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Basis functions and weak ODE solution
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$begingroup$
Given some linear differential operator $L$, I'm trying to solve the eigenvalue problem $L(u) = lambda u$. Given basis functions, call them $phi_i$, I use a variational procedure and the Ritz method to approximate $lambda$ via the associated weak formulation
$$langle L(phi_i),phi_jrangle = lambda langle phi_i,phi_jrangle.$$
As you can see, this expression is now a matrix equation, solutions to which are straightforward. For my particular problem, the basis functions are $$phi_j = cosleft( frac{pi j}{2}(x+1) right) coshleft( frac{pi j}{2}(y+h) right).$$
However, this solution, when inputted into the weak formulation equation, does not output correct eigenvalues. However, $phi_j$ can be split into even and odd components:
$$ phi_j^o = sin left( pi(j-1/2)x right)coshleft( pi(p-1/2)(y+h) right)\
phi_j^e = cos left( pi j x right)coshleft( pi j(y+h) right)
$$
Now to obtain eigenvalues I solve two separate equations, one for even eigenvalues and one for odd:
$$langle L(phi_i^e),phi_j^erangle = lambda langle phi_i^e,phi_j^erangle\
langle L(phi_i^o),phi_j^orangle = lambda langle phi_i^o,phi_j^orangle.$$
This latter approach gives correct solutions: why? Any insight or direction is greatly appreciated.
functional-analysis ordinary-differential-equations change-of-basis
New contributor
$endgroup$
add a comment |
$begingroup$
Given some linear differential operator $L$, I'm trying to solve the eigenvalue problem $L(u) = lambda u$. Given basis functions, call them $phi_i$, I use a variational procedure and the Ritz method to approximate $lambda$ via the associated weak formulation
$$langle L(phi_i),phi_jrangle = lambda langle phi_i,phi_jrangle.$$
As you can see, this expression is now a matrix equation, solutions to which are straightforward. For my particular problem, the basis functions are $$phi_j = cosleft( frac{pi j}{2}(x+1) right) coshleft( frac{pi j}{2}(y+h) right).$$
However, this solution, when inputted into the weak formulation equation, does not output correct eigenvalues. However, $phi_j$ can be split into even and odd components:
$$ phi_j^o = sin left( pi(j-1/2)x right)coshleft( pi(p-1/2)(y+h) right)\
phi_j^e = cos left( pi j x right)coshleft( pi j(y+h) right)
$$
Now to obtain eigenvalues I solve two separate equations, one for even eigenvalues and one for odd:
$$langle L(phi_i^e),phi_j^erangle = lambda langle phi_i^e,phi_j^erangle\
langle L(phi_i^o),phi_j^orangle = lambda langle phi_i^o,phi_j^orangle.$$
This latter approach gives correct solutions: why? Any insight or direction is greatly appreciated.
functional-analysis ordinary-differential-equations change-of-basis
New contributor
$endgroup$
add a comment |
$begingroup$
Given some linear differential operator $L$, I'm trying to solve the eigenvalue problem $L(u) = lambda u$. Given basis functions, call them $phi_i$, I use a variational procedure and the Ritz method to approximate $lambda$ via the associated weak formulation
$$langle L(phi_i),phi_jrangle = lambda langle phi_i,phi_jrangle.$$
As you can see, this expression is now a matrix equation, solutions to which are straightforward. For my particular problem, the basis functions are $$phi_j = cosleft( frac{pi j}{2}(x+1) right) coshleft( frac{pi j}{2}(y+h) right).$$
However, this solution, when inputted into the weak formulation equation, does not output correct eigenvalues. However, $phi_j$ can be split into even and odd components:
$$ phi_j^o = sin left( pi(j-1/2)x right)coshleft( pi(p-1/2)(y+h) right)\
phi_j^e = cos left( pi j x right)coshleft( pi j(y+h) right)
$$
Now to obtain eigenvalues I solve two separate equations, one for even eigenvalues and one for odd:
$$langle L(phi_i^e),phi_j^erangle = lambda langle phi_i^e,phi_j^erangle\
langle L(phi_i^o),phi_j^orangle = lambda langle phi_i^o,phi_j^orangle.$$
This latter approach gives correct solutions: why? Any insight or direction is greatly appreciated.
functional-analysis ordinary-differential-equations change-of-basis
New contributor
$endgroup$
Given some linear differential operator $L$, I'm trying to solve the eigenvalue problem $L(u) = lambda u$. Given basis functions, call them $phi_i$, I use a variational procedure and the Ritz method to approximate $lambda$ via the associated weak formulation
$$langle L(phi_i),phi_jrangle = lambda langle phi_i,phi_jrangle.$$
As you can see, this expression is now a matrix equation, solutions to which are straightforward. For my particular problem, the basis functions are $$phi_j = cosleft( frac{pi j}{2}(x+1) right) coshleft( frac{pi j}{2}(y+h) right).$$
However, this solution, when inputted into the weak formulation equation, does not output correct eigenvalues. However, $phi_j$ can be split into even and odd components:
$$ phi_j^o = sin left( pi(j-1/2)x right)coshleft( pi(p-1/2)(y+h) right)\
phi_j^e = cos left( pi j x right)coshleft( pi j(y+h) right)
$$
Now to obtain eigenvalues I solve two separate equations, one for even eigenvalues and one for odd:
$$langle L(phi_i^e),phi_j^erangle = lambda langle phi_i^e,phi_j^erangle\
langle L(phi_i^o),phi_j^orangle = lambda langle phi_i^o,phi_j^orangle.$$
This latter approach gives correct solutions: why? Any insight or direction is greatly appreciated.
functional-analysis ordinary-differential-equations change-of-basis
functional-analysis ordinary-differential-equations change-of-basis
New contributor
New contributor
New contributor
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Josh McCraneyJosh McCraney
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