Section 9

Connections with the geometry: from surfaces to equations

James V. Herod*

Web page maintained by Evans M. Harrell, II, harrell@math.gatech.edu.

This section is not about how to find analytic solutions for first order
partial differential equations. Rather, it is about connections between the
geometry of surfaces in R^{3} and these first order equations. The
intention with the section is to build intuition, not techniques.

We consider a function F from R^{3} to the numbers which is smooth
enough to have continuous partial derivatives. We also suppose that the
equation

F(x, y, z) = 0

defines z as a function of x and y. The function z will have a graph that is a
surface in R^{3}.

Figure 9.1: Graph of a surface F(x,y,z) = 0.

Recall that the equation for the plane tangent to the graph of z at {a,b,c} is

0 = F([[partialdiff]]F,[[partialdiff]]x)(a,b,c).(x-a) + F([[partialdiff]]F,[[partialdiff]]y)(a,b,c).(y-b) + F([[partialdiff]]F,[[partialdiff]]x)(a,b,c).(z-c)

or

z = F([[partialdiff]]F/[[partialdiff]]x,[[partialdiff]]F/[[partialdiff]]z) (x-a) + F([[partialdiff]]F/[[partialdiff]]y,[[partialdiff]]F/[[partialdiff]]z) (y-b) + c

What is important is that the vector

grad(F)(a,b,c) = { F([[partialdiff]]F,[[partialdiff]]x)(a,b,c), F([[partialdiff]]F,[[partialdiff]]y)(a,b,c), F([[partialdiff]]F,[[partialdiff]]x)(a,b,c)}

is perpendicular to the graph of z at z(a,b) = c.

**Example 9.1**

If F(x,y,z) = 12 - (x-2)^{2} - 2 (y-2)^{2} - z then

grad(F) = {-2 x + 4, -4 y + 8, -1}.

At x = 1, y = 1, we have F(1,1,9) = 0 and grad(F) = {2, 4, -1}. The tangent plane to the surface F(x,y,z) = 0 is given by

z = 2x + 4y + 3.

Suppose that the function u defined by F(x, y, u(x,y) ) = 0 also satisfies the partial differential equation

p ux + q uy = r.

Because

F([[partialdiff]]F,[[partialdiff]]x) + F([[partialdiff]]F,[[partialdiff]]z) F([[partialdiff]]z,[[partialdiff]]x) = 0

and

F([[partialdiff]]F,[[partialdiff]]y) + F([[partialdiff]]F,[[partialdiff]]z) F([[partialdiff]]z,[[partialdiff]]y) = 0.

solving for zx and zy and substituting into the partial differential equation, we have

p F([[partialdiff]]F,[[partialdiff]]x) + q F([[partialdiff]]F,[[partialdiff]]y) + r F([[partialdiff]]F,[[partialdiff]]z) = 0.

Thus, the vector {p(x,y,z),q(x,y,z), r(x,y,z)} is perpendicular to grad(F)(x,y,z):

dotproduct( {p, q, r}, grad(F) ) = 0.

We distinguish between the characteristic curve determined and the trajectory turning along the surface above the characteristic curve. The characteristic curves are given by {x(t), y(t), 0} and the characteristic trajectories are given by {x(t), y(t), z(t)}. Since the characteristic trajectories are made so that they move in the direction of {p, q, r} they define a trajectory on the surface. They are also tangent to the plane, which is in turn tangent to the surface and normal to the gradient of F.

The path in the x-y plane which is the "characteristic curve" has the property that x[[minute]] = p and y[[minute]] = q. Above it, on the surface defined by

F(x,y,z) = 0,

there lies the characteristic trajectory pointing in the direction {p, q, r}:

x[[minute]] = p, y[[minute]] = q, z[[minute]] = r.

In the last section, we considered the role of the trajectory [[Gamma]] = (f(s), g(s), h(s)) on which the initial conditions are given.We found that at no point of curve {f(s), g(s), 0} can be tangent to the characteristic curve at some point {x(t), y(t), 0}. That is, there is no [[lambda]] with

f(s) = x(t) and g(s) = y(t)

and

[[lambda]] f[[minute]](s) = x[[minute]](t) and [[lambda]] g[[minute]](s) = y[[minute]](t).

Figure 9.2: Characteristic curves and an initial curve.

In some sense, a natural grid on the surface is composed of the characteristic trajectories defined by

x[[minute]](t) = p(x,y,z), y[[minute]](t) = q(x,y,z), and z[[minute]](t) = r(x,y,z)

and the trajectories orthogonal to these defined by

{u[[minute]](t), v[[minute]](t), w[[minute]](t)} = grad(F) x {p, q, r}

where "x" denotes the cross product. Only lack of imagination would dismiss this natural grid on the surface as having no utility.

**Example 9.2 ** Consider a surface u(x,y) that satisfies the partial
differential equation

x ux + y uy = u.

From our work to this point, we can get the characteristic curves in the x-y place for this surface. They are given by y = m x for all possible values of m. To define exactly one fuction z to satisfy the differential equation, we need an initial curve [[Gamma]] = {f(s), g(s), h(s)} on the surface. We take {f(s), g(s), h(s)} = {cos(s), sin(s), 1}.

Figure 9.3: Plot of [[Gamma]] = {cos(s), sin(s), 1}.

In the x-y plane, we note that {f(s), g(s), 0} is not tangent to any of the characteristic curves.

Figure 9.4: Characteristic curves and [[Gamma]].

We next follow our procedure to determine x, y, z and u.

x(t) = cos(s) exp(t), y(t) = sin(s) exp(t), and z(t) = exp(t).

We ask if the equations

x(s,t) = a

y(s,t) = b

can be solved for a and b in terms of s and t. We have only to compute the Jacobian for H(s,t) = (x(s,t), y(s,t)). It is

Jacobian(H(s,t)) = B(ACO2( - sin(s) e^{t}, cos(s) e^{t},
cos(s) e^{t}, sin(s) e^{t})).

Since this has determiant e^{t}, we expect to be able to solve the two
equations (at least close aroung [[Gamma]]). The result of that inversion is

t = log(a^{2} + b^{2})/2

s = arctan(a/b).

Now

u(a,b) = z(t) = exp(t) = R(a^{2} + b^{2}).

One should check this solution. The graph of the solution looks right, yes?

Figure 9.5: plot of u(x,y) = R(a^{2} + b^{2}).

Can you visualize the natural grid on the surface?

**Exercise.**

9.1 Solve the partial differential equation

x ux + y uy = - u, subject to u(x,y) = 1 if x^{2} + y^{2}
= 1.

Draw the characteristic curves, the initial curve, and the surface.

9.2 Find a partial differential equation and initial curve to describe these surfaces:

(a) z = x^{2} + y^{2}.

(b) x^{2} + y^{2} + z^{2} = 1.

(c) z = 2x + 3y + 36.

(d) z = (2x + 3y)^{2}

9.3 Draw the "natural grid" for Example 9.2.