Part 3: Boundary Conditions

Partial differential equations often occur with boundary conditions, which are constraints on the solution at different points in space. To illustrate how boundary conditions arise in applications, let us suppose that u( x,t) is the displacement at x in [0,l] of a string of length l at time t: If the string has enough tension to vibrate, then the force of the string's tension can be shown to be of the form

force  of  Tension = k  ¶2u ¶x2

for some constant k. Thus, if all forces other than tension in the string are ignored, then force equal to mass m times acceleration yields

m  ¶2u ¶t2 = force  of  Tension

so that if we let a² = k/m, then the partial differential equation describing the motion of the string is

¶2u ¶t2 = a2  ¶2u ¶x2

(5)

which is the one-dimensional wave equation.

Moreover, since the string is fixed at x = 0 and x = l, we also have the boundary conditions

u(0,t) = 0        and        u( l,t) = 0

(6)

for all times t. If we avoid the trivial solution u = 0, then these boundary conditions can be used to determine some of the arbitrary constants in the separated solution.       

EXAMPLE 5    Find the solution of the one dimensional wave equation (5) subject to the boundary conditions (6).       

Solution: To do so, we substitute u( x,t) = f(x) T( t) into the equation to obtain

¶2 ¶t2 ( f( x) T(t) ) = a2  ¶2 ¶x2 ( f( x) T( t) )         Þ    f( x) T '' (t) = a2T(t) f'' (x)

To separate the variables, we then divide throughout by a²f(x) T( t) :

f( x) T '' (t) a2f( x) T( t) =  a2T( t) f'' (x) a2f( x) T( t)

This in turn simplifies to

T'' (t) a2T( t) =  f'' (x) f( x)

As a result, there must be a constant l such that

T ''  a2T = l        and         f''  f = l

These in turn reduce to the differential equations

T '' = la2T        and        f''  = lf

If l > 0, however, then the oscillations would become arbitrarily large in amplitude, which is not physically possible. Thus, we assume that l is negative, which is to say that l = -w² for some number w. As a result, we have

T''  = -a2w2T        and        f''  = -w2f

Both equations are harmonic oscillators, so that the general solutions are

T( t) = A1cos( awt) +B1sin(awt)         and        f( x) = A2cos( wx) +B2sin( wx)

where A₁,B₁,A₂, and B₂ are arbitrary constants.

        Let's now concentrate on f( x) . The boundary conditions (6) imply that

u( 0,t) = f( 0) T( t) = 0       and        u( l,t) = f( l) T( t) = 0

If we let T( t) = 0, then we will obtain the solution u(x,t) = 0 for all t. This is called the trivial solution since it is the solution corresponding to the string not moving at all. To avoid the trivial solution, we thus assume that

f( 0) = 0        and        f( l) = 0

However, f( x) = A₂cos( wx) +B₂sin( wx) , so that f( 0) = 0 implies that

0 = A2cos( 0) +B2sin( 0) = A2

Thus, A₂ = 0 and f( x) = B₂sin( wx). The boundary condition f( l) = 0 then implies that

B2sin( wl) = 0

If we let B₂ = 0, then we again obtain the trivial solution. To avoid the trivial solution, we let sin( wl) = 0, which in turn implies that

wl = np

for any integer n. Thus, there is a solution for w_n = np/l for each value of n, which means that

fn( x) = B2sin æ è  np l x ö ø

is a solution to the vibrating string equation for each n. Consequently, for each integer n there is a separated solution of the form

un( x,t) = é ë A1cos æ è  anp l t ö ø +B1sin æ è  anp l t ö ø ù û B2sin æ è  np l x ö ø

(7)