Introduction

## PART I: FIRST ORDER EQUATIONS Section 5 Complications: Nonlinear Equations James V. Herod*

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

It's no surprise that things can get more complicated. That happened in sophomore differential equations, too. There, the linear situation was not so hard. The linear equations contained nth order, constant coefficient equations and matrix systems. The complications for the linear ordinary differential equation were how to factor polynomials and how to do the necessary linear algebra.

You will remember that a differential equations course typically starts with a few special tricks to do special nonlinear problems and then turns to linear equations. These often begin as constant coefficient equations and moves to linear systems. Most of the course is spent studying these linear systems. It is well that time should be spent on these, for a most common idea for non-linear systems is to approximate the non-linear system with a linear one, solve the linear approximation, and fuss over the accuracy.

We find ourselves ready to take a peek at non-linear equations in these first order, partial differential equations. Our first complication will be those of quasi-linear systems. Even for this special quasi-linear case, you are an optimist if you think that you, or even your favorite computer algebra system, will be able to get the solutions in closed form. You would be lucky to be able to draw graphs, to draw surfaces for solutions of the general quasi-linear equations.

We try a few examples and, with modifications, stay in the context of what has come before. Initially, our systems take this form:

ux + g(u) uy = f(x,y)

with u(0,y) = [[phi]](y).

We begin by looking at an example and then try to see that there are general patterns that sometimes work for this form.

Example 5.1.

The following nonlinear example is not really so bad. And, it is already bad enough! Here's the equation:

ux + u(x,y) uy = 0

u(0,y) = y.

Following the patterns that have come before, define

z(s) = u(x(s),y(s)).

We have three differential equations:

x[[minute]](s) = 1, with x(0) = 0,

y[[minute]](s) = z(s), with y(0) = [[xi]], (5.1)

and z[[minute]](s) = 0, with z(0) = [[xi]].

Because x(s) = s is an easy solution for the first equation, we have effectively two remaining ordinary differential equations:

F(dy,ds) = z(s), y(0) = [[xi]]

and F(dz,ds) = 0, z(0) = [[xi]].

This pair of equations holds on the characteristic curve

{x(s), y(s)} = {s, y(s)}.

We have on that curve that z(s) is the constant [[xi]] . Also, y(s) is not hard to obtain:

y(s) = [[xi]] s + [[xi]] = (s +1) [[xi]] = (x +1) [[xi]].

From a practical perspective, you specify [[xi]] and determine how u changes for all x and y on the characteristic that goes through the point {0,[[xi]]}:

u(x,y) = [[xi]] = y/(x +1).

Better check it: u(0,y) = y

and ux + u uy = - F(y,(x +1)2) + F(y,(x +1)) F(1,(x +1)) = 0.

We can not leave this example without noting that equations (5.1) could have been written in the form

B(A(x[[minute]],y[[minute]],z[[minute]])) = B(ACO3( 0, 0, 0,0, 0, 1, 0, 0, 0)) B(A(x,y,z)) + B(A(1,0,0))

B(A(x,y,z))(0) = B(A(0,[[xi]],[[xi]])).

Example 5.2.

Here's a new equation:

ux + 3 u(x,y) uy = 0

u(0,y) = exp(-y2).

The pattern seems to be to take

z(s) = u(x(s), y(s)).

Then

F(dx,ds) = 1, x(0) = 0,

F(dy,ds) = 3 z(s), y(0) = [[xi]],

and F(dz,ds) = 0, z(0) = exp(-[[xi]]2).

The function z is constant on the characteristic:

z(s) = exp(-[[xi]] 2)

and y defines the line (whose slope depends on [[xi]] ) given by

y(s) = 3 exp(-[[xi]]2) s + [[xi]].

Also x(s) = s. We have this relation between x, y, s, and [[xi]]:

x = s

y = 3 s exp(-[[xi]]2) + [[xi]].

At this point in the previous example, we solved for [[xi]] and s in terms of x and y. Good luck in this example! We can solve for y explicitly in terms of x:

y = 3 x exp(-[[xi]]2) + [[xi]].

Here's how we might draw the surface: Choose [[xi]]. Above the line

y = 3 x exp(-[[xi]]2) + [[xi]]

one has the level curve u = exp(-[[xi]]2).

Here are the characteristic curves:

Figure 5.1. Characteristic plots, [[xi]] = 0, 1/2, 1, & 3/2.

We should have should have seen that the characteristic curves can cross! What are the implications of this phenomena? Which value of [[xi]] do we use when the characteristic curves cross? Surely there must be something interesting here.

We will come back later and examine this problem in some generality and see how we should have predicted this possibility and what are the implications.

In summary, we have considered the equation

ux + f(u) uy = 0, with u(0,y) = g(y).

This leads to the three ordinary differential equations

F(dx,ds) = 1, x(0) = 0,

F(dy,ds) = f(z(s)), y(0) = [[xi]] ,

F(dz,dt) = 0, z(0) = g([[xi]] ).

This is a triple of coupled equations on the characteristic curves. The third equation says that z(t) is constant on the characteristic curve and the constant is g([[xi]]). Thus, the second equation says the derivative of y must be the constant f(g([[xi]])). Hence, y(s) = f(g([[xi]] )) s + [[xi]]. This produces a family of straight lines having slopes that change with [[xi]].

Here's where we are: we have the characteristics

x(s) = s,

y(s) = s f(g([[xi]])) + [[xi]] .

or y = x f(g([[xi]])) + [[xi]] .

At every point {x,y} on the line, we have the value u(x,y) sitting above it with u(x,y) constantly g([[xi]]) along that characteristic. Go back and see if this does not check with the above two examples.

Seems too easy? You're right. Try to draw an example! It's the geometry. The geometry!

The nonlinear problems are not always so bad. Look at these two examples before we wade into the geometry.

The next example reminds us that all the trouble with the previous example came with the exponential function.

Example 5.3.

The equation

ut + u ux = 0

with u(0,x) = x

has solution given implicitly by

u = x - t u

or u(t,x) = F(x,1 + t).

Example 5.4. Here is a similar example with MAPLE syntax.

The equation

ut - u ux = 0

with u(0,x) = x.

* z:=t->eta; x:=t->-eta*t+eta;

* x1:=t-> -t+1: x2:=t->-2*t+2:

x3:=t->-3*t+3: x4:=t->-4*t+4:

* plot({[t,x1(t),t=0..2],[t,x2(t),t=0..2],

[t,x3(t),t=0..2],[t,x4(t),t=0..2]});

* u:=(t,x)->x/(1-t);

* plot3d(u(t,x),t=0..6/7,x=0..1,orientation=[-160,75],axes=NORMAL);

* char:=(t,x)->(1-t)*x;

* plot3d([t,char(t,x),u(t,x)],t=0..6/7,x=0..1,

orientation=[-160,75],axes=NORMAL);

Finally, here is a reminder that things can be good, and bad.

Example 5.5.

The equation is

x ux + y u uy = -x y, u(x,y) = 7 if xy = 2.

Suppose that {x(s),y(s)} describe the characteristic curves and consider

z(s) = u(x(s),y(s)).

We have the equation F(dz,ds) = -x(s) y(s).

This leads to the triple of equations

F(dx,ds) = x, x(0) = [[xi]],

F(dy,ds) = y z, y(0) = 2/[[xi]],

F(dz,ds) = -x y, z(0) = 7.

The question then is how to solve this nonlinear system.

Well,

(x y)[[minute]] = x[[minute]] y + x y[[minute]] = x y + x y z = - z[[minute]] - z[[minute]] z = -(z + F(z2,2))[[minute]]

Thus

x(s) y(s) = - z - z2/2 + C

where C is independent of s.

In fact, x(0) y(0) = [[xi]].2/[[xi]]. It follows that

2 = - (7 + 49/2)+C

or C = 67/2.

Hence,

u2 + 2u + 2x y - 67 = 0.

Using that the initial condition is 7 (instead of -9), we get that

u(x,y) = -1+R(68-2xy).

Exercises

5.1. Solve these partial differential equations.

(a) ut + (1+ 2u) ux = 0, with u(0,x) = x.

(b) ut + u2 ux = 0, with u(0,x) = x.

(c) ut + u ux = 0, with u(0,x) = x2.

5.2. Which of the following graphs goes with which equation in the previous exercise?

Answer: Graphs are c, a, and b respectively.