0
$\begingroup$

How to solve this equation? Should I use method of characteristics? Question states: find the solution that satisfies this condition: \begin{aligned} xu_{x}-yu_y+u &= x\\ u&= x^2 \ when \ y=x \end{aligned}

I just plugged "u" and solved accordingly, at the end I got x=0 and x=1/3.

$\endgroup$
0
$\begingroup$

$$xu_x-yu_y=x-u$$ Charpit-Lagrange system of characteristic OEDs : $$\frac{dx}{x}=\frac{dy}{-y}=\frac{du}{x-u}$$ First characteristic equation from $\frac{dx}{x}=\frac{dy}{-y}$ : $$xy=c_1$$ Second characteristic equation from $\frac{dx}{x}=\frac{du}{x-u}$ : $$xu-\frac12 x^2=c_2$$ General solution of the PDE on the form of implicit equation : $$xu-\frac12 x^2=F(xy)$$ where $F$ is an arbitrary function, to be determined according to the boundary condition. $$u(x,y)=\frac12 x+\frac{1}{x} F(xy)$$ Condition : $u(x,x)=x^2=\frac12 x+\frac{1}{x} F(x^2)$ $$F(x^2)=x^3-\frac12 x^2$$ This determines the function $F$ : $$F(X)=X^{3/2}-\frac12 X$$ We put this function into the general solution where $X=xy$ : $$u(x,y)=\frac12 x+\frac{1}{x}\left((xy)^{3/2}-\frac12 (xy) \right)$$ $$u(x,y)=\frac{x-y}{2}+x^{1/2}y^{3/2}$$

$\endgroup$
  • $\begingroup$ First of all, thanks a lot, but I couldnt get how did you use c1 and c2? $\endgroup$ – Karl Markov Mar 9 at 8:26
  • $\begingroup$ This basic in the method of characteristics. See the theory. The general solution on the form of implicit equation is $c_2=F(c_1)$ or equivalently $c_1=G(c_2)$ with $F$ and $G$ arbitrary functions (in fact one being the inverse of the other). $\endgroup$ – JJacquelin Mar 9 at 8:32
  • $\begingroup$ Thanks a lot, now everything is clear $\endgroup$ – Karl Markov Mar 9 at 8:54
0
$\begingroup$

Solving the PDE we get

$$ u(x,y) = \frac{x^2+2\phi(x y)}{2x} $$

and the condition

$$ u(x,x) = \frac{x^2+2\phi(x^2)}{2x} = x^2 $$

gives

$$ \phi(x^2) = x^3-\frac 12 x^2 = x^2\left(\sqrt{x^2}-\frac 12\right) $$

and finally

$$ \phi(xy) = x y\left(\sqrt{x y}-\frac 12\right) $$

$\endgroup$

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.