Special type III: Separable PDEs

Definition 4 (Separable PDE)   A first order PDE is in separable form, if it can be written in form

$$f(x, u_x) = g(y, u_y).$$

Since LHS is function of $x$ only, which is equal to RHS which is a function of $y$ only. It must be equal to constant say $a.$

$$f(x,u_x)=g(y,u_y)=a.$$

Solve for $f(x, u_x) = a$ for $u_x$ which gives $u_{x} = \varphi(x, a)$ and solve for $u_y$ : $g(y, u_y) = a \mapsto u_{y} = \psi(y, a).$ Since $du = u_x dx + u_y dy$, one have

$$du = \varphi(x, a) dx + \psi(y,a) dy$$

$$\therefore u = \int \varphi(x, a) dx + \int \psi(y, a) dy + b$$

Example 1.6.5   Solve $u_{x}^2 + u_{y}^2 = x+y$.

Solution The given equation is in separable form.

$$u_{x}^2 - x = -u_{y}^2 + y = a \hspace{1cm} (a ~$$is constant$$)$$

$$\therefore~u_{x} = \pm \sqrt{x+a},$$    and $$u_y = \pm \sqrt{y-a},$$

$$\therefore~du = \pm \sqrt{x+a} dx \pm \sqrt{y-a}dy,$$

$$u = \pm \frac{2}{3} (x+a)^{\frac{3}{2}} \pm \frac{2}{3} (y-a)^{\frac{3}{2}} + b,~ (b~$$ is constant$$).$$