Conceptual question with first-order linear PDEs.

[Eclair_de_XII]

Homework Statement

"Show that a solution of the homogeneous PDE $au_x+bu_y+cu=0$ cannot be zero at one, and only one point in the plane."

My interpretation of this is that $u(x,y)$ is zero everywhere on the plane except on that point $(x_0,y_0)$.

Homework Equations

$w=bx-ay$
$z=y$

$x=\frac{1}{b}(w+az)$
$y=z$

The Attempt at a Solution

$v(w,z)=u(\frac{1}{b}(w+az),z)$
$v_z(w,z)=\frac{a}{b}u_x+u_y$
$bv_z(w,z)=au_x+bu_y$
$bv_z+cv=0$ or $v_z+\frac{b}{c}v=0$

I solve this and come up with the solution: $v(w,z)=e^{-\frac{b}{c}v}g(w)$ or:

$u(x,y)=e^{-\frac{b}{c}y}⋅g(bx-ay)$

Then I set $u(x,y)=0$ and now I have $g(bx_0-ay_0)=0$ since $e^{-\frac{b}{c}y}≠0,∀y∈ℝ$. My thinking is that the solution curve is never zero when it intersects the characteristic line. And that intersection is the point at which $u(x,y)≠0$. But I'm not very sure that my thinking is correct; that's like the only point I can think of where $u(x,y)$ is non-zero.

It's like, in general, I need to define a specific side condition such that the solution intersects each characteristic line exactly once. So I guess the solution is only defined on the characteristic line and is zero everywhere else? And if it is tangent to one of those lines, or is parallel to them, there is either no solution or infinitely many? Help me understand this.

*There's also a part two to this problem, but I don't want to post it until the first part is solved.*

 

[Mark44]

"Show that a solution of the homogeneous PDE $au_x+bu_y+cu=0$ cannot be zero at one, and only one point in the plane."

My interpretation of this is that $u(x,y)$ is zero everywhere on the plane except on that point $(x_0,y_0)$.

I don't see that this necessarily follows. Couldn't u(x, y) be zero at, say, two or three points?

 

[hilbert2]

The equation basically says that the directional derivative of $u(x,y)$ to the direction of vector $a\mathbf{i} + b\mathbf{j}$ is equal to $cu(x,y)$. If $u(x,y)=0$ at some point, what can you say about the values of $u$ on a line passing through that point and having direction given by $a\mathbf{i} + b\mathbf{j}$ ?

 

[Eclair_de_XII]

The equation basically says that the directional derivative of $u(x,y)$ to the direction of vector $a\mathbf{i} + b\mathbf{j}$ is equal to $cu(x,y)$. If $u(x,y)=0$ at some point, what can you say about the values of $u$ on a line passing through that point and having direction given by $a\mathbf{i} + b\mathbf{j}$ ?

I'm guessing that the line is perpendicular to $a\mathbf{i} + b\mathbf{j}$?