Introduction

## PART I: FIRST ORDER EQUATIONS 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 R3 and these first order equations. The intention with the section is to build intuition, not techniques.

We consider a function F from R3 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 R3.

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).

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) et, cos(s) et, cos(s) et, sin(s) et)).

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

t = log(a2 + b2)/2

s = arctan(a/b).

Now

u(a,b) = z(t) = exp(t) = R(a2 + b2).

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

Figure 9.5: plot of u(x,y) = R(a2 + b2).

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 x2 + y2 = 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 = x2 + y2.

(b) x2 + y2 + z2 = 1.

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

(d) z = (2x + 3y)2

9.3 Draw the "natural grid" for Example 9.2.