Why must the directional derivative $au_x + bu_y$ always be zero?

Question

I am currently studying the textbook Partial Differential Equations: An Introduction, second edition, by Walter A. Strauss. Chapter 1.2 First-Order Linear Equations says the following:

Let us solve

$$au_x + bu_y = 0,$$

where $a$ and $b$ are constants not both zero.

Geometric Method The quantity $au_x + bu_y$ is the directional derivative of $u$ in the direction of the vector $\mathbf{V} = (a, b) = a \mathbf{i} + b \mathbf{j}$. It must always be zero. This means that $u(x, y)$ must be constant in the direction of $\mathbf{V}$.

Why must it always be zero?

I would appreciate it if someone would please take the time to clarify this.

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EDIT:

The authors go on to say the following:

Coordinate Method Chang variables (or "make a change of coordinates"; Figure 2) to

$$x^\prime = ax + by \ \ \ \ \ \ \ \ \ \ y^\prime = bx - ay. \tag{3}$$

Replace all $x$ and $y$ derivatives by $x^\prime$ and $y^\prime$ derivatives. by the chain rule,

$$u_x = \dfrac{\partial{u}}{\partial{x}} = \dfrac{\partial{u}}{\partial{x^\prime}} \dfrac{\partial{x^\prime}}{\partial{x}} + \dfrac{\partial{u}}{\partial{y^\prime}} \dfrac{\partial{y^\prime}}{\partial{x}} = au_{x^\prime} + bu_{y^\prime}$$

$$u_y = \dfrac{\partial{u}}{\partial{y}} = \dfrac{\partial{u}}{\partial{y^\prime}} \dfrac{\partial{y^\prime}}{\partial{y}} + \dfrac{\partial{u}}{\partial{x^\prime}} \dfrac{\partial{x^\prime}}{\partial{y}} = bu_{x^\prime} - au_{y^\prime}.$$

Hence $au_x + bu_y = a(au_{x^\prime} + bu_{y^\prime}) + b(bu_{x^\prime} - au_{y^\prime}) = (a^2 + b^2)u_{x^\prime}$. So, since $a^2 + b^2 \not= 0$, the equation takes the form $u_{x^\prime} = 0$ in the new (primed) variables. Thus the solution is $u = f(y^\prime) = f(bx - ay)$, with $f$ an arbitrary function of one variable.

Again, it is not clear here why $a^2 + b^2 \not= 0$ implies that the equation takes the form $u_{x^\prime} = 0$ in the new (primed) variables.

Answer 1

$au_x + bu_y = 0\\ \implies \left (\hat i\frac{\partial u}{\partial x}+\hat j\frac{\partial u}{\partial y}\right ) \cdot (a\hat i+b\hat j)=0\\ \implies (\vec{\nabla u}) \cdot (a\hat i+b\hat j)=0$

Thus the directional derivative $(\vec{\nabla u})$ in the direction of vector $\vec V=(a,b)$ is zero.

Answer 2

The P.D.E is satisfied by some $u$ if

$\vec{\nabla} u$ is always perpendicular to $a\hat{i}+b\hat{j}$.

This means,

$\vec{\nabla}u=F(x,y)(-b\hat{i}+a\hat{j})$, for some scaler function $F(x,y)$.

$\Rightarrow \frac{\partial u}{\partial x}=-bF(x,y)$ and $\frac{\partial u}{\partial y}=aF(x,y)$.....(1)

If $\frac{\partial^2 u}{\partial x\partial y}=\frac{\partial^2 u}{\partial y\partial x}$,

then $F(x,y)$ is itself solution of the given differential equation.

Such solutions are $u(x,y)=-bx+ay$; $u(x,y)=e^{-bx+ay}$

So, from one solution you get another and this continues.

For example if we use, $F(x,y)=-bx+ay$, then we get $u(x,y)=\frac{b^2x^2}{2}-abxy+\frac{a^2x^2}{2}$.

We get another $u(x,y)$, from here. If the infinite solution set is $\mathcal T$. Then it follows some properties....

For any $u ,v \in \mathcal T$, then

1) $cu+dv \in \mathcal T, (c,d) \in \mathbb R^2$.

2) And most interestingly $u(x,y)v(x,y) \in \mathcal T$.

[ As$(a\hat{i}+b\hat{j}).\vec{\nabla}uv \\ =(a\hat{i}+b\hat{j}).(v\vec{\nabla}u+u\vec{\nabla}v)$. ]

Hence, $\mathcal T$ is an infinite set which is closed under multiplication and addition.