The homogeneous part of the solution is easy

I have a idea but not sure if it is correct.

non-homogeneous part of the solution of $a(x,y)u_x+b(x,y)u_y=f(x,y)$ is:$\int_C f(x,y/(a^2+b^2)ds$

where the integral is along the characteristic curve C

I do not know if it is correct.

Also can some one guide me to show $(1/x)u_x+(1/y)u_y=1/y$ subject to $u(x,1)=(3-x^2)/2$


i am trying to solve $$\frac1x u_x + \frac1y u_y = \frac1y, \,u(x,1)=\frac{3-x^2}2.$$

define the characteristic curve $C$ parametrized by $s$ by $$\frac{dx}{ds} = \frac1x,\, x(0) = a, \frac{dy}{ds} = \frac1y,\, y(0) = 1$$ the solutions are $$x = \sqrt{a^2+2s}, \,y = \sqrt{1+2s}.\tag 1 $$

along the characteristic $C,$ we have $$\frac{du}{ds}=\frac1y=\frac1{\sqrt{1+2s}},\, u = \frac{(3-a^2)}2 \text{ at } s = 0. $$ this has solution $$u(s) = \sqrt{1+2s}+\frac{1-a^2}2 \tag 2$$

we can solve $(1)$ for $a, s$ as $$a= \sqrt{x^2-y^2-1}, s = \sqrt{\frac{y^2-1}2} \tag 3 $$

using $(3)$ in $(2),$ we get $$ u = y + \frac{1-(x^2-y^2-1)}2 = y+\frac12(y^2-x^2).$$

  • $\begingroup$ but can we have a general formula for it? $\endgroup$ – mnmn1993 Feb 21 '15 at 14:20
  • $\begingroup$ if you mean the solution of $au_x + bu_y = c,$ then i don't you can write down the solution explicitly. the same method works, even there the characteristics must fill the domain in a $1$-$1$ fashion so that you can go between the characteristic coordinates and the $x,y$ coordinates. $\endgroup$ – abel Feb 21 '15 at 14:26

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