Linear homogeneous equation $u_x + u_y + u = 0$

Question

Equation: $u_x + u_y + u = 0$

I'm stuck on how to solve this linear homogeneous equation. As I study from the book, it only mentioned the general solution is $u(x,y) = f(ay - bx)$ without the additional $u$ term. How should I start a good approach?

Answer 1

$$u_x+u_y+u=0$$ Use Lagrange characteristic method to solve the first order PDE: $$\frac {dx}{1}=\frac {dy}{1}=-\frac {du}{u}$$ First differential equation is easy to solve: $$ {dx}={dy}$$ $$\implies x = y +c_1$$ $$ \implies c_1=x-y$$ The second DE gives us:

$$\frac {dy}{1}=-\frac {du}{u}$$ $$ \implies y = -\ln u +c $$ $$\implies e^yu=c_2$$ Now the solution opf the PDE is $$f(c_1)=c_2 \implies e^yu=f \left (x-y\right )$$ $$\boxed {u(x,y)=e^{-y}f \left (x-y \right )}$$

Answer 2

Setting $v=e^{x}u$, we have the partial derivatives $v_x=e^{x}(u_x+u)$ and $v_y=e^{x}u_y$, so that $$ v_x + v_y = e^{x}(u_x+u_y+u) = 0 . $$ Now the result of the book $v(x,y) = f(y-x)$ can be used. Finally, $$u(x,y) = e^{-x}f(y-x)\, ,$$ where $f$ is a function to be determined by using the boundary conditions.

NB. The coordinates $x$ and $y$ play symmetric roles, and thus, we may write $u(x,y) = e^{-y}g(x-y)$.

Answer 3

Separation of variables provides some traction here, to wit:

Set

$u(x, y) = v(x)w(y); \tag 1$

then

$u_x = v_x w, \tag 2$

and

$u_y = v w_y; \tag 3$

then the given equation

$u_x + u_y + u = 0 \tag 4$

becomes

$v_x w + v w_y + vw = 0; \tag 5$

in any region of the $xy$-plane where

$v w \ne 0, \tag 6$

we may write this as

$\dfrac{v_x}{v} + \dfrac{w_y}{w} + 1 = 0. \tag 7$

Since $v$ depends only on $x$ and $w$ only on $y$, it follows that each of the functions $v_x/v$ and $w_y/w$ is constant; this may be more rigorously seen by differentiating (7) with respect to $x$, obtaining

$\left ( \dfrac{v_x}{v} \right )_x = 0, \tag 8$

implying

$\dfrac{v_x}{v} = \lambda, \; \text{a constant}, \tag 9$

with a similar result holding for $w_y/w$; in light of (7) we thus have

$\dfrac{w_y}{w} = -1 - \lambda. \tag{10}$

(9) and (10) are easily solved for $v$ and $w$, provided we supply initial conditions of the form

$v(x_0) = v_0, \; w(y_0) = w_0 \tag{11}$

at some point $(x_0, y_0)$; of course we must ensure that

$v_0 w_0 = u_0 = u(x_0, y_0) \tag{12}$

if we are given $u$ at this point. The solutions to (9) and (10) are then

$v(x) = v_0 e^{\lambda(x - x_0)} \tag{13}$

and

$w(y) = w_0 e^{(-1 - \lambda)(y - y_0)}; \tag{14}$

thus

$u(x, y) = v_0 w_0 e^{\lambda(x - x_0) + (-1 - \lambda)(y - y_0)}$ $= u_0 e^{\lambda(x - x_0) + (-1 - \lambda)(y - y_0)}. \tag{15}$