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you would not expect well-posedness of the backward-in-time problem for the diffusion
equation.
As a third example, consider solving a matrix equation instead of a PDE: namely,
Au = b, where A is an m × n matrix and b is a given m-vector. The ”data” of this
problem comprise the vector b. If m > n, there are more rows than columns and the
system is overdetermined. This means that no solution can exist for certain vectors b;
that is, you don’t necessarily have existence. If, on the other hand, n > m, there are
more columns than rows and the system is underdetermined. This means that there are
lots of solutions for certain vectors b; that is, you can’t have uniqueness. Now suppose
that m = n but A is a singular matrix; that is, det A = 0 or A has no inverse. Then the
problem is still ill-posed (neither existence nor uniqueness). It is also unstable. To illustrate
the instability further, consider a nonsingular matrix A with one very small eigenvalue.
The solution is unique but if b is slightly perturbed, then the error will be greatly magnified
in the solution u. Such a matrix, in the context of scientific computation, is called ill-
conditioned. The ill-conditioning comes from the instability of the matrix equation with
a singular matrix.
As a fourth example, consider Laplace’s equation u xx + u yy = 0 in the region D =
(−∞ < x < ∞, 0 < y < ∞). It is not a well-posed problem to specify both u and u y on
the boundary of D, for the following reason. It has the solutions

1 √
u n (x, y) = e − n sin nx sin ny. (5.10)
n
√
Notice that they have boundary data u n (x, 0) = 0 and ∂u n /∂y(x, 0) = e − n sin nx,
which tends to zero as n → ∞. But for y 6= 0 the solutions u n (x, y) do not tend to zero
as n → ∞. Thus the stability condition (iii) is violated.
We present the slightly more realistic examples.

Example 5.1 A string with fixed endpoints. Consider a string fixed at x = 0 and
x = a, as in Fig. 5.4

x = 0 x = a

Figure 5.4: A string with fixed endpoints

It satisfies the wave equation
2
2
1 ∂ u ∂ u
= , 0 > x > w
2
c ∂t 2 ∂x 2
with boundary conditions

u(0, t) = u(a, t) = 0, t > 0,

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