Is this PDE not solvable using characteristic lines?

Question

My PDE book solved the following equation using the method of characteristic lines: $3u_x - 2u_y + u = x, u = u(x,y).$ I then encountered the following problem in the exercises:

$u_x + u_y - u = 0$. Since this problem looks similar in structure to the one in the example, I thought I would try to solve it similarly:

The characteristic lines have slope 1. They constitute the family of lines $x - y = d$. Hence we make a change of variables: $w = x - d$ and $z = y$. Therefore $x = w - d$ and $y = z$. Setting $v(w, z) = u(x, y)$, we have $u_x + u_y = (v_ww_x + v_zz_x) + (v_ww_y + v_zz_y) = (v_1\cdot1 + v_z\cdot1) = v_w + v_z$. Then $u_x + u_y - u = 0$ becomes $v_w + v_z - v = 0$.

So we arrive at the exact same DE when we try to use the method of characteristic lines. Did I do something incorrectly or will this DE require a different method?

Answer 1

The change of variables should make one of the new variables constant on the characteristic lines. Thus try $w = x - y$, $z = y$.

Answer 2

Of course it is!

The method of characteristics for your linear 1st order PDE reads:

$$\frac{\mathrm{d}x}{1} = \frac{\mathrm{d}y}{1}= \frac{\mathrm{d}u}{u},$$

which leads to $x-y = \xi$ (from 1st and 2nd identities) and $u = \eta \, e^x$ (from 1st and 3rd identities). Put $\eta = \eta(\xi ) = \eta(x-y)$ and you are done.

• Note the following funny fact of your solution: if you set $\eta(x-y) = f(x-y) e^{ -x+y}$ your solution then becomes $f(x-y)e^y$ which is nothing but to make use of 2nd and 3rd identity instead of 1st and 3rd (symmetry).

Hope this helps.

Cheers!