Section 4
Stepping up the Dimension
James V. Herod*

*(c) Copyright 1993,1994,1995 by James V. Herod, All rights reserved.

Web page maintained by Evans M. Harrell, II,

Stepping up the Dimension

In Sections 1 and 2, we considered partial differential equations that led to solutions u(x,y). Here we will think of problems in higher dimensions. One might have u(t, x, y) and develop techniques for such systems. In this stepping up the dimensions, u is a function of three variables -- t, x, and y -- not just x and y. An equation that could lead to such a system might look like

a(t, x, y) u_t + b(t, x, y) u_x + c(t, x, y) u_y = d(t, x, y) u.

Alternatively, one might step up the deminsion by having a coupled system with two unknowns functions -- u(t,x) and v(t,x). A couple of systems could be connected by two partial differential equations. For example, u(t,x) and v(t,x) might be related by

u_t - 2u_x + v_x = 0

v_t - 4u_x + 3v_x = 0

We consider both these types in this section.

First, we consider an equation for u(t, x, y) such as

au_t + bu_x + cu_y = du

with appropriate initial conditions.

A modification of the methods that we have seen may be used to solve this equation. An example shows that the techniques are not substantially harder than those in the previous section.

Example 4,1

We find a solution u(t, x, y) for the differential equation

u_t + 2u_x - u_y = 0

with initial condition

u(0,x,y) = exp(-(x2+y2)).

There are these two ordinary differential equations to solve:

x'(t) = 2, x(0) = [[xi]],

and y'(t) = -1, y(0) = [[eta]].

Solutions are x(t) = 2t + [[xi]], and y(t) = -t+[[eta]]. We have that if u is a solution of the partial differential equation then u(t, x(t), y(t)) satisfies

F([[partialdiff]] ,[[partialdiff]]t)u(t,x(t),y(t)) = 0, with u(0, [[xi]], [[eta]]) = exp(-([[xi]]2+[[eta]]2)).

This function of t is z from the previous section and must be constant since it has derivative zero: if

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


F(dz,dt) = 0.

As before, we solve for [[xi]] and [[eta]] as functions of t, x, and y. This implies that

u(t, x, y) = exp(-([x-2t]2 + [y+t]2)).

(Some comments by Evans Harrell.)

The problem of drawing graphs is strikingly different from that in Sections 1 and 2. There, one had an initial distribution along a line and watched as the line and distribution above it evolved as x increased. Now the initial distribution is a surface. To watch it evolve, we find it easier to draw "snapshots" at several times. The alternative would be to draw surfaces in 4 dimensions: t, x, y, and u ! Or, we could have animation, changing in time.

As we start to draw graphs, we first predict how the initial distribution should change. It will not change in size for u(t,x,y) is the same as the initial distribution, only shifted by replacing x by x - 2t and y by y + t. The shifting process will follow the characteristics. The characteristic curves can be drawn with a graphing routine in MAPLE.

These characteristic curves are best                                           
thought of as following the graphs of                                           
{x(t),y(t)} parametrically. Looking                                             
at the equations for x(t) and y(t)                                              
gives direction to the picture of the                                           
characteristic curves drawn to the                                              
right.  These plots, together with an                                           
understanding of direction, enables                                             
one to predict the following plots of                                           
the "snapshots" of u(t,x,y) for t =                                             
0, t = 1/2, and t = 1                                                           
                                                  Figure 4,1: Characteristic    

t = 0 t = 1/2 t = 1

Figure 4,2 Snapshots of the solution for Example 4.1

The plots of u(t,x,y) are viewed from the first quadrant. Exercises 1 and 2 at the end of this section use these methods. Syntax that might plot these graphs is such as what follows.

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


Here is the second type problem in stepping up the dimension where we consider a coupled system. Recalling the methods for solving linear systems of equation in undergraduate ordinary differential equations, it will be no surprise that the techniques for solving this equation might involve the use of eigenvalues and eigenvectors. In this case, it is beneficial to recall that if A is a matrix, there is a matrix K such that A can be transformed to "diagonal form", actually Jordan form, as

A = K-1 D K or KAK-1 = D.

Often this procedure works: if A is an nxn matrix, then there may be n eigenvectors, {k1, k2, k3, ..., kn} such that the matrix K-1 with columns these n vectors has the property that

K A K-1

is diagonal. Using this idea, if we wish to solve

B(A(u_t,v_t)) = A B(A(u_x,v_x)) (4.1)

then the system B(A(U_t,V_t)) = KA K-1 B(A(U_x,V_x)) = B(ACO2( d1, 0, 0, d2)) B(A(U_x,V_x)) (4.2)

will be easier to solve. We can use the methods of the previous section. The solution of the original system can be obtained from this transformed problem as follows. The solution will be {u(t,x), v(t,x)} defined by

B(A(u(t,x),v(t,x))) = K-1 B(A(U(t,x),V(t,x))) or KB(A(u,v)) = B(A(U,V))

BecauseB(A(U,V)) solves (4,2) we have

B(A(u_t,v_t)) = K-1 B(A(U_t,V_t))

= K-1 [K A K-1] B(A(U_x,V_x)) = A K-1 B(A(U_x,V_x)) = A B(A(u_x,v_x)).

We will carry out this procedure with MAPLE. This program will get the Jordan form and produce the matrix K even if there are not as many eigenvectors for a single eigenvalue as the multiplicity of that eigenvalue.

Example 4.2

Consider the equations:

u_t - 2u_x + v_x = 0

v_t - 4u_x + 3v_x = 0

with initial conditions

u(0,x) = sin(x),

v(0,x) = cos(x).

We re-write the equation as a system

F([[partialdiff]] ,[[partialdiff]]t) B(A(u,v)) = B(ACO2( 2, -1, 4, -3))B(A(u_x,v_x)). (4.3)


B(A(u,v))(0,x) = B(A(sin(x),cos(x))).

A= B(ACO2( 2, -1, 4, -3))

This A can be transformed to a diagonal form. We use MAPLE to carry out these operations. We first call up the linear algebra package.

* with(linalg):

* A:=matrix([[2,-1],[4,-3]]);

* J:=jordan(A,K);

* evalm(K);

Eigenvalues and eigenvectors for the matrix A are {1,{1,1}} and {-2, {1,4}}. MAPLE chooses to make K and K-1 in its own way.

We have K = B(ACO2( 1, -1/4, 1, -1)), K-1 = B(ACO2(4/3, -1/3, 4/3, -4/3))

and KAK-1 = B(ACO2( 1, 0, 0, -2)).

The new form for the equations is

U_t = U_x and V_t = -2 V_x. (4.4)

This is the new pair of equations to solve. Since the initial value for the equation (4.3) is

u(0,x) = sin(x) and v(0,x) = cos(x),

then the equation (4.4) should have initial value

B(A(U(0,x),V(0,x))) = K B(A( sin(x), cos(x))).

* evalm(K &* vector([sin(x),cos(x)]));

which yields

U(0,x) = sin(x ) - cos(x)/4

and V(0,x) = sin(x ) - cos(x ).

This means that the solution for (4.4) is

U(t,x) = sin(x + t) - cos(x + t)/4

and V(t,x) = sin(x - 2t) - cos(x - 2t).


* f(t,x):=(t,x)->sin(x+t)-cos(x+t)/4; g:=(t,x)->sin(x-2*t)-cos(x-2*t);

We can now define u(t,x) and v(t,x) from U and V.

* sol:=evalm(inverse(K) &* vector([f(t,x),g(t,x)])));

The above MAPLE exploration leads to

u(t,x) = F(4 sin(x+t) - cos(x+t) - sin(x-2t) + cos(x-2t),3)


v(t,x) = F(4 sin(x+t) - cos(x+t) - 4sin(x-2t) + 4cos(x-2t),3).

Here is a check to be sure our answer is correct.

* diff(u(t,x),t)-2*diff(u(t,x),x)+diff(v(t,x),x);

* diff(v(t,x),t)-4*diff(u(t,x),x)+3*diff(v(t,x),x);

Both these last MAPLE calculations yields 0. This finishes the problem.

In all our work, we want more than analytic solutions. We seek geometric understanding for how u and v evolve with time. We need to draw pictures. Two pictures to draw would be graphs of u(t,x) and of v(t,x). However, it might be best to see the parametric curve {x,u(t,x),v(t,x)} change as t evolves. It would be fantastic to see the animated plots using MAPLE In this document, several "snapshots" of {x, u(t,x), v(t,x)} are taken at advancing times.

First, here are plots of u(t,x) and v(t,x).

u(t,x) v(t,x)

Figure 4.3 a Solution survaces for Example 4.2

Now we draw graphs of {x, u(0,x), v(0,x)} from two perspectives:

{x. u(0,x),v(0,x)} viewed from [[theta]] = 55 and [[theta]] = 15.

Figure 4.3b Parametric plots of solutions for Example 4.2, t = 0

The next pair of graphs are {x, u(1/8,x), v(1/8,x)} from two perspectives.

{u(1/8,x),v(1/8,x)} viewed from [[theta]] = 55 and [[theta]] = 15.

Figure 4.3c Parametric plots of solutions for Example 4.2, t = 1/8

Is it impossible to resist trying some of these?

Example 4.3: This problem is interesting because of the special structure of the matrix that is involved.

F([[partialdiff]] ,dt)B(A(u,v,w)) = B(ACO3(-2, 1, 0, 0,-2, 0, 0, 0,-3)) F([[partialdiff]] ,[[partialdiff]]x)B(A(u,v,w)),

u(0,x) = cos(x), v(0,x) = sin(x), w(0,x) = exp(-x).

Example 4.4: Here is an alternate example with MAPLE syntax. The example is the same as the previous, only the associated matrix is

B(ACO3( 8, 4, 0, -1, 4, 0, 0, 0, 1)).

* with(linalg):

* A:=matrix([[8,4,0],[-1,4,0],[0,0,1]]);

* J:=jordan(A,K);

* evalm(K);evalm(inverse(K));

* init:=evalm(K &*vector([0,cos(x),sin(x)]));

* f:=(t,x)->sin(x+t): h:=(t,x)->2*cos(x+6*t):

* diff(f(t,x),t)-diff(f(t,x),x); f(0,x);

diff(h(t,x),t)-6*diff(h(t,x),x); h(0,x);

* diff(h(t,x),x);

* z:=t->-2*sin(eta)*t-cos(eta):



* g:=(t,x)-> -2*sin(x+6*t)*t-cos(x+6*t);

* diff(g(t,x),t)-6*diff(g(t,x),x)-diff(h(t,x),x); g(0,x);

* sol:=evalm(inverse(K) &* vector([f(t,x),g(t,x),h(t,x)])):

* u:=(t,x)->sol[1]: v:=(t,x)->sol[2]: w:=(t,x)->sol[3]:

* diff(u(t,x),t) - 8*diff(u(t,x),x) - 4*diff(v(t,x),x);

diff(v(t,x),t) + diff(u(t,x),x) - 4*diff(v(t,x),x);


* subs(t=0,u(t,x)); subs(t=0,v(t,x)); subs(t=0,w(t,x));


4.1. Solve these partial differential equations: use the initial value u(0,x,y) = exp(-x2 - y2) for each of these exercises:

(a) u_t + u_x + 2u_y = 0,

(b) u_t + xu_x - u_y = 0

4.2. Contrast the solutions for

u_t + 2u_x - u_y = 0,

u_t + 2u_x - u_y + u = 0, and

u_t + 2u_x - u_y - u = 0.

4.3: Solve these partial differential equations with u(0,x) = f(x) and v(0,x) = g(x):

F([[partialdiff]] ,[[partialdiff]]t)B(A(u,v)) + B(ACO2( b, c, -1, 0))B(A(u_x,v_x)) = 0,

where b = 4, c = 3,

where b = 2, c = -3, and

where b = -4, c = 4.

4.4. Solve these partial differential equations: use the initial value u(0,x) = cos(x), v(0,x) = sin(x) for each of the exercises.

(a) u_t + u_x -3v_x = 0, v_t -2u_x +2v_x = 0,

(b) u_t + 4v_x = 0, v_t + u_x = 0.

4.5. Solve these two problems with u(0,x) = f(x) and v(0,x) = g(x):

(a) F([[partialdiff]] ,[[partialdiff]]t)B(A(u,v)) + B(ACO2( 1, 1, 4,1))B(A(u_x,v_x)) = 0.

(b) F([[partialdiff]] ,[[partialdiff]]t)B(A(u,v)) + B(ACO2( 0, 8, 2, 0))B(A(u_x,v_x)) = 0.

4.5. Solve these partial differential equations:

(a) F([[partialdiff]] ,[[partialdiff]]t)B(A(u,v)) + B(ACO2(-3, 1, -1, -1))B(A(u_x,v_x)) = 0.

(b) F([[partialdiff]] ,[[partialdiff]]t)B(A(u,v,w)) + B(ACO3( 1, 1,-1,-1, 3,-1,-1, 1, 1))B(A(u_x,v_x,w_x)) = 0.

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