1
$\begingroup$

Consider the following PDE: $$xu_x+(y+1)u_y=u-1.$$

Using this formula: $$\frac{dx}{x}=\frac{dy}{y+1}=\frac{du}{u-1}.$$

This yields $c_1=\frac{y+1}{x}$ and $c_2=\frac{u-1}{x}.$

We have: $$F\left(\frac{y+1}{x}\right)=\frac{u-1}{x}.$$

Given the following Cauchy condition - $u(x,2x-1)=e^x$. This yields $$xF(2)+1=e^x.$$

Am I on right right track? I'm a little confused because we have an $F(2)$.

$\endgroup$
  • $\begingroup$ I don't follow the line "This yields..." Aren't you supposed to integrate $\frac{dx}{x}=\frac{dy}{y+1}=\frac{du}{u-1}$? $\endgroup$ – user147263 Oct 13 '14 at 17:59
  • $\begingroup$ You integrate two at a time. For example, I did the left most and middle, and then the left most and the right most one. That's how I yielded $c_1$ and $c_2$. $\endgroup$ – emka Oct 13 '14 at 18:23
  • $\begingroup$ But integration should produce something like $\ln x$, $\ln (y+1)$, $\ln (u-1)$... $\endgroup$ – user147263 Oct 13 '14 at 20:29
0
$\begingroup$

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

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

$\dfrac{dy}{dt}=y+1$ , we have $y+1=y_0e^t=y_0x$

$\dfrac{du}{dt}=u-1$ , we have $u(x,y)=F(y_0)e^t+1=xF\left(\dfrac{y+1}{x}\right)+1$

$u(x,2x-1)=e^x$ :

$xF(2)+1=e^x$ , which is impossible.

$\therefore$ There is no solution.

$\endgroup$

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.