1
$\begingroup$

I am trying to use the method of characteristics to solve the following first order PDE in three variables:

$$u_x + x \,y \,u_y + 2 x^2 \,z\,\ln z\,u_z = 0 $$

I have begun with the following: $ \frac{dx}{dt} = 1, \frac{dy}{dt}=x\,y, \frac{dz}{dt}=2x^2\,z\,\ln z$

I note that $x = t \,$ (if $\, x(0)=x_0$), and thus y and z are not independent of t, so I think I need to introduce another variable. Here is where I am confused.

$\frac{dy}{ds}=\frac{dy^2}{d^2t}=y, \frac{dz}{ds}=\frac{dz^2}{d^2t}=4\,t\,z\ln z$

This leads to $\ln y = s, y = e^s$ and $\frac{dz}{z\ln z}=4t \,ds, \ln (\ln z)=4ts$ and $z = e^{e^{4ts}}$

so $$x=t, y=s, z = e^{e^{4ts}}$$

$\endgroup$
  • $\begingroup$ where is the boundary data? $\endgroup$ – qbert Nov 30 '17 at 5:54
  • $\begingroup$ I got no boundary data, I am supposed to write them as an autonomous system and express my answer as a generic function of the first integrals. $\endgroup$ – NCAD Nov 30 '17 at 6:48
1
$\begingroup$

$$u_x + x \,y \,u_y + 2 x^2 \,z\,\ln z\,u_z = 0 $$ $$\frac{dx}{1}=\frac{dy}{xy}=\frac{dz}{2x^2z\ln|z|}=\frac{du}{0}$$

First family of characteristic curves from $\quad du=0 \quad\to\quad u=c_1$

Second family of characteristic curves from $\quad \frac{dx}{1}=\frac{dy}{xy} \quad\to\quad \frac{x^2}{2}-\ln|y|=c_2$

Third family of characteristic curves from $\quad\frac{dx}{1}=\frac{dz}{2x^2z\ln|z|} \quad\to\quad \frac{2}{3}x^3-\ln|\ln|z||=c_3 $

General solution of the PDE expressed on the form of implicit equation : $$\Phi\left(u\:\:,\:\: \frac{x^2}{2}-\ln|y| \:\:,\:\: \frac{2}{3}x^3-\ln|\ln|z|| \right)=0$$ $\Phi$ is any differentiable function of three variables.

Or equivalently, general solution of the PDE on explicit form :

$$u(x,y,z)=\text{F}\left( \frac{x^2}{2}-\ln|y| \:\:,\:\: \frac{2}{3}x^3-\ln|\ln|z|| \right)$$ F is any differentiable function of two variables.

$\endgroup$
  • 2
    $\begingroup$ I don't agree with the doraemonpaul's result. I think that $z^{-e^\frac{2x^3}{3}}$ is not correct and should be $z^{ e^{-\frac{2}{3}x^3}}$. May be a typo ? In fact, after correction : $$u=f\left(ye^{-\frac{x^2}{2}},z^{ e^{-\frac{2}{3}x^3} }\right)=\text{F}\left( \frac{x^2}{2}-\ln|y| \:\:,\:\: \frac{2}{3}x^3-\ln|\ln|z|| \right)$$ where $f(\alpha,\beta)=F(X,Y)$ with $\alpha=e^{-X}$ and $\beta=e^{e^{-Y}}$. Except the discrepancy, both results are consistent and are two equivalent form of function. $\endgroup$ – JJacquelin Nov 30 '17 at 15:09
0
$\begingroup$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$

$\dfrac{dy}{dt}=xy=ty$ , letting $y(0)=y_0$ , we have $y=y_0e^\frac{t^2}{2}=y_0e^\frac{x^2}{2}$

$\dfrac{dz}{dt}=2x^2z\ln z=2t^2z\ln z$ , letting $z(0)=z_0$ , we have $z=z_0^{e^\frac{2t^3}{3}}=z_0^{e^\frac{2x^3}{3}}$

$\dfrac{du}{dt}=0$ , letting $u(0)=f(y_0,z_0)$ , we have $u(x,y,z)=f(y_0,z_0)=f(ye^{-\frac{x^2}{2}},z^{e^{-\frac{2x^3}{3}}})$

$\endgroup$
  • $\begingroup$ So I'm mistaken in my understanding of the theory of autonomous systems in that $\frac{dy}{dt}$ and $\frac{dz}{dt}$ do not have to be independent of t? $\endgroup$ – NCAD Nov 30 '17 at 13:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.