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Solved Problems

Since u−v = 0, we conclude that the solution of the Dirichlet problem is unique.

Exercise 200. Not all combinations of boundary conditions/initial conditions lead
to so called well-posed problems. Essentially, a well posed problem is one where the
solutions depend continuously on the boundary data. Otherwise it is considered “ill

posed”.
Consider Laplace’s equation on the unit-square

u xx + u yy = 0,

with u(0, y) = u(1, y) = 0 and u(x, 0) = 0, u (x, 0) = sin(nπx).
y

(a) Show that even as → 0, you can find n so that the solution can attain any
finite value for any y > 0. Use this to then show that this problem is ill posed.

(b) Contrast this with the case where u(0, y) = u(1, y) = 0 and u(x, 0) = 0,

u(x, 1) = sin(nπx). Is this well posed?
Solution:

(a) We seek a solution of the form u(x, y) = sin(nπx)Y (y). This form satisfies
the boundary conditions at x = 0, 1.

u xx + u yy = 0

00
2
−(nπ) Y + Y = 0, Y (0) = 0
Y = c sinh(nπy)

Now we apply the inhomogeneous boundary condition.

u (x, 0) = sin(nπx) = cnπ sin(nπx)
y

u(x, y) = sin(nπx) sinh(nπy)
nπ

For = 0 the solution is u = 0. Now consider any > 0. For any y > 0 and
any finite value M, we can choose a value of n such that the solution along

y = 0 takes on all values in the range [−M . . . M]. We merely choose a value
of n such that
sinh(nπy) M
≥ .
nπ
Since the solution does not depend continuously on boundary data, this prob-
lem is ill posed.

(b) We seek a solution of the form u(x, y) = c sin(nπx) sinh(nπy). This form
satisfies the differential equation and the boundary conditions at x = 0, 1 and
at y = 0. We apply the inhomogeneous boundary condition at y = 1.

u(x, 1) = sin(nπx) = c sin(nπx) sinh(nπ)

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