PDE characteristics.

Question

I have solved the following equation using characteristics: $$u_{xx}+2u_{xy}-3u_{yy}=0$$ and obtained the characteristics: $$\xi=x+y$$ $$\eta=x-\frac{1}{3}y$$ I have determined the general solution to be of the form: $$u(x,y)=F(x+y)+G(x-\frac{1}{3}y)$$ Given the conditions: $$u(x=0,y)=y, u_x(x=0,y)=y$$ I am having a hard time finding the particular solution. Applying the first condition I obtained: $$F(y)+G(-\frac{1}{3}y)=y$$ I am not quite sure how to correctly differentiate $u(x,y)$ wrt $x$ and also how to combine both conditions to yield the particular solution. I'd be thankful for some guidance.

Answer 1

Please use the general solution of the form $u(x,y)=F(x+y)+G(y-3x)$ instead for convenience.

$u(0,y)=y$ :

$F(y)+G(y)=y$

$u_x(0,y)=y$ :

$F'(y)-3G'(y)=y$

$F(y)-3G(y)=\dfrac{y^2}{2}+c$

$\therefore F(y)=\dfrac{y^2+6y+2c}{8},G(y)=\dfrac{y-y^2-2c}{8}$

$\therefore u(x,y)=\dfrac{(x+y)^2+6(x+y)+y-3x-(y-3x)^2}{8}$