Examples of boundary value problems

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We will use k to denote the square root of the absolute value of λ.

If λ = 0 then

y(x) = Ax + B\,

solves the ODE.

Substituted boundary conditions give that both A and B are equal to zero.

For positive λ we obtain that

y(x) = A e^{kx} + B e^{-kx}\,

solves the ODE.

Substitution of boundary conditions again yields A = B = 0.

For negative λ it is easy to show that

y(x) = A \sin(kx) + B \cos(kx)\,

solves the ODE.

From the first boundary condition,

0 = y(0) = A \sin(0) + B \cos(0) = B\,.

Now, after the cosine is gone, we will substitute the second boundary condition:

y(\pi) = A \sin(k \pi) = 0\,.

So either A = 0 or k is an integer.

Thus we get that the eigenfunctions which solve the "boundary value problem" are

y_n(x) = A \sin(nx) \quad n = 0,1,2,3,....

One may easily check that they satisfy the boundary conditions.

[edit] Example (partial)

Consider the elliptic eigenvalue problem (boundary value problem)

\nabla^2 v + \lambda v = {\partial^2 v \over \partial x^2} + {\partial^2 v \over \partial y^2} + \lambda v = 0,\,0<x<1,\,0<y<1

with boundary conditions

v(x,0) = v(x,1) = 0,\,0<x<1,
{\partial\over\partial x}v(0,y) = {\partial\over\partial x}v(1,y) = 0,\,0<y<1.

We suppose the solution is of the form

v = X(x)Y(y)\,

substituting,

{\partial^2\over\partial x^2}(X(x)Y(y))+{\partial^2\over\partial y}(X(x)Y(y))+\lambda X(x)Y(y)\,
= X''(x)Y(y)+X(x)Y''(y)+\lambda X(x)Y(y)= 0\,.

Divide throughout by X(x):

= {X''(x)Y(y) \over X(x)}+{X(x)Y''(y)\over X(x)}+{\lambda X(x)Y(y)\over X(x)}
= {X''(x)Y(y) \over X(x)}+Y''(y)+\lambda Y(y) = 0

and then by Y(y):

= {X''(x)\over X(x)}+{Y''(y)+\lambda Y(y)\over Y(y)} = 0.

Now X′′(x)/X(x) is a function of x only, as is (Y′′(y) + λY(y))/Y(y), so there are separation constants so

{X''(x)\over X(x)} = k = {Y''(y)+\lambda Y(y)\over Y(y)}.

From our boundary conditions we have

v(x,0) = X(x)Y(0) = 0,\ v(x,1) = X(x)Y(1) = 0,
{\partial\over\partial x}v(0,y) = X'(0)Y(y) = 0,\ {\partial\over\partial x}v(1,y)=X'(1)Y(y)=0

we want

Y(0) = 0,\ Y(1) = 0,\ X'(0) = 0,\ X'(1) = 0

which splits up into ordinary differential equations

{X''(x)\over X(x)} = k,
X''(x) - k X(x) = 0\,

and

{Y''(y)+\lambda Y(y)\over Y(y)} = k,
Y''(y)+(\lambda-k) Y(y) = 0\,

which we can evaluate the boundary conditions and solutions accordingly.

[edit] See also