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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

user127096
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clarkson
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1 Answers1

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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 $\langle b,-a\rangle$.

user127096
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  • I know the parametric scaler and parametric vector forms.I don't know the other two.In implicit vector form how can the dot product of a vector and a normal to it be a constant. Shouldn't it be zero?As vector and a normal to it are perpendicular the angle between them is 90$^0$ – clarkson Mar 16 '14 at 18:36
  • @clarkson General form of equation of a line is something you ought to have seen. Its conversion to vector form amounts to noticing that $\alpha x+\beta y $ is the same as $\left<\alpha,\right>\cdot \vec r$ where $\vec r=\left< x,y\right> $ is the position vector. No, there is no need for $\vec r$ to be orthogonal to $\vec n$. The normal vector $\vec n$ is orthogonal to the direction vector of the line; it need not be orthogonal to position vectors of points on the line. – user127096 Mar 16 '14 at 19:21
  • I found this article link. In it I don' t understand what it means by "" That is a straight line is a locus of points whose radius-vector has a fixed scalar product with a given vector n, normal to the line." – clarkson Mar 16 '14 at 19:22
  • @clarkson The sentence you quoted attempts to translate the formula that precedes it into "plain English". As usual, the translation is harder to grasp than the formula itself. It can be rephrased as: "A line can be described by the equation: (radius-vector)$\cdot$(a given vector n) = constant". – user127096 Mar 16 '14 at 19:28
  • Thanks a lot for the explanation – clarkson Mar 16 '14 at 19:37