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We will use k to denote the square root of the absolute value of λ. In mathematics, a square root of a number x is a number r such that , or in words, a number r whose square (the result of multiplying the number by itself) is x. ...
If λ = 0 then solves the ODE. Substituted boundary conditions give that both A and B are equal to zero. For positive λ we obtain that solves the ODE. Substitution of boundary conditions again yields A = B = 0. For negative λ it is easy to show that solves the ODE. From the first boundary condition, - .
Now, after the cosine is gone, we will substitute the second boundary condition: - .
So either A = 0 or k is an integer. Thus we get that the eigenfunctions which solve the "boundary value problem" are - .
One may easily check that they satisfy the boundary conditions.
Example (partial)
Consider the elliptic eigenvalue problem (boundary value problem) In mathematics, an Elliptic operator is a major type of differential operator P defined on spaces of complex-valued functions, or some more general function-like objects, such that the coefficients of the highest-order derivatives satisfy a positivity condition. ...
In mathematics, a number is called an eigenvalue of a matrix if there exists a nonzero vector such that the matrix times the vector is equal to the same vector multiplied by the eigenvalue. ...
with boundary conditions We suppose the solution is of the form substituting, - .
Divide throughout by X(x): and then by Y(y): - .
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 From our boundary conditions we have - ,
we want which splits up into ordinary differential equations In mathematics, an ordinary differential equation (or ODE) is a relation that contains functions of only one independent variable, and one or more of its derivatives with respect to that variable. ...
and which we can evaluate the boundary conditions and solutions accordingly.
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