Examples of boundary value problems

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,,0

with 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 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,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


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