Can this Helmholtz PDE with Robin boundary conditions be solved analytically?
$begingroup$
Consider the following Helmholtz problem in the infinite triangle $y>0,;x>y$ with parameters $Q<0$, $Pge0$, $P<|Q|$.
$$left{begin{align}
&psi^{(2,0)}(x,y)+psi^{(0,2)}(x,y)+Epsi(x,y)=0,\
&psi^{(0,1)}(x,0)-frac Q2psi(x,0)=0,\
&psi^{(1,0)}(x,x)-psi^{(0,1)}(x,x)-frac P{sqrt2}psi(x,x)=0\
&|psi(x,y)|<infty.
end{align}right.tag1
$$
I'm mainly interested in the solution with lowest $E$. In the case of $P=0$ it's easy to see that
$$psi_0(x,y)=expleft(frac Q2 (x+y)right)tag2$$
with eigenvalue
$$E=-frac{Q^2}2tag3$$
solves the problem. But what about $P>0$? Can $(1)$ still be solved analytically (i.e. in terms of elementary or special functions)? If not, can the solution be given in the form of an integral or a series with explicitly specified terms?
pde closed-form boundary-value-problem eigenfunctions
asked Jan 7 at 10:18
RuslanRuslan
3,72721533
$endgroup$
add a comment |
$begingroup$
Consider the following Helmholtz problem in the infinite triangle $y>0,;x>y$ with parameters $Q<0$, $Pge0$, $P<|Q|$.
$$left{begin{align}
&psi^{(2,0)}(x,y)+psi^{(0,2)}(x,y)+Epsi(x,y)=0,\
&psi^{(0,1)}(x,0)-frac Q2psi(x,0)=0,\
&psi^{(1,0)}(x,x)-psi^{(0,1)}(x,x)-frac P{sqrt2}psi(x,x)=0\
&|psi(x,y)|<infty.
end{align}right.tag1
$$
I'm mainly interested in the solution with lowest $E$. In the case of $P=0$ it's easy to see that
$$psi_0(x,y)=expleft(frac Q2 (x+y)right)tag2$$
with eigenvalue
$$E=-frac{Q^2}2tag3$$
solves the problem. But what about $P>0$? Can $(1)$ still be solved analytically (i.e. in terms of elementary or special functions)? If not, can the solution be given in the form of an integral or a series with explicitly specified terms?
pde closed-form boundary-value-problem eigenfunctions
asked Jan 7 at 10:18
RuslanRuslan
3,72721533
$endgroup$
add a comment |
$begingroup$
Consider the following Helmholtz problem in the infinite triangle $y>0,;x>y$ with parameters $Q<0$, $Pge0$, $P<|Q|$.
$$left{begin{align}
&psi^{(2,0)}(x,y)+psi^{(0,2)}(x,y)+Epsi(x,y)=0,\
&psi^{(0,1)}(x,0)-frac Q2psi(x,0)=0,\
&psi^{(1,0)}(x,x)-psi^{(0,1)}(x,x)-frac P{sqrt2}psi(x,x)=0\
&|psi(x,y)|<infty.
end{align}right.tag1
$$
I'm mainly interested in the solution with lowest $E$. In the case of $P=0$ it's easy to see that
$$psi_0(x,y)=expleft(frac Q2 (x+y)right)tag2$$
with eigenvalue
$$E=-frac{Q^2}2tag3$$
solves the problem. But what about $P>0$? Can $(1)$ still be solved analytically (i.e. in terms of elementary or special functions)? If not, can the solution be given in the form of an integral or a series with explicitly specified terms?
pde closed-form boundary-value-problem eigenfunctions
asked Jan 7 at 10:18
RuslanRuslan
3,72721533
$endgroup$
Consider the following Helmholtz problem in the infinite triangle $y>0,;x>y$ with parameters $Q<0$, $Pge0$, $P<|Q|$.
$$left{begin{align}
&psi^{(2,0)}(x,y)+psi^{(0,2)}(x,y)+Epsi(x,y)=0,\
&psi^{(0,1)}(x,0)-frac Q2psi(x,0)=0,\
&psi^{(1,0)}(x,x)-psi^{(0,1)}(x,x)-frac P{sqrt2}psi(x,x)=0\
&|psi(x,y)|<infty.
end{align}right.tag1
$$
I'm mainly interested in the solution with lowest $E$. In the case of $P=0$ it's easy to see that
$$psi_0(x,y)=expleft(frac Q2 (x+y)right)tag2$$
with eigenvalue
$$E=-frac{Q^2}2tag3$$
solves the problem. But what about $P>0$? Can $(1)$ still be solved analytically (i.e. in terms of elementary or special functions)? If not, can the solution be given in the form of an integral or a series with explicitly specified terms?
pde closed-form boundary-value-problem eigenfunctions
pde closed-form boundary-value-problem eigenfunctions
asked Jan 7 at 10:18
RuslanRuslan
3,72721533
asked Jan 7 at 10:18
RuslanRuslan
3,72721533
asked Jan 7 at 10:18
RuslanRuslan
3,72721533
asked Jan 7 at 10:18
RuslanRuslan
3,72721533
asked Jan 7 at 10:18
RuslanRuslan
3,72721533
3,72721533
add a comment |
add a comment |
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