1
$\begingroup$

I need help with the following question: The given equation is:

$$yu_x+xu_y=u^2$$

Show that the characteristic lines of the equation are: $$\begin{cases} x(t)=C_1e^t+C_2e^{-t}&\\ y(t)=C_1e^t-C_2e^{-t}&\\ u(t)=\frac{1}{C_3-t} &&or&& u(t)\equiv0 \end{cases}$$ My idea is to solve the Following ODE's system: $$\begin{cases} x_t=y&\\ y_t=x&\\ u_t=u^2\end{cases}$$ i still not figure out how i need to to that.that's not a classic ODE. tried to integrae

$x_t=y$ and $y_t=x$ separately but then i don't get the right answer. tried to solve ODE first-order system of X and Y but I found that the eigenvalue is equal to 0.

Any ideas? thanks:)

$\endgroup$
2
$\begingroup$

Write ( I use s instead of t) : $$\frac {dx}{ds}=y$$ Differentiate wrt s: $$\frac {d^2x}{ds^2}=\frac {dy}{ds}$$ Note that we have $$\frac {dy}{ds}=x$$ So that we have, $$\frac {d^2x}{ds^2}-x=0$$ It's linear of second order

$$r^2-1=0 \implies r=\pm 1 \implies x(s)=c_1e^s+c_2e^{-s}$$ for $y(s)$ $$\frac {dx}{ds}=y \implies y(s)=c_1e^s-c_2e^{-s}$$

For the last one

For $u=0$ as @holo pointed out in the comment we have $u=c$ And for $u \ne 0$ we have that $$\frac {du}{ds}=u^2 \implies \int \frac {du}{u^2}=\int ds$$ $$\implies u(s)=\frac 1 {c_3-s}$$

$\endgroup$
  • 2
    $\begingroup$ In the last line you should note that there exists constant solution($u\equiv 0$) $\endgroup$ – ℋolo Oct 4 '18 at 16:02
  • $\begingroup$ Oh yes @holo thanks a lot $\endgroup$ – Isham Oct 4 '18 at 16:05
  • 1
    $\begingroup$ @Holo Corrected thank you $\endgroup$ – Isham Oct 4 '18 at 16:07
  • 1
    $\begingroup$ while finding $y(s)$ we know that $y_t=x$ and we found x so we can just integrate both sides with respect to s. and get the answer.we get that$C_1=C_4,-C_2=C_5$ Thanks:) $\endgroup$ – Omer Ben Oct 4 '18 at 17:50
  • $\begingroup$ @OmerBen you're welcome...yes thats correct for the coefficients $\endgroup$ – Isham Oct 4 '18 at 18:19
0
$\begingroup$

You can see that $$ x_t+y_t=x+y $$ and $$ y_t-x_t=-(y-x) $$ which are both scalar equations that lead to the solution of the first two equation.

Or you could consider that $$ x_{tt}=y_t=x $$ which again gives the given solution for $x$ and $y=x_t$.


Choosing the way of the Lagrange equations $$ \frac{dx}y=\frac{dy}x=\frac{du}{u^2} $$ results in the constants along characteristics $y^2-x^2=c_1$ and $\ln|x+y|+\frac1u=c_2$ which are dependent, $c_2=\Phi(c_1)$, which implies the general solution form $$ u=\frac1{\Phi(y^2-x^2)-\ln|x+y|}. $$

$\endgroup$

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.