Web page maintained by Evans M. Harrell, II, harrell@math.gatech.edu.
It's too much to hope that the initial values will lie along the simple straight line that is the y-axis in the x-y plane, or the x-axis in the t-x plane. How different can initial curves be? What happens if the initial values coincide with a characteristic curve? Sounds like trouble!
A first understanding of the nature of the characteristic curves is with the following idea.
Consider the following partial differential eqaution:
p(x,y) ux + q(x,y) uy = r(x,y). (8.1)
Consider also parametric curves given by {x(t),y(t)} where
F(dx,dt) = x'(t) and F(dy,dt) = y'(t).
If we let z(t) = u(x(t), y(t)), then
z' = x' ux + y'uy.
Under what conditions do the two equations
A(p ux + q uy = r, x' ux + y' uy = z') (8.2)
uniquely determine ux and uy? The answer is that p y' - q x' should be not zero. That is,
p F(dy,dt) - q F(dx,dt) != 0,
or F(dy,dx) != F(q,p) (8.3)
holds on the initial curve.
This leads to this first observation:
Observation1.
The characteristic curves are parametric curves in the plane precisely where equations (8.2) do not uniquely determine ux and uy.
Example 8.1: Here is partial differential equation with initial conditions and which has many different solutions.
x ux + y uy = u, with u(x,x) = x.
Each of u(x,y) = x, u(x,y) = y, and u(x,y) = R(xy) is a solution.
Why is there more than one solution? The reason we predict there should be more than one solution is the characteristic curve and the initial curve have slope y/x = 1 on the initial curve: a contradiction to the conclusion of (8.3). Initial curves are
x(t) = [[xi]] et, y(t) = [[xi]] et, and y/x = 1.
Also, q/x = y/x = 1.See Exercise 8.1 to obtain three solutions for this equation.
We have in these note already come to understand this second observation.
Observation 2.
Suppose that {x(s), y(s)} is a parametric curve in the plane and h(s) = u(x(s),y(s)). Then these are equivalent:
(a) x' = p, y' = q, and z' = r,
(b) p ux + q uy = r.
To address the next observation, we again review a notion from the calculus. The question is: under what conditions can we guarantee that solutions of the equation
x(t,[[xi]]) = a, y(t,[[eta]]) = b (8.4)
can be inverted to solve for [[xi]] and [[eta]] in terms of a, b, and t?
Before answering the question, recall why the answer is important. Suppose we want to solve (8.1) subject to the initial condition that u([[xi]],[[eta]]) is specified as [[phi]]([[xi]],[[eta]]) along some curve [[Gamma]]([[xi]],[[eta]]) = 0.
Suppose also that x and y are characteristics made up to satisfy
F(dx,dt) = p(x,y) and F(dy,dt) = q(x,y)
and that z(t) satisfies
z'(t) = u(x(t), y(t)).
These three functions are subject to x(0) = [[xi]] y(0) = [[eta]], and z(0) = [[phi]]([[xi]],[[eta]]).
We want to define u at some point {a,b}. We define it to be
u(a,b) = z(x(s,[[xi]]),y(s,[[eta]]))
for exactly the right [[xi]] and [[eta]] so that the characteristic that starts at {[[xi]],[[eta]]} goes through {a,b}. The problem is how to solve for [[xi]] and [[eta]].
This then is the question: how do we know that one can invert (8.3), at least in theory? We address that issue now.
Here is the result that we want to know:
Theorem: Suppose that H has continuous partial derivatives on a neighborhood of a point c and the matrix H'(c) has determinant not zero. The there is a positive number d such that the restriction of H to U = Dd(c) has a continuous inverse with domain f(U).
Suppose that the initial curve is paramaterized as {f(s), g(s)}. For example, if the initial values were specified along the y axis in the x-y plane, then perhaps
f(s) = 0 and g(s) = s.
Or, if the initial values were specified along the line y = x, then
f(s) = g(s).
In the general situation, suppose the characteristic is parameterized by {x(t), y(t)} and
F(dx,dt) = p(t,x,y), F(dy,dt) = q(t,x,y)
and {x(0), y(0)} is a point on that initial curve, say {[[xi]], [[eta]]}. There must be a number s such that {f(s), g(s)} = {[[xi]], [[eta]]}. The functions x and y of equation (8.4) depend both on t and on the initial value, so we denote x with x(t,s) and y with y(t,s). Thus the parameterization for one characteristic curve is
{x(t,s), y(t,s)}.
We ask, is there a region where the equations
A(x(t,s) = a, y(t,s) = b)
can be inverted to find t and s as functions of a and b?
The answer is yes if and only if the the linear approximation to the two equations can be inverted. That is, the answer is yes if and only if the determinant of the Jacobian of H(t,s) = {x(t,s), y(t,s)} is not zero:
det(H'(t,s)) [[equivalence]] det(Jacobian(H(t,s))) != 0.
We compute the Jacobian:
H'(t,s) = B(ACO2( F([[partialdiff]]x,[[partialdiff]]t), F([[partialdiff]]x,[[partialdiff]]s), F([[partialdiff]]y,[[partialdiff]]t), F([[partialdiff]]y,[[partialdiff]]s))) .
We don't want the Jacobian at just any place; we want the Jacobian at t = 0.
H'(0,s) = B(ACO2( p, f'(s), q, g'(s))) .
This has determinant
p g'(s) - q f'(s) = q g'(s) B(F(p,q) - F(f'(s),g'(s))).
Thus, we ask that the initial curve given parametrically by (f(s), g(s)} satisfy
F(f'(s),g'(s)) != F(p(f(s), g(s)),q(f(s), g(s))).
Observation 3: The initial curve cannot be tangent to any characteristic curve. That is, if the initial curve is given parametrically by {f(s), g(s)} and a characteristic curve is given by {x(t),y(t)}, then for all s and t with {f(s), g(s)} = {x(t), y(t)}, we have
F(f'(s),g'(s)) != F(x'(t),y'(t)).
We give some examples.
Example 8.2.
We contrast these two examples:
ux + 3uy = 5 ux + 3uy = 5
u(-[[xi]],[[xi]]) = 7 u([[xi]],3[[xi]]) = 7
In both examples, we see that we define x(s) = s+c and y(s) = 3s+d.
In the first example, we take x(0) = -[[xi]] and y(0) = [[xi]] so that x(s) = s-[[xi]] and y(s) = 3s+[[xi]]. Moreover, if z(s) = u(x(s),y(s)) then z'(s) = 5 so that z(s) = 5s+h(0) = 5s + 7 and u(x(s),y(s)) = 5s + 7. The solution is obtained by solving for s in terms of x and y, so that
u(x,y) = 5F(x+y,4) + 7.
In the second example, take x(0) = [[xi]] and y(0) = 3[[xi]], so that x(s) = s+[[xi]] and y(s) = 3s+3[[xi]]. As in the first example, u(x(s),y(s)) = 5s + 7. Here is the problem: try to solve for s in terms of x and y.
We have x = s+[[xi]] and y = 3s + 3[[xi]]. We cannot solve this pair of equations for s and [[xi]] in terms of x and y. A geometric understanding for this algebraic problem is that a "characteristic curve" coincides with the "initial curve." (Draw the characteristics and the initial curve to see this better for yourself.)
In the following example, characteristics do not cross, and do not coincide with the initial curve. Yet, the solution surface is approaching a discontinuity in the solution. Such a solution might be expected to fail to obey the defining equation in an "real" situation. In a fluid, one might expect breaking waves!
Example 8.2. ("Return to the Surf")
We consider the partial differential equation
ux + F(y,10) uy = 0 with u(0,y) = exp(-(y+5)2).
The characteristics are defined by the two functions x and y defined by
F(dx,ds) = 1 with x(0) = 0, and F(dy,ds) = y /10 with y(0) = [[xi]].
Then x(s) = s and y(s) = exp(s/10)[[xi]]. It follows that U(x(s),y(s)) is constant in s and is exp(-(y+5)2). We invert to get x and y in terms of s and [[xi]] to find that
u(t,x) = exp( y exp(-x/10)+5)2).
Here is the surface:
Figure 8.1
Exercise
8.1a. Solve the partial differential equation in the first quadrant.
x ux + y uy = u
subject to the boundary condition
u(x,x) = x.
Ans: u = x or y or R(xy).
8.1b Consider the same equation subject to the boundary condition
u(x,x2) = x3.
Ans: x=c et, y = c2et, z = y2/x.
8.2 Solve the ones you can. If you can't solve them, explain why you could have predicted this.
(a) 2 u ux + uy = 0, with u(x,0) = x,
(b) u ux + 2 uy = 0, with u(x,0) = x.
(c) u ux + uy = 0, with u(x,x) = x,
(d) u ux + uy = 0, with u(x,-x) = x.
(e) 2 u ux + uy = 0, with u(0,y) = y,
(f) u ux + 2 uy = 0, with u(0,y) = y.
8.3 Find solutions for these two equations to see the effect of changing the sign of the right side:
x ux + y uy = u, with u(cos(t),sin(t)) = 1.
and x ux + y uy = -u, with u(cos(t),sin(t)) = 1.