# Boundary value problem

In mathematics, a boundary value problem consists of a differential equation to be satisfied at all points in the interior of an interval or a region and a set of boundary conditions specifying the values of the solution or some of its derivatives everywhere on the boundary of the interval or region. Boundary value problems may be posed for ordinary differential equations as well as partial differential equations.

Boundary value problems arise in several branches of physics. Problems involving the wave equation, such as the determination of normal modes, are often stated as boundary value problems.

A large class of important boundary value problems are the Sturm-Liouville problems. The analysis of these problems involves the eigenfunctions of a differential operator.

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## Example (ordinary)

We wish to find a function y(x) which solves the following Sturm-Liouville problem:

[itex]y'' = \lambda y\,[itex]

and satisfy the boundary conditions

[itex]y(0)=0, \quad y(\pi) = 0[itex].

We will use k to denote the square root of the absolute value of [itex]\lambda[itex].

If [itex]\lambda = 0[itex] then

[itex]y(x) = Ax + B\,[itex]

solves the ODE.

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

For positive [itex]\lambda[itex] we obtain that

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

solves the ODE.

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

For negative [itex]\lambda[itex] it is easy to show that

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

solves the ODE.

From the first boundary condition,

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

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

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

So either A = 0 or k is an integer.

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

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

One may easily check that they satisfy the boundary conditions.

## Example (partial)

Consider the elliptic eigenvalue problem (boundary value problem)

[itex]\nabla^2 v + \lambda v = {\partial^2 v \over \partial x^2} + {\partial^2 v \over \partial y^2} + \lambda v = 0,\,0

with boundary conditions

[itex]v(x,0) = v(x,1) = 0,\,0
[itex]{\partial\over\partial x}v(0,y) = {\partial\over\partial x}v(1,y) = 0,\,0

We suppose the solution is of the form

[itex]v = X(x)Y(y)\,[itex]

substituting,

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

Divide throughout by X(x):

[itex]= {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[itex]

and then by Y(y):

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

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

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

From our boundary conditions we have

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

we want

[itex]Y(0) = 0,\ Y(1) = 0,\ X'(0) = 0,\ X'(1) = 0[itex]

which splits up into ordinary differential equations

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

and

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

which we can evaluate the boundary conditions and solutions accordingly.

## Bibliography

• A. D. Polyanin, Handbook of Linear Partial Differential Equations for Engineers and Scientists, Chapman & Hall/CRC Press, 2002.de:randwertproblem

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