s,t)=s$$

Solving for u:

$$u(x,y)=\pm\sqrt{\frac{x}{y}}$$

To what extent can I rely upon this method to solve more complex non-linear PDEs? Guess that would involve some analysis of sorts limited by the ability to affect some integration or another. Some limitations I can think of include:

1. I just glossed-over the need sometimes to solve a set of 5 equations instead of the simpler 3 sets like above. Seems though should be able to numerically integrate the 5 equations (or 3) no matter how complex and thus arrive at least in principle to a parametric solution described above.

2. u(x,y) cannot allways be determined explicitly in terms of x and y although the parametric forms are equally valid albeit a bit more difficult to manipulate. Suppose I can work on them a bit but I mean, that could take a whole semester to fully analyze:

$$F(x,y,u,p,q)=0$$

2. Oct 5, 2005

### qbert

sorry,
$$u = \pm \sqrt{ \frac{x}{y}}$$
isn't a solution to
$$u^2 \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} =0, \: u(x,0)=x.$$

the problem is in the u(x,0)=x part.
$$u = \frac{-1 + \sqrt{1+4xy}}{2y}$$

3. Oct 5, 2005

### saltydog

Good for you Qbert. Yea, I thought about not meeting the initial conditions last night. Problem lies in:

$$\frac{dx(s,t)}{dt}=u^2=s^2;\quad x(s,0)=s$$

Thus:

$$x(s,t)=s^2t+s$$

and solving for u(x,y):

$$u(x,y)=\frac{-1\pm\sqrt{1+4xy}}{2y}$$

However, the negative one does not converge to u(x,0)=x but the positive one does.

Dang it! I hate when that happens. :yuck: And I'm not surprised you detected it. Thanks! Guess I gotta' go and update the one in the HW section now.

4. Oct 5, 2005

### saltydog

Alright, suppose I glossed-over something again. This:

$$\mathop\lim\limits_{y\to 0}\left(\frac{-1+\sqrt{1+4xy}}{2y}\right)=x$$

Well, it's not obvious to me anyway. :surprised .

Anyway, since they both go to zero, I can justify using . . . wait, let me look up the spelling . . . L'Hopital's rule. That give me:

$$\frac{x}{\sqrt{1+4xy}}$$ which then goes to x as y goes to zero.

5. Oct 5, 2005

### saltydog

I have a question:

So for:

$$u^2\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=0;\quad u(x,0)=x$$

we got the solution in parametric form:

$$x(s,t)=s^2t+s$$

$$y(s,t)=t$$

$$u(s,t)=s$$

Now suppose I couldn't solve for u explicitly in terms of x and y and just had the parametric equations above.

I don't know how to algebraically verify the solution using only the parametric equations. Can someone show me or should I know this one too?

Thanks,
Salty

6. Oct 6, 2005

### saltydog

Alright I think I have it thanks to . . . well something else I worked on in the forum:

So we have:

$$u^2\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=0$$

with:

$$x=s^2t+s$$

$$y=t$$

$$u=s$$

so, lets start calculating partials:

$$\frac{\partial u}{\partial s}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+ \frac{\partial u}{\partial y}\frac{\partial y}{\partial s}=\frac{\partial u}{\partial x}(2st+1)$$

$$\frac{\partial u}{\partial t}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial u}{\partial y}\frac{\partial y}{\partial t}=s^2\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}$$

thus:

$$\frac{\partial u}{\partial x}=\frac{1}{2st+1}\frac{\partial u}{\partial s}$$

$$\frac{\partial u}{\partial y}=\frac{\partial u}{\partial t}-\frac{s^2}{2st+1}\frac{\partial u}{\partial s}$$

Substituting these into the original PDE:

$$\frac{u^2}{2st+1}\frac{\partial u}{\partial s}+\frac{\partial u}{\partial t}-\frac{s^2}{2st+1}\frac{\partial u}{\partial s}=0$$

Now, since u=s, the two partials with respect to s cancel leaving us with:

$$\frac{\partial u}{\partial t}=0$$

but since u=s, that partial is zero, thus satisfying the relation.

Last edited: Oct 6, 2005