Question

Show that v(x, y;a, b)= (x+y)[2xy+(a-b)(x-y)+2ab]/(a+b)³ is the Reimann function for second order pde u_xy + (2/x+y)(u_x + u_y)=0

Answer 1

$L(u)=u_{xy}+a(x,y)u_x+b(x,y)u_y+c(x,y)u=f(x,y)$

$L^*(v)=v_{xy}-(av)_x-(bv)_y+cv$

Condition of Riemann's function:

$L^*(v)=0$

$v_x=\frac{2xy+(a-b)(x-y)+2ab}{(a+b)^3}+\frac{(x+y)[2y+(a-b)]}{(a+b)^3}$

$v_{xy}=\frac{2x-(a-b)}{(a+b)^3}+\frac{2y+(a-b)}{(a+b)^3}+\frac{2(x+y)}{(a+b)^3}=\frac{4(x+y)}{(a+b)^3}$

$a(x,y)v=b(x,y)v=2/(x+y)\cdot (x+y)[2xy+(a-b)(x-y)+2ab]/(a+b)^3=$

$=2[2xy+(a-b)(x-y)+2ab]/(a+b)^3$

$cv=0$

$(av)_x=\frac{2(2y+(a-b))}{(a+b)^3}$

$(bv)_y=\frac{2(2x-(a-b))}{(a+b)^3}$

$L^*(v)=\frac{4(x+y)}{(a+b)^3}-\frac{2(2y+(a-b))}{(a+b)^3}-\frac{2(2x-(a-b))}{(a+b)^3}=0$