Question

Consider the equation yu_x - xu_y = 0 (y>0). Check for each of the following initial conditions whether the problem is solvable. If it is solvable, find a solution. If it is not, explain why:
(a) u(x,0) = x^2
(b) u(x,0) = x
(c) u(x,0) = x, x>0

Answer 1

Task:

Consider the equation $y \cdot u_x - x \cdot u_y = 0, (y \geq 0)$. Check for each of the following initial conditions whether the problem is solvable. If it is solvable, find a solution. If it is not, explain why:

(a) $u(x,0) = x^2$

(b) $u(x,0) = x$

(c) $u(x,0) = x, x > 0$

Solution:

$$y \cdot u_x - x \cdot u_y = 0, (y \geq 0)$$ $$\frac{dx}{y} = \frac{dy}{-x}$$ $$- x \, dx = y \, dy$$ $$y \, dy + x \, dx = 0$$ $$\varphi(x, y) = \frac{y^2}{2} + \frac{x^2}{2}$$ $$u(x, y) = c_1 \left(\frac{y^2}{2} + \frac{x^2}{2}\right) = c_2 (y^2 + x^2)$$

(a)

$$u(x, 0) = x^2$$ $$u(x, 0) = c_2 (0^2 + x^2)$$ $$c_2 = 1$$ $$u(x, y) = y^2 + x^2, y > 0$$

(b)

$$u(x, 0) = x$$ $$u(x, 0) = c_2 (0^2 + x^2) = c_2 x^2$$ $$c_2 = \frac{1}{x}$$ $$u(x, y) = \frac{y^2}{x} + x, y \geq 0$$

(c)

$$u(x, y) = \frac{y^2}{x} + x, x > 0, y \geq 0$$

Answer:

(a) $u(x,y) = y^{2} + x^{2}, y > 0$

(b) $u(x,y) = \frac{y^2}{x} + x, y \geq 0$

(c) $u(x,y) = \frac{y^2}{x} + x, x > 0, y \geq 0$