Quasi-linear hyperbolic PDE help

[Dustinsfl]

I am using the book Elementary Partial Differential Equations by Berg and McGregor.

However, the book neglected to discuss problems of the this form, $$uu_{xy}-u_xu_y=0$$.

How do I approach this problem?

Thanks.

 

[hunt_mat]

This is a quasi-linear hyperbolic PDE if that's any help. Use characteristic co-ordinates to put it in canonical form

 

[Dustinsfl]

Unfortunately, that doesn't help since I am only in Chapter 1 section 2.

 

[hunt_mat]

There is a trick:
$$u\partial_{x}\partial_{y}u=\partial_{x}u\partial_{y}u\Rightarrow\frac{\partial_{x}\partial_{y}u}{\partial_{x}u}=\frac{\partial_{y}u}{u}$$
We notice that this is the same as:
$$\frac{\partial}{\partial y}\left(\ln (\partial_{x}u\right) =\frac{\partial}{\partial y}\ln u$$
From here it is easy to continue.

 

[Dustinsfl]

Thanks.

I understand the left side.

But I am having trouble seeing that

$$<br /> \frac{\partial}{\partial y}\left(\ln (\partial_{x}u\right)$$

How is that the ln?

Thanks.

 

[hunt_mat]

If you understand one side then you understand the other, you know from basic calculus that:
$$\frac{f'(x)}{x}=(\ln f(x))'$$
So for the LHS $$f(x)=\partial_{x}u$$ and the RHS $$f(x)=u$$. See now?

 

[Dustinsfl]

Yes, but isn't the partial derivative of $$\frac{\partial}{\partial x}\Rightarrow \frac{\partial ^2}{\partial x^2}$$?

Which would lead to $$\frac{u_y u_{xx}}{u_x}$$

 

[hunt_mat]

Your differentiating w.r.t. y

 

[Dustinsfl]

Then shouldn't the derivative of a function of x be zero w.r.t y?

 

[hunt_mat]

u=u(x,y)

 

[Dustinsfl]

Ok, I am going to think about this for a moment. Thanks for your help.

 

[hunt_mat]

Consider:
$$\frac{\partial_{x}\partial_{y}u(x,y)}{\partial_{x}u(x,y)}=\frac{\partial_{y}\partial_{x}u(x,y)}{\partial_{x}u(x,y)}=\partial_{y}\ln (\partial_{x}u(x,y))$$
Does this make more sense?

 

[Strants]

If you understand one side then you understand the other, you know from basic calculus that:
$$\frac{f'(x)}{x}=(\ln f(x))'$$
So for the LHS $$f(x)=\partial_{x}u$$ and the RHS $$f(x)=u$$. See now?

As far as I can see,
$$\frac{f'(x)}{f(x)} = (\ln f(x))'$$

I assume that was what you meant.

 

[hunt_mat]

yes, it was. Cheers.

 

[Dustinsfl]

$\int \frac{u_{xy}}{u_x}dy$ Should this dy be this $\partial y\mbox{?}$

$$\int \frac{u_{xy}}{u_x}dy=\int\frac{u_y}{u}dy\Rightarrow ln(u_x)=ln(u)+f(x)$$

Every time I integrate from here, I don't obtain the solution $$u(x,y)=f(x)g(y)$$

Should I exponentiate or move everything to left side?

 

[Dustinsfl]

Solved, Thanks.

 

[hunt_mat]

You should get a solution in the form:
$$u(x,y)=e^{A(x)}e^{B(y)}$$
Where A(x) and B(y) are arbitraty function obtained via integration.