Solving PDE using method of characteristic equation

Question

When solving the PDE $aU_x + bU_y = 0$ where $a,b$ are constants, how is the line parallel to vector $\left<a,b\right>$ is found to be $bx-ay=\text{constant}$?

Is it by the method of finding the equation of a line that passes through a given point with position vector $\vec r_0$ and parallel to a vector $\vec v$ is $\vec r =\vec r_0+t \vec v$, $t$ is a parameter?

Orthogonal vector is $\left<b,-a\right>$. A parallel vector to the line is $ \left<a,b\right> $. If $(x,y)$ is an arbitrary point on the line, $\left<x,y\right> = \left<b,-a\right>+t\left<a,b\right>$ .

Thus $\frac {x-b}a=\frac{y+a}b=t$.
Which gives $bx-ay=a^2-b^2$
Thus $bx-ay=C$, $C$ is a constant

Answer 1

An equation of a line in the plane can be written in

• parametric scalar form, $x=x_0+at$, $y=y_0+ bt$
• parametric vector form, $\vec r=\vec r_0+t\vec v$. Here $\vec v$ is a direction vector, its components are $a$ and $b$ from the preceding equation.
• implicit scalar form, $\alpha x+\beta y=c$
• implicit vector form, $\vec r \cdot \vec n = c$. Here $\vec n$ is a normal vector, its components are $ \alpha $ and $\beta$ from the preceding equation.

With characteristic curves, we have the direction vector (components $a,b$) but we want an implicit form of equation. For this we need a normal vector. Solution: rotate the direction vector by 90 degrees, using the rotation matrix
$$\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$ This transforms $\langle a,b\rangle$ into $\langle b,-a\rangle$.