How to solve linear differential-difference equation?
$begingroup$
Given a linear differential-difference equation:
$$A_{n+2}+partial A_{n+1}+partial^2 A_n=0,$$
where $A$ is a function of $n$ and $x$, and $partial$ represents the derivative about $x$.
How to solve this equation? The general case is this form
$$A_{n+2}+P_1 A_{n+1}+P_2 A_n=0,$$
where $P_1,P_2$ are differential operator depending on function of $x$.
I have tried to set $A_n=B^nA_0$, where $B$ is a pseudo differential operator and $A_0$ is a function of $x$, then I obtain
$$B^2+P_1B+P_2=0.$$
After solving $B$, we have the solution of $A_n$. I don't know whether this is correct and how to do next.
ordinary-differential-equations recurrence-relations differential-operators
asked Dec 22 '18 at 2:20
W. muW. mu
755310
$endgroup$
add a comment |
$begingroup$
Given a linear differential-difference equation:
$$A_{n+2}+partial A_{n+1}+partial^2 A_n=0,$$
where $A$ is a function of $n$ and $x$, and $partial$ represents the derivative about $x$.
How to solve this equation? The general case is this form
$$A_{n+2}+P_1 A_{n+1}+P_2 A_n=0,$$
where $P_1,P_2$ are differential operator depending on function of $x$.
I have tried to set $A_n=B^nA_0$, where $B$ is a pseudo differential operator and $A_0$ is a function of $x$, then I obtain
$$B^2+P_1B+P_2=0.$$
After solving $B$, we have the solution of $A_n$. I don't know whether this is correct and how to do next.
ordinary-differential-equations recurrence-relations differential-operators
asked Dec 22 '18 at 2:20
W. muW. mu
755310
$endgroup$
1
$begingroup$
Try $A_n(x) = be^{rx}c^n$. You can get a relationship between the constants $c$ and $r$.
$endgroup$
– Michael
Dec 22 '18 at 3:03
add a comment |
$begingroup$
Given a linear differential-difference equation:
$$A_{n+2}+partial A_{n+1}+partial^2 A_n=0,$$
where $A$ is a function of $n$ and $x$, and $partial$ represents the derivative about $x$.
How to solve this equation? The general case is this form
$$A_{n+2}+P_1 A_{n+1}+P_2 A_n=0,$$
where $P_1,P_2$ are differential operator depending on function of $x$.
I have tried to set $A_n=B^nA_0$, where $B$ is a pseudo differential operator and $A_0$ is a function of $x$, then I obtain
$$B^2+P_1B+P_2=0.$$
After solving $B$, we have the solution of $A_n$. I don't know whether this is correct and how to do next.
ordinary-differential-equations recurrence-relations differential-operators
asked Dec 22 '18 at 2:20
W. muW. mu
755310
$endgroup$
Given a linear differential-difference equation:
$$A_{n+2}+partial A_{n+1}+partial^2 A_n=0,$$
where $A$ is a function of $n$ and $x$, and $partial$ represents the derivative about $x$.
How to solve this equation? The general case is this form
$$A_{n+2}+P_1 A_{n+1}+P_2 A_n=0,$$
where $P_1,P_2$ are differential operator depending on function of $x$.
I have tried to set $A_n=B^nA_0$, where $B$ is a pseudo differential operator and $A_0$ is a function of $x$, then I obtain
$$B^2+P_1B+P_2=0.$$
After solving $B$, we have the solution of $A_n$. I don't know whether this is correct and how to do next.
ordinary-differential-equations recurrence-relations differential-operators
ordinary-differential-equations recurrence-relations differential-operators
asked Dec 22 '18 at 2:20
W. muW. mu
755310
asked Dec 22 '18 at 2:20
W. muW. mu
755310
asked Dec 22 '18 at 2:20
W. muW. mu
755310
asked Dec 22 '18 at 2:20
W. muW. mu
755310
asked Dec 22 '18 at 2:20
W. muW. mu
755310
755310
1
$begingroup$
Try $A_n(x) = be^{rx}c^n$. You can get a relationship between the constants $c$ and $r$.
$endgroup$
– Michael
Dec 22 '18 at 3:03
add a comment |
1
$begingroup$
Try $A_n(x) = be^{rx}c^n$. You can get a relationship between the constants $c$ and $r$.
$endgroup$
– Michael
Dec 22 '18 at 3:03
1
1
$begingroup$
Try $A_n(x) = be^{rx}c^n$. You can get a relationship between the constants $c$ and $r$.
$endgroup$
– Michael
Dec 22 '18 at 3:03
$begingroup$
Try $A_n(x) = be^{rx}c^n$. You can get a relationship between the constants $c$ and $r$.
$endgroup$
– Michael
Dec 22 '18 at 3:03
add a comment |
1 Answer
1
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$begingroup$
Given any infinitely differentiable functions $A_0(x), A_1(x)$, we can proceed to find $A_n(x)$ for all $n in {0, 1, 2, ...}$.
Given $A_0(x)=f(x), A_1(x)=g(x)$, where $f, g$ are given infinitely differentiable functions, a general solution is
$$ A_n(x) = c_nf^{(n)}(x) + d_ng^{(n-1)}(x) quad forall n in {0, 1, 2, ...} $$
where $f^{(n)}(x)$, $g^{(n)}(x)$ represent the $n$th derivative (and we use the convention $f^{(0)}= f$, $g^{(0)}=g$, $g^{(-1)}= 0$), and $c_n, d_n$ are solutions to the linear recurrence relation
begin{align*}
c_{n+2} &=-c_{n+1} - c_n quad forall n in {0, 1, 2, ...} \
d_{n+2} &=-d_{n+1} - d_n quad forall n in {0, 1, 2, ...}
end{align*}
with initial conditions $(c_0,c_1)=(1,0)$, $(d_0,d_1) = (0,1)$.
answered Dec 22 '18 at 4:56
MichaelMichael
13.2k11429
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add a comment |
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$begingroup$
Given any infinitely differentiable functions $A_0(x), A_1(x)$, we can proceed to find $A_n(x)$ for all $n in {0, 1, 2, ...}$.
Given $A_0(x)=f(x), A_1(x)=g(x)$, where $f, g$ are given infinitely differentiable functions, a general solution is
$$ A_n(x) = c_nf^{(n)}(x) + d_ng^{(n-1)}(x) quad forall n in {0, 1, 2, ...} $$
where $f^{(n)}(x)$, $g^{(n)}(x)$ represent the $n$th derivative (and we use the convention $f^{(0)}= f$, $g^{(0)}=g$, $g^{(-1)}= 0$), and $c_n, d_n$ are solutions to the linear recurrence relation
begin{align*}
c_{n+2} &=-c_{n+1} - c_n quad forall n in {0, 1, 2, ...} \
d_{n+2} &=-d_{n+1} - d_n quad forall n in {0, 1, 2, ...}
end{align*}
with initial conditions $(c_0,c_1)=(1,0)$, $(d_0,d_1) = (0,1)$.
answered Dec 22 '18 at 4:56
MichaelMichael
13.2k11429
$endgroup$
add a comment |
$begingroup$
Given any infinitely differentiable functions $A_0(x), A_1(x)$, we can proceed to find $A_n(x)$ for all $n in {0, 1, 2, ...}$.
Given $A_0(x)=f(x), A_1(x)=g(x)$, where $f, g$ are given infinitely differentiable functions, a general solution is
$$ A_n(x) = c_nf^{(n)}(x) + d_ng^{(n-1)}(x) quad forall n in {0, 1, 2, ...} $$
where $f^{(n)}(x)$, $g^{(n)}(x)$ represent the $n$th derivative (and we use the convention $f^{(0)}= f$, $g^{(0)}=g$, $g^{(-1)}= 0$), and $c_n, d_n$ are solutions to the linear recurrence relation
begin{align*}
c_{n+2} &=-c_{n+1} - c_n quad forall n in {0, 1, 2, ...} \
d_{n+2} &=-d_{n+1} - d_n quad forall n in {0, 1, 2, ...}
end{align*}
with initial conditions $(c_0,c_1)=(1,0)$, $(d_0,d_1) = (0,1)$.
answered Dec 22 '18 at 4:56
MichaelMichael
13.2k11429
$endgroup$
add a comment |
$begingroup$
Given any infinitely differentiable functions $A_0(x), A_1(x)$, we can proceed to find $A_n(x)$ for all $n in {0, 1, 2, ...}$.
Given $A_0(x)=f(x), A_1(x)=g(x)$, where $f, g$ are given infinitely differentiable functions, a general solution is
$$ A_n(x) = c_nf^{(n)}(x) + d_ng^{(n-1)}(x) quad forall n in {0, 1, 2, ...} $$
where $f^{(n)}(x)$, $g^{(n)}(x)$ represent the $n$th derivative (and we use the convention $f^{(0)}= f$, $g^{(0)}=g$, $g^{(-1)}= 0$), and $c_n, d_n$ are solutions to the linear recurrence relation
begin{align*}
c_{n+2} &=-c_{n+1} - c_n quad forall n in {0, 1, 2, ...} \
d_{n+2} &=-d_{n+1} - d_n quad forall n in {0, 1, 2, ...}
end{align*}
with initial conditions $(c_0,c_1)=(1,0)$, $(d_0,d_1) = (0,1)$.
answered Dec 22 '18 at 4:56
MichaelMichael
13.2k11429
$endgroup$
Given any infinitely differentiable functions $A_0(x), A_1(x)$, we can proceed to find $A_n(x)$ for all $n in {0, 1, 2, ...}$.
Given $A_0(x)=f(x), A_1(x)=g(x)$, where $f, g$ are given infinitely differentiable functions, a general solution is
$$ A_n(x) = c_nf^{(n)}(x) + d_ng^{(n-1)}(x) quad forall n in {0, 1, 2, ...} $$
where $f^{(n)}(x)$, $g^{(n)}(x)$ represent the $n$th derivative (and we use the convention $f^{(0)}= f$, $g^{(0)}=g$, $g^{(-1)}= 0$), and $c_n, d_n$ are solutions to the linear recurrence relation
begin{align*}
c_{n+2} &=-c_{n+1} - c_n quad forall n in {0, 1, 2, ...} \
d_{n+2} &=-d_{n+1} - d_n quad forall n in {0, 1, 2, ...}
end{align*}
with initial conditions $(c_0,c_1)=(1,0)$, $(d_0,d_1) = (0,1)$.
answered Dec 22 '18 at 4:56
MichaelMichael
13.2k11429
answered Dec 22 '18 at 4:56
MichaelMichael
13.2k11429
answered Dec 22 '18 at 4:56
MichaelMichael
13.2k11429
answered Dec 22 '18 at 4:56
MichaelMichael
13.2k11429
13.2k11429
add a comment |
add a comment |
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$begingroup$
Try $A_n(x) = be^{rx}c^n$. You can get a relationship between the constants $c$ and $r$.
$endgroup$
– Michael
Dec 22 '18 at 3:03