Solving basic first order PDEs + Method of Characteristics

[StopWatch]

Homework Statement

How were the integral lines dt/a = dx/b derived from the PDE aUt + bUx = 0 where Ut is the partial derivative with respect to time and Ux with respect to x and a, b are constants.

Homework Equations

I honestly have no idea. I may be unprepared for this course as Strauss's book feels like it's skipping through a lot of steps.

If anyone could by chance explain or guide me to an explanation of how the Method of Characteristics (which I think is related here?) works that would be amazing. I just don't see how the system of ODEs comes from a PDE in the manner, for example, described here:

http://www.stanford.edu/class/math220a/handouts/firstorder.pdf

In particular where it says "...Curve C will satisfy the following system..."

Again, any help would be much appreciated. I'm more looking for a reason why all this should work.

 

[voko]

That statement and the system of the equations (2.2) follows from the basic properties of curves. If a curve is represented as $\mathbf{R} = \mathbf{R}(s)$ where s is the parameter, then $\mathbf{\tau} = \frac {d\mathbf{R}} {ds}$ is its tangent vector. System (2.2) states just that.

If you don't understand why this vector is tangent, then you can draw some curve and consider two points on it very close to each other. Then $\mathbf{R}(s + \Delta s) - \mathbf{R}(s)$ will be the chord connecting these two points, which, as $\Delta s$ gets smaller, approaches the tangent line. Dividing that by $\Delta s$ does not change that, so $\frac {\mathbf{R}(s + \Delta s) - \mathbf{R}(s)} {\Delta s}$ also approximates the tangent, so taking the limit we obtain the property stated above.

 

[StopWatch]

Thanks voko.

So in essence we have a tangent vector <dx/ds,dy/ds,dz/ds> that is TANGENT to <a(x,y), b(x,y),c(x,y)> and therefore perpendicular to the unit normal?

I feel like I'm still not getting this. Should I know those from systems of ODEs? Should I review that maybe?

 

[voko]

So in essence we have a tangent vector <dx/ds,dy/ds,dz/ds> that is TANGENT to <a(x,y), b(x,y),c(x,y)> and therefore perpendicular to the unit normal?

$(a(x,y), b(x,y),c(x,y))$ is a vector we have, and we want to find a curve whose tangent vector is this vector.

We want to find such a curve because we know that the solution of the PDE is some surface z = u(x, y), whose normal is $(-\frac {\partial u}{\partial x}, -\frac {\partial u}{\partial y}, 1)$, and the PDE really means that this normal (it is not unit, by the way) is orthogonal to $(a(x,y), b(x,y),c(x,y))$. Which means this vector is in the tangent plane of the surface, and we construct this surface by finding all the curves with this tangent vector.

I feel like I'm still not getting this. Should I know those from systems of ODEs? Should I review that maybe?

It is not exactly clear what "those" are. If that's normals and tangents to curves and surfaces, then those are studied either in calculus or, in more details, in differential geometry. But all you really need to understand is that $\frac {d\mathbf{R}(s)}{ds}$ is a tangent vector to a curve, and that $(-\frac {\partial u}{\partial x}, -\frac {\partial u}{\partial y}, 1)$ is a normal vector to a surface.

 

[StopWatch]

Good explanation.

But with respect to this statement:

"and we construct this surface by finding all the curves with this tangent vector."

Won't some curves to this tangent vector be outside the surface? Like if we imagine a plane on a unit sphere with a vector tangent at one point on the sphere wouldn't the curves corresponding to the sphere be a subset of the the total number of curves that vector is tangent to?

 

[StopWatch]

I'm probably being a bit dense by the way, I appreciate the help.

 

[voko]

Good explanation.

But with respect to this statement:

"and we construct this surface by finding all the curves with this tangent vector."

Won't some curves to this tangent vector be outside the surface? Like if we imagine a plane on a unit sphere with a vector tangent at one point on the sphere wouldn't the curves corresponding to the sphere be a subset of the the total number of curves that vector is tangent to?

The surface is DEFINED by the PDE to have that tangent vector. So every curve that has that tangent vector EVERYWHERE must be part of that surface. EVERYWHERE is a very serious restriction. For the case of a sphere, its equation is $yu_x - xu_y = 0$, so the tangent vector is $(y, -x, 0)$. You could say that, for example, at x = y = 0, the tangent vector is simply (0, 0, 0), and there are many curves that might have such a tangent vector there. But it is only circles that have $(y, -x, 0)$ as their tangent vector everywhere, not just at (0, 0, 0). And a sphere is a "collection" of circles (as are some other surfaces, e.g., a circular cone).

 

[StopWatch]

Thanks! That makes some sense.