0
$\begingroup$

I have problem with solving following task:

Using characteristic method find solution $ u(x,y) $ of

$$ \begin{aligned} x^{2} u_{x}+y^{2} u_{y} &=(x+y) u \quad \text { for }(x, y) \in \mathbb{R}^{+} \times \mathbb{R}^{+} \\ u\left(x, \frac{x}{2}\right) &=1 \quad \text { for } x>0 \end{aligned} $$

I found out that $$ y(s) = \frac{-1}{s} + C \\ x(s) = \frac{-1}{s} + C \\ z(s) = -2s + C$$

where $z(s)$ is function $u(x(s), y(s))$ How can i find boundary conditions? Thanks for help.

$\endgroup$
0
$\begingroup$

You solved the differential equations incorrectly, as $-\frac1{s+C}\ne-\frac1s+C$, and you did not use different integration constants in (as of that point) unconnected integrations.

$ds=\frac{dx}{x^2}$ integrates to $s=-\frac{1}{x}+\frac1{x_0}$ so that $$ x(s)=\frac{x_0}{1-x_0s}. $$ Similarly, $$ y(s)=\frac{y_0}{1-y_0s}. $$ Using these solutions in $$ (x+y)ds=dz $$ gives $$ -\ln|1-x_0s|-\ln|1-y_0s|=z-z_0 $$ Now insert $y_0=\frac{x_0}2$, $z_0=1$ and eliminate $s$.


Alternatively, find that $$\frac{dx}{x^2}=\frac{dy}{y^2}\implies y^{-1}-x^{-1}=c_1$$ and $$ \frac{dy-dx}{y^2-x^2}=\frac{dz}{x+y}\implies\frac{d(y-x)}{y-x}=dz \\~\\ \implies z-\ln|y-x|=c_2=\phi(c_1) \\~\\ \implies u(x,y)=z=\ln|y-x|+\phi(\frac1y-\frac1x) $$ and with the initial conditions $$ 1=z_0=\ln|\frac{x_0}2|+\phi(\frac1{x_0})\implies \phi(t)=1+\ln|2t| $$

$\endgroup$

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.