Section 2

PDE's using ODE's

James V. Herod*

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

In the previous section, we recalled some techniques about systems of ordinary differential equations. In this section we will use those ideas. In the process, we make a connection between the partial differential equation

p(x,y) u_x + q(x,y) u_y + r(x,y) u = 0 (2.1)

and three ordinary differential equations:

A(F(dx,dt) = p(x,y),F(dy,dt) = q(x,y)) (2.2)

F(dz,dt) + s(t) z = 0 (2.3)

where s(t) is to be defined below.

To make the connection with the previous section and between (2.1) and (2.2), suppose that x(t) and y(t) are functions from [0,*) to the reals that satisfy the system of equations (2.2) with initial conditions:

F(dx,dt) = p(x,y), x(0) = 0

F(dy,dt) = q(x,y), y(0) = [[eta]].

These are equations such as those we solved in the last section. Suppose the functions p and q have the properties required in the previous section. Choose a point {a,b} in the region where the equations

x(t) = a

y(t) = b

can be solved for [[eta]] and t uniquely in terms of a and b. Geometrically,
this locates the initial point {0,[[eta]]} on the y axis so that a point
following along the trajectory given parametrically by the functions x and y
and beginning at {0,[[eta]]} arrives at {a,b} in time t. This is a
*characteristic curve* for (2.1) beginning at {0,[[eta]]}.

We derive the solution surface by thinking of the smooth surface u(x(t),y(t)) above the trajectories {x(t), y(t)}. Here's a way to describe its evolution there. We choose r from (2.1) and solve (2.3) for z where

s(t) = r(x(t), y(t)). (2.4)

Then define u implicitely:

u(x(t), y(t)) = z(t). (2.5)

It follows that

p(x,y) u_x + q(x,y) u_y + r(x,y) u

= x[[minute]] u_x + y[[minute]] u_y + r(x,y) u

(2.6)

= F(d ,dt)u(x(t),y(t)) + r(x(t),y(t)) u(x(t),y(t))

= F(dz,dt) + s(t) z = 0.

Because z(0) = u(x(0), y(0)) = u(0,[[eta]]), we know how the initial condition for [[eta]] should be made:

z(0) = u(0,[[eta]]).

The function z depends on t and [[eta]]. One could write z = z(t,[[eta]]).

(Some comments by Evans Harrell.)

**Summary:**

We have the recipe for providing a solution for (2.1):

(1) Solve for the characteristics as in (2.2) with initial conditions. (2) Define s(t) from (2.4). (3) Solve (2.3) with initial conditions z(0) = u(0,[[eta]]). (4) Invert the equations x = x(t,[[eta]]) and y = y(t,[[eta]]) to solve for t and [[eta]] in terms of x and y: t = t(x,y), [[eta]] = [[eta]](x,y). (5) Compose u(x,y) = z(t,[[eta]]) = z(t(x,y), [[eta]](x,y)).

**Example 2.1:**

We generate a solution to the partial differential equation

1 u_x + 2 u_y + 3 u = 0 (2.7)

with initial conditions

U(0,y) = exp(-y^2).

Following the procedure suggested above, we solve the system of ordinary differential equations

x[[minute]](t) = 1, x(0) = 0,

y[[minute]](t) = 2, y(0) = [[eta]].

This has an easy solution

x(t) = t and y(t) = 2t + [[eta]]. (2.8)

We must now solve the differential equation that arises from the context of the partial differential equation

F(dz,dt) + 3z = 0, z(0) = exp(-[[eta]]^2).

Of course, this has solution

z(t) = exp(-3t) exp(-[[eta]]^2). (2.9)

It remains to "invert" the relation (2.6) between {x,y} and {t,[[eta]]}. In fact,

[[eta]] = y- 2x

t = x. (2.10)

One should check now to see that substitution of (2.7) into (2.5) provides a solution for the differential equation. The substitution yields

u(x,y) = exp(-3x) exp(-[y-2x]^2). (2.11)

Here is a Maple verification that this answer is correct:

> u:=(x,y) -> exp(-3*x)*exp(-(y-2*x)^2);

> diff(u(x,y),x)+2*diff(u(x,y),y)+3*u(x,y);

> simplify(");

> u(0,y);

Because we now have the powerful mathematical and graphing tool Maple , we can not resist looking at the graph of this solution.

> plot3d(u(x,y),x=0..3,y=-3..3);

` Figure 2.1: Graph of the solution for Example 2.1`

` It might help intution to look at the above graph and imagine the
lines y = 2x + [[eta]] and to see how the initial distribution decays along
those curves.`

**Summary for Example 2.1:** The characteristics for

1 u_x + 2 u_y + 3 u = 0

are the lines x(t) = t, y(t) = 2 t + [[eta]]

or y = 2 x + [[eta]].

The determining differential equation along the characteristic is

F(dz,dt) + 3 z = 0, z(0) = exp(-3 [[eta]]^2)

which has solution exp(-3t) exp(- [[eta]]^2). After inverting the relationship between {x, y} and {t, [[eta]]}, we find the solution for the partial differential equation is

u(x,y) = exp(-3x) exp(-[y - 2x]^2).

**Example 2.2:**

We find solutions for the differential equation

u_x + y u_y + x u = 0 for -* < y < *, 0 < x (2.12)

subject to the initial condition

u(0,y) = sin(y).

We also draw graphs of the parameteric curves x(t,[[eta]]) and y(t,[[eta]]) and of the solution surface u(x,y).

The parameteric curves x and y are determined by the equations

F(dx,dt) = 1, with x(0) = 0,

F(dy,dt) = y, with y(0) = [[eta]].

This system of equations has solution

x(t,[[eta]]) = t,

y(t,[[eta]]) = [[eta]] exp(t).

We invert and find [[eta]] and t as functions of x and y:

t = x and [[eta]] = y exp(-x).

The determining differential equation for u that is associated with this system is

F(dz,dt) + t z = 0, with z(0) = sin([[eta]]).

This equation has solution

z(t) = exp(-t^2/2) sin([[eta]]).

Finally, we substitute the inversion to x and y coordinates:

u(x,y) = exp(-x^2/2) sin(y e^-x). (2.13)

Figure 2.2 a. Characteristic Curves

Figure 2.2 b. Solution Surface

The Figure 2.2a has the characteristic curves drawn as trajectories {x(t),y(t)}. Figure 2.2b shows the graph of u. Again, it is of value to visualize in this three dimensional graph how the initial value decays and spreads along the trajectories {x,y}.

The trajectories {x(t),y(t)} are called the *characteristics curves
associated with the partial differential equation.* One speaks of solving
the partial differential equation by *the method of characteristics. *

Surely there are interesting questions remaining here:

(1) How robust is this method? (2) How bizarre might the characteristic curves be? (3) Since one curve carries information from the initial value, what happens if a curve from a point [[eta]] where the initial data was small gets close to a curve from another point [[eta]] where the data was large. (One might imagine riding gently along one characteristic curve and looking up to see a huge wave bearing down along another characteristic curve that started at some different initial value!)

**Example 2.3. "Surfer's Delight"**

Let p(x,y) = 1 and

q(x,y) = BLC{(A(-y if y >= 0, 0 if y <= 0)).

Consider u that satisfies

p(x,y) u_x + q(x,y) u_y = 0, with u(0,y) = [[pi]]/2+ arctan(y). (2.14)

This equation has solution u(x,y) = [[pi]]/2+ arctan(y e^x) if y > 0. The graph of the solution suggests an approach to discontinuity of the solution for x large and |y| small.

Figure 2.3. Graph of [[pi]]/2+ arctan(y e^x).

**Summary for visualization **

**of **

**first order, quasilinear partial differential equations.**

We will explain this summary with the example

x u_x - xy u_y = u for x > 1, u(1,y) = f(y) for all y.

For the purpose of analyzing this problem, it makes the calculus easier to change the equation to

u_x - y u_y = u/x for x > 1.

**Visualization of the characteristics.**

There are two methods to use Maple to visualize the characteristics.

(1) Use the numerical integration procedures, or

(2) solve for the characteristic curves analytically and graph the solutions.

**Implementation of Method 1:**

This method is more appropriate when the characteristics can not be found analytically.

> with(DEtools):

> eqns:=[diff(x(t),t)=1,diff(y(t),t)+ y(t)= 0];

inits:={[0,1,-2],[0,1,-1],[0,1,1],[0,1,2],[0,1,3]};

> phaseportrait(eqns,[x,y],0..2,inits,stepsize=0.1,x=-1/2..3,y=-1..3);

**Implementation of method 2:**

The characteristics solve the equations

x' = 1 with x(0) = 1 and y' = -x y with y(0) = eta.

These two equations can be solved with Maple .

> dsolve({diff(x(t),t)=1,diff(y(t),t)+ y(t)= 0, x(0)=1, y(0) = eta},

{x(t),y(t)});

> assign(");

Having the solutions, we draw their graphs.

> plot({[t+1,-2*exp(-t),t=0..2],[t+1,-1*exp(-t),t=0..2],

[t+1,1*exp(-t),t=0..2],[t+1,2*exp(-t),t=0..2],

[t+1,-1*exp(-t),t=0..2],[t+1,3*exp(-t),t=0..2]},

x=-1/2..3,y=-1..3);

**Visualization of the solution surfaces.**

There are three methods to use Maple to visualize the solution surfaces.

(1) Use the numerical integration procedures, or

(2) Graph the solutions parametrically, or

(3) Graph the solutions numerically.

**Implementation of method 1:**

This method uses the numerical procedures in Maple . It is more appropriate when the characteristics can not be found analytically.

> restart: with(DEtools);

> PDEplot([1,-y,-u/x],[x,y,u],[1,s,exp(-s^2)],s=-1..3,numchar=20,

axes=NORMAL, orientation = [-50,40]);

**Implementation of Method 2.**

This method plots the surface parametrically. It is more appropriate when the equations for the characteristics can not be inverted. First the characteristics are determined; as above,

x(t) = t+1 and y(t) = eta*exp(-t).

The function z(t) satisfies the differential equation

z'(t) = z(t)/x(t) = z(t)/(t+1), with z(0) = exp(-eta^2).

We solve this equation.

> dsolve({diff(z(t),t) = z(t)/(t+1),z(0) = exp(-eta^2)},z(t));

Instead of inverting the equations here, we plot {x(t),y(t),z(t)} parametrically.

> plot3d([t+1,eta*exp(-t),exp(-eta^2)*(t+1)],t=0..2,eta=-1..3,

axes=NORMAL, orientation = [-5,45]);

**Implementation of Method 2.**

This method plots the surface analytically. We must invert the characteristic equations.

> restart;solve({t+1=a,eta*exp(-t)=b},{t,eta});

> assign(");

> u:=(a,b)->exp(-eta^2)*(t+1);

> u(a,b);

> plot3d(u(a,b),a=1..3,b=-1..3,axes=NORMAL);

**Exercises**

(2.1) Solve these partial differential equations by the method of characteristics. Sketch the characteristics, and the solution's surface.

(a) u_x + u_y = 0, -* < y < *, 0 < x with u(0,y) = cos(y).

(b) 1u_x + yu_y + 2u = 0, -* < y < *, 0 < x with u(0,y) = sin(y).

(2.2) Solve these partial differential equations:

(a) x u_x +y u_y + x = 0, u(1,y) =y.

(b) u_x = 3x^2, u(0,y) = f(y).

(c) x u_x + y u_y = x y (x^2 + 1), u(1,y)=y.

(d) x^2 u_x + y^2 u_y = u, u(1,y) = y.

(2.3) Solve these partial differential equations:

(a) xu_x + yu_y = -5u, u(1,y) = sin(y).

(b) u_x + u_y + x u = 0, u(0,y) = f(y).

(c) u_x + 2 u_y +2 u = 0, u(x,x) = f(x). (Attention: the initial condition is on the line y = x.)

(d) x u_x + y u_y = u+1, u(x,x^2) = x^2.

(2.4) Solve these partial differential equations

(a) x u_x - y u_y = y, with u(1,y) = f(y)

(b) x u_x + y u_y = 2 u, with u(0,y) = f(y)

(c) x^2 u_x - y^2 u_y = y^2 - x^2 , with u(1,y) = f(y).

(2.5) Solve these partial differential equations

(a) u_x + u_y = 0, u(1,y) = cos(y)

(b) x^2 u_x + y^2 u_y - (x+y) u = 0, u(1,y) =exp(-y^2).

(2.6) Make four partial differential equations, each with initial value u(0,y) = exp(-y^2) and each having one of these properties:

a. the distribution moves along characteristics with slope 1 and has no change in magnitude.

b. the distribution moves along characteristics with slope - 1 and has no change in magnitude.

c. the distribution moves along characteristics with slope 1 and decreases in magnitude.

d, the distribution moves along characteristics with slope 1 and increases in magnitude.