Mixing
PDE with some variables absent
$begingroup$
Suppose we have a homogeneous linear PDE of order $n$ with the form
$$frac{partial^nf(x,y)}{partial x^n}+p_{n-1}(x)frac{partial^{n-1}f(x,y)}{partial x^{n-1}}+cdots+p_0(x)f(x,y)=0$$
where the $p_i$'s are smooth. Must it be true that $f(x,y)=X_1(x)Y_1(y)+cdots+X_n(x)Y_n(y)$ for some functions $X_i$ and $Y_i$?
I can prove that this is true for $n=1$:
Let $q(x)$ be such that $q'(x)=p_0(x)$, then
$$e^{q(x)}f_x(x,y)+p_0(x)e^{q(x)}f(x,y)=0$$
$$frac{partial}{partial x}(e^{q(x)}f(x,y))=0$$
$$f(x,y)=Y(y)e^{-q(x)}$$
For $n=2$, assuming that $f(x,y_0)$ is non-zero everywhere, we can set $f(x,y)=f(x,y_0)g(x,y)$ and use reduction of order to get a similar conclusion. But I don't know if that assumption can be dropped.
This question arose from solving an equation in spherical coordinates, so the domain of the equation is $(0,infty)times[0,pi]times[0,2pi]$. (I believe the number of absent variables should not matter, so I used an equation with two variables only.) But more general answers are also welcomed.
ordinary-differential-equations pde linear-pde
asked Jan 12 at 7:55
Poon LeviPoon Levi
42137
$endgroup$
add a comment |
$begingroup$
Suppose we have a homogeneous linear PDE of order $n$ with the form
$$frac{partial^nf(x,y)}{partial x^n}+p_{n-1}(x)frac{partial^{n-1}f(x,y)}{partial x^{n-1}}+cdots+p_0(x)f(x,y)=0$$
where the $p_i$'s are smooth. Must it be true that $f(x,y)=X_1(x)Y_1(y)+cdots+X_n(x)Y_n(y)$ for some functions $X_i$ and $Y_i$?
I can prove that this is true for $n=1$:
Let $q(x)$ be such that $q'(x)=p_0(x)$, then
$$e^{q(x)}f_x(x,y)+p_0(x)e^{q(x)}f(x,y)=0$$
$$frac{partial}{partial x}(e^{q(x)}f(x,y))=0$$
$$f(x,y)=Y(y)e^{-q(x)}$$
For $n=2$, assuming that $f(x,y_0)$ is non-zero everywhere, we can set $f(x,y)=f(x,y_0)g(x,y)$ and use reduction of order to get a similar conclusion. But I don't know if that assumption can be dropped.
This question arose from solving an equation in spherical coordinates, so the domain of the equation is $(0,infty)times[0,pi]times[0,2pi]$. (I believe the number of absent variables should not matter, so I used an equation with two variables only.) But more general answers are also welcomed.
ordinary-differential-equations pde linear-pde
asked Jan 12 at 7:55
Poon LeviPoon Levi
42137
$endgroup$
add a comment |
$begingroup$
Suppose we have a homogeneous linear PDE of order $n$ with the form
$$frac{partial^nf(x,y)}{partial x^n}+p_{n-1}(x)frac{partial^{n-1}f(x,y)}{partial x^{n-1}}+cdots+p_0(x)f(x,y)=0$$
where the $p_i$'s are smooth. Must it be true that $f(x,y)=X_1(x)Y_1(y)+cdots+X_n(x)Y_n(y)$ for some functions $X_i$ and $Y_i$?
I can prove that this is true for $n=1$:
Let $q(x)$ be such that $q'(x)=p_0(x)$, then
$$e^{q(x)}f_x(x,y)+p_0(x)e^{q(x)}f(x,y)=0$$
$$frac{partial}{partial x}(e^{q(x)}f(x,y))=0$$
$$f(x,y)=Y(y)e^{-q(x)}$$
For $n=2$, assuming that $f(x,y_0)$ is non-zero everywhere, we can set $f(x,y)=f(x,y_0)g(x,y)$ and use reduction of order to get a similar conclusion. But I don't know if that assumption can be dropped.
This question arose from solving an equation in spherical coordinates, so the domain of the equation is $(0,infty)times[0,pi]times[0,2pi]$. (I believe the number of absent variables should not matter, so I used an equation with two variables only.) But more general answers are also welcomed.
ordinary-differential-equations pde linear-pde
asked Jan 12 at 7:55
Poon LeviPoon Levi
42137
$endgroup$
Suppose we have a homogeneous linear PDE of order $n$ with the form
$$frac{partial^nf(x,y)}{partial x^n}+p_{n-1}(x)frac{partial^{n-1}f(x,y)}{partial x^{n-1}}+cdots+p_0(x)f(x,y)=0$$
where the $p_i$'s are smooth. Must it be true that $f(x,y)=X_1(x)Y_1(y)+cdots+X_n(x)Y_n(y)$ for some functions $X_i$ and $Y_i$?
I can prove that this is true for $n=1$:
Let $q(x)$ be such that $q'(x)=p_0(x)$, then
$$e^{q(x)}f_x(x,y)+p_0(x)e^{q(x)}f(x,y)=0$$
$$frac{partial}{partial x}(e^{q(x)}f(x,y))=0$$
$$f(x,y)=Y(y)e^{-q(x)}$$
For $n=2$, assuming that $f(x,y_0)$ is non-zero everywhere, we can set $f(x,y)=f(x,y_0)g(x,y)$ and use reduction of order to get a similar conclusion. But I don't know if that assumption can be dropped.
This question arose from solving an equation in spherical coordinates, so the domain of the equation is $(0,infty)times[0,pi]times[0,2pi]$. (I believe the number of absent variables should not matter, so I used an equation with two variables only.) But more general answers are also welcomed.
ordinary-differential-equations pde linear-pde
ordinary-differential-equations pde linear-pde
asked Jan 12 at 7:55
Poon LeviPoon Levi
42137
asked Jan 12 at 7:55
Poon LeviPoon Levi
42137
asked Jan 12 at 7:55
Poon LeviPoon Levi
42137
asked Jan 12 at 7:55
Poon LeviPoon Levi
42137
asked Jan 12 at 7:55
Poon LeviPoon Levi
42137
42137
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let ${X_1,X_2,dots,X_n}$ be a fundamental set of solutions for the $n$-th order linear ordinary differential equation
begin{equation}
X^{(n)} + p_{n-1} X^{(n-1)} + cdots + p_0 X = 0.
end{equation}
Then the function
begin{equation}
f(x,y) := left( C_1 X_1(x) + C_2 X_2(x) + cdots + C_n X_n(x) right) Y(y)
end{equation}
is a solution of your linear partial differential equation, for any numbers $C_1, C_2, dots, C_n in mathbb{R}$ and for any function $Y$.
Define functions $Y_i(y) := C_i Y(y)$, $i =1,2,dots,n$, to obtain the desired form
begin{equation}
f = X_1 Y_1 + X_2 Y_2 + cdots + X_n Y_n.
end{equation}
answered Jan 13 at 0:22
ChristophChristoph
58116
$endgroup$
$begingroup$
I understand that these are solutions, but can we prove that these are the only solutions?
$endgroup$
– Poon Levi
Jan 13 at 3:29
$begingroup$
I think I got it now: if $f$ is a solution, then for all $y_0$, $g(x)=f(x,y_0)$ will be a solution to the corresponding linear ODE. So, it can be written uniquely as $g=a_1X_1+cdots+a_nX_n$ for $a_1,dots,a_ninmathbb{R}$. These $a_i$'s are constant with respect to $x$ but will in general depend on $y_0$. Renaming the variable, we have $f(x,y)=a_1(y)X_1(x)+cdots+a_n(y)X_n(x)$. Does this look fine?
$endgroup$
– Poon Levi
Jan 13 at 5:30
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let ${X_1,X_2,dots,X_n}$ be a fundamental set of solutions for the $n$-th order linear ordinary differential equation
begin{equation}
X^{(n)} + p_{n-1} X^{(n-1)} + cdots + p_0 X = 0.
end{equation}
Then the function
begin{equation}
f(x,y) := left( C_1 X_1(x) + C_2 X_2(x) + cdots + C_n X_n(x) right) Y(y)
end{equation}
is a solution of your linear partial differential equation, for any numbers $C_1, C_2, dots, C_n in mathbb{R}$ and for any function $Y$.
Define functions $Y_i(y) := C_i Y(y)$, $i =1,2,dots,n$, to obtain the desired form
begin{equation}
f = X_1 Y_1 + X_2 Y_2 + cdots + X_n Y_n.
end{equation}
answered Jan 13 at 0:22
ChristophChristoph
58116
$endgroup$
$begingroup$
I understand that these are solutions, but can we prove that these are the only solutions?
$endgroup$
– Poon Levi
Jan 13 at 3:29
$begingroup$
I think I got it now: if $f$ is a solution, then for all $y_0$, $g(x)=f(x,y_0)$ will be a solution to the corresponding linear ODE. So, it can be written uniquely as $g=a_1X_1+cdots+a_nX_n$ for $a_1,dots,a_ninmathbb{R}$. These $a_i$'s are constant with respect to $x$ but will in general depend on $y_0$. Renaming the variable, we have $f(x,y)=a_1(y)X_1(x)+cdots+a_n(y)X_n(x)$. Does this look fine?
$endgroup$
– Poon Levi
Jan 13 at 5:30
add a comment |
$begingroup$
Let ${X_1,X_2,dots,X_n}$ be a fundamental set of solutions for the $n$-th order linear ordinary differential equation
begin{equation}
X^{(n)} + p_{n-1} X^{(n-1)} + cdots + p_0 X = 0.
end{equation}
Then the function
begin{equation}
f(x,y) := left( C_1 X_1(x) + C_2 X_2(x) + cdots + C_n X_n(x) right) Y(y)
end{equation}
is a solution of your linear partial differential equation, for any numbers $C_1, C_2, dots, C_n in mathbb{R}$ and for any function $Y$.
Define functions $Y_i(y) := C_i Y(y)$, $i =1,2,dots,n$, to obtain the desired form
begin{equation}
f = X_1 Y_1 + X_2 Y_2 + cdots + X_n Y_n.
end{equation}
answered Jan 13 at 0:22
ChristophChristoph
58116
$endgroup$
$begingroup$
I understand that these are solutions, but can we prove that these are the only solutions?
$endgroup$
– Poon Levi
Jan 13 at 3:29
$begingroup$
I think I got it now: if $f$ is a solution, then for all $y_0$, $g(x)=f(x,y_0)$ will be a solution to the corresponding linear ODE. So, it can be written uniquely as $g=a_1X_1+cdots+a_nX_n$ for $a_1,dots,a_ninmathbb{R}$. These $a_i$'s are constant with respect to $x$ but will in general depend on $y_0$. Renaming the variable, we have $f(x,y)=a_1(y)X_1(x)+cdots+a_n(y)X_n(x)$. Does this look fine?
$endgroup$
– Poon Levi
Jan 13 at 5:30
add a comment |
$begingroup$
Let ${X_1,X_2,dots,X_n}$ be a fundamental set of solutions for the $n$-th order linear ordinary differential equation
begin{equation}
X^{(n)} + p_{n-1} X^{(n-1)} + cdots + p_0 X = 0.
end{equation}
Then the function
begin{equation}
f(x,y) := left( C_1 X_1(x) + C_2 X_2(x) + cdots + C_n X_n(x) right) Y(y)
end{equation}
is a solution of your linear partial differential equation, for any numbers $C_1, C_2, dots, C_n in mathbb{R}$ and for any function $Y$.
Define functions $Y_i(y) := C_i Y(y)$, $i =1,2,dots,n$, to obtain the desired form
begin{equation}
f = X_1 Y_1 + X_2 Y_2 + cdots + X_n Y_n.
end{equation}
answered Jan 13 at 0:22
ChristophChristoph
58116
$endgroup$
Let ${X_1,X_2,dots,X_n}$ be a fundamental set of solutions for the $n$-th order linear ordinary differential equation
begin{equation}
X^{(n)} + p_{n-1} X^{(n-1)} + cdots + p_0 X = 0.
end{equation}
Then the function
begin{equation}
f(x,y) := left( C_1 X_1(x) + C_2 X_2(x) + cdots + C_n X_n(x) right) Y(y)
end{equation}
is a solution of your linear partial differential equation, for any numbers $C_1, C_2, dots, C_n in mathbb{R}$ and for any function $Y$.
Define functions $Y_i(y) := C_i Y(y)$, $i =1,2,dots,n$, to obtain the desired form
begin{equation}
f = X_1 Y_1 + X_2 Y_2 + cdots + X_n Y_n.
end{equation}
answered Jan 13 at 0:22
ChristophChristoph
58116
answered Jan 13 at 0:22
ChristophChristoph
58116
answered Jan 13 at 0:22
ChristophChristoph
58116
answered Jan 13 at 0:22
ChristophChristoph
58116
58116
$begingroup$
I understand that these are solutions, but can we prove that these are the only solutions?
$endgroup$
– Poon Levi
Jan 13 at 3:29
$begingroup$
I think I got it now: if $f$ is a solution, then for all $y_0$, $g(x)=f(x,y_0)$ will be a solution to the corresponding linear ODE. So, it can be written uniquely as $g=a_1X_1+cdots+a_nX_n$ for $a_1,dots,a_ninmathbb{R}$. These $a_i$'s are constant with respect to $x$ but will in general depend on $y_0$. Renaming the variable, we have $f(x,y)=a_1(y)X_1(x)+cdots+a_n(y)X_n(x)$. Does this look fine?
$endgroup$
– Poon Levi
Jan 13 at 5:30
add a comment |
$begingroup$
I understand that these are solutions, but can we prove that these are the only solutions?
$endgroup$
– Poon Levi
Jan 13 at 3:29
$begingroup$
I think I got it now: if $f$ is a solution, then for all $y_0$, $g(x)=f(x,y_0)$ will be a solution to the corresponding linear ODE. So, it can be written uniquely as $g=a_1X_1+cdots+a_nX_n$ for $a_1,dots,a_ninmathbb{R}$. These $a_i$'s are constant with respect to $x$ but will in general depend on $y_0$. Renaming the variable, we have $f(x,y)=a_1(y)X_1(x)+cdots+a_n(y)X_n(x)$. Does this look fine?
$endgroup$
– Poon Levi
Jan 13 at 5:30
$begingroup$
I understand that these are solutions, but can we prove that these are the only solutions?
$endgroup$
– Poon Levi
Jan 13 at 3:29
$begingroup$
I understand that these are solutions, but can we prove that these are the only solutions?
$endgroup$
– Poon Levi
Jan 13 at 3:29
$begingroup$
I think I got it now: if $f$ is a solution, then for all $y_0$, $g(x)=f(x,y_0)$ will be a solution to the corresponding linear ODE. So, it can be written uniquely as $g=a_1X_1+cdots+a_nX_n$ for $a_1,dots,a_ninmathbb{R}$. These $a_i$'s are constant with respect to $x$ but will in general depend on $y_0$. Renaming the variable, we have $f(x,y)=a_1(y)X_1(x)+cdots+a_n(y)X_n(x)$. Does this look fine?
$endgroup$
– Poon Levi
Jan 13 at 5:30
$begingroup$
I think I got it now: if $f$ is a solution, then for all $y_0$, $g(x)=f(x,y_0)$ will be a solution to the corresponding linear ODE. So, it can be written uniquely as $g=a_1X_1+cdots+a_nX_n$ for $a_1,dots,a_ninmathbb{R}$. These $a_i$'s are constant with respect to $x$ but will in general depend on $y_0$. Renaming the variable, we have $f(x,y)=a_1(y)X_1(x)+cdots+a_n(y)X_n(x)$. Does this look fine?
$endgroup$
– Poon Levi
Jan 13 at 5:30
add a comment |
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