Obtain the solution of
(x^2)*u_x + (y^2)*u_y = (x+y)u
(Hint: to obtain a general solution you may need to use the fact that dy/y^2 = (dx - dy)/ (x^2 - y^2)
Obtain the solution of
(x^2)*u_x + (y^2)*u_y = (x+y)u
(Hint: to obtain a general solution you may need to use the fact that dy/y^2 = (dx - dy)/ (x^2 - y^2)
Characteristic equation
$\displaystyle \frac{dx}{x^2} = \frac{dy}{y^2} = \frac{du}{(x+y)u}$
so
$\displaystyle 1)\;\;\;\frac{dx}{x^2} = \frac{dy}{y^2}\;\;\;\text{giving}\;\;\; I_1 = \frac{1}{x} - \frac{1}{y}$
$\displaystyle 2)\;\;\;\frac{dx}{x^2} - \frac{dy}{y^2} = \frac{du}{(x+y)u}$ or $\displaystyle \frac{d(x-y)}{x-y} = \frac{du}{u}$ giving $\displaystyle I_2 = \frac{u}{x+y}$
Solution: $\displaystyle u = (x+y) f \left( \frac{1}{x} - \frac{1}{y}\right)$
Are you able to solve this question?
Consider u_t + (v.grad)u = 0 , where grad = upside down triangle, and v is a vector, dotted with the grad symbol.
where v = (-ax, ay) represents 2D stagnation point flow and u = u(x,y,t). Obtain a solution of the Cauchy problem given that u = u0(x,y) at t = 0
(u0 is read as "u naught")