- Examples of boundary value problems
We will use k to denote the
square root of the absolute value of lambda.If lambda = 0 then
:y(x) = Ax + B,
solves the ODE.
Substituted boundary conditions give that both "A" and "B" are equal to zero.
For positive lambda 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 lambda 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.
Example (partial)
Consider the elliptic
eigenvalue problem (boundary value problem):abla^2 v + lambda v = {partial^2 v over partial x^2} + {partial^2 v over partial y^2} + lambda v = 0,,0with boundary conditions
:v(x,0) = v(x,1) = 0,,0
:partialoverpartial x}v(0,y) = {partialoverpartial x}v(1,y) = 0,,0
Using the
separation of variables , we suppose the solution is of the form:v = X(x)Y(y),
substituting,
:partial^2overpartial x^2}(X(x)Y(y))+{partial^2overpartial 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") + λ"Y"("y"))/"Y"("y"), so there are separation constants so
:X"(x)over X(x)} = k = {Y"(y)+lambda Y(y)over Y(y)}.
which splits up into
ordinary differential equation s: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,From our boundary conditions we have
:v(x,0) = X(x)Y(0) = 0, v(x,1) = X(x)Y(1) = 0,
:partialoverpartial x}v(0,y) = X'(0)Y(y) = 0, {partialoverpartial x}v(1,y)=X'(1)Y(y)=0
we want
:Y(0) = 0, Y(1) = 0, X'(0) = 0, X'(1) = 0
for which we can evaluate the boundary conditions and solutions accordingly.
ee also
*
differential equation
*boundary value problem
*initial value problem
*Sturm-Liouville theory
Wikimedia Foundation. 2010.