Mixing
Green function for solving ODE: continuity and discontinuity of derivatives
$begingroup$
I am studying Green functions and I want to find the Green function $G(x,x_0)$ of the linear operator
begin{equation}
mathcal{L}=left(frac{d^n}{dx^n}+a_1(x)frac{d^{n-1}}{dx^{n-1}}+dots+a_n(x) right)
end{equation}
This means solving the following problem
begin{equation}
frac{d^n G}{dx^n}+a_1(x)frac{d^{n-1}G}{dx^{n-1}}+dots+a_n(x)G=delta(x-x_0)
tag{1}
end{equation}
I understand that I have to impose continuity in $x_0$ for all the derivatives, up to $n-2$, and impose the jump discontinuity in $x_0$ for the $(n-1)$-th derivative. I don't exactly get why this is the case. I have searched in some lecture notes and found the following clues:
First approach
They proceed by integrating equation $(1)$, and they say "we obtain$G^{(n-1)}=H(x-x_0)$ + some continuous functions. The $(n-1)$-th derivative is not continuous, but suffers a discontinuous jump there. Integrating again shows that $G^{(n−2)}$ is continuous".
My problem here is the following:
how can I say that the result of integrating equation $(1)$ gets me "some continuous functions" if I don't know anything about the continuity of $G(x, x_0)$ and its derivatives?
Second approach
I found it for a $2^{nd}$ order ODE but I assume it can be generalized. It proceeds by saying "Suppose first that $G(x,x_0)$ was discontinuous at $x=x_0$, with the discontinuity modelled by a step function. Then $G' propto delta(x-x_0)$ and consequently $G''propto delta(x−x_0)$. However, the form of the ODE shows that $mathcal{L}G$ involves no generalized functions beyond $delta(x-x_0)$, thus $G(x, ξ)$ must be continuous." He THEN proceeds in integrating the ODE in a small neighborhood of $x_0$, and he asserts that "Also, since $G$ is continuous, $G'$ must be bounded so the term $G'$ also cannot contribute as the integration region shrinks to zero size". He then concludes that it's only the second derivative that contributes to the magnitude 1 jump.
My problems here:
1) we started by modeling the discontinuity of $G$ by a step function, noticed that there was a contradiction and consequently stated that "so the function must be continuous". But isn't it a bit restrictive, to just assume a jump discontinuity?
2) Can this be generalized for a generic $n$-order equation?
My questions
To resume, I would like to know:
1) What is the correct way to proceed to demonstrate that $G$ and its derivatives up to the $n-2$-th are continuous in $x_0$ and its $(n-1)$-th derivative has a jump discontinuity
2) As a side question: if the $(n-1)$-th derivative of the function is discontinuous in $x_0$ how can $G^{n}$ exist in $x_0$?
I'm a physicist not a mathematician, so please be kind =)
ordinary-differential-equations greens-function
asked Jan 13 at 2:00
LuthienLuthien
1448
$endgroup$
add a comment |
$begingroup$
I am studying Green functions and I want to find the Green function $G(x,x_0)$ of the linear operator
begin{equation}
mathcal{L}=left(frac{d^n}{dx^n}+a_1(x)frac{d^{n-1}}{dx^{n-1}}+dots+a_n(x) right)
end{equation}
This means solving the following problem
begin{equation}
frac{d^n G}{dx^n}+a_1(x)frac{d^{n-1}G}{dx^{n-1}}+dots+a_n(x)G=delta(x-x_0)
tag{1}
end{equation}
I understand that I have to impose continuity in $x_0$ for all the derivatives, up to $n-2$, and impose the jump discontinuity in $x_0$ for the $(n-1)$-th derivative. I don't exactly get why this is the case. I have searched in some lecture notes and found the following clues:
First approach
They proceed by integrating equation $(1)$, and they say "we obtain$G^{(n-1)}=H(x-x_0)$ + some continuous functions. The $(n-1)$-th derivative is not continuous, but suffers a discontinuous jump there. Integrating again shows that $G^{(n−2)}$ is continuous".
My problem here is the following:
how can I say that the result of integrating equation $(1)$ gets me "some continuous functions" if I don't know anything about the continuity of $G(x, x_0)$ and its derivatives?
Second approach
I found it for a $2^{nd}$ order ODE but I assume it can be generalized. It proceeds by saying "Suppose first that $G(x,x_0)$ was discontinuous at $x=x_0$, with the discontinuity modelled by a step function. Then $G' propto delta(x-x_0)$ and consequently $G''propto delta(x−x_0)$. However, the form of the ODE shows that $mathcal{L}G$ involves no generalized functions beyond $delta(x-x_0)$, thus $G(x, ξ)$ must be continuous." He THEN proceeds in integrating the ODE in a small neighborhood of $x_0$, and he asserts that "Also, since $G$ is continuous, $G'$ must be bounded so the term $G'$ also cannot contribute as the integration region shrinks to zero size". He then concludes that it's only the second derivative that contributes to the magnitude 1 jump.
My problems here:
1) we started by modeling the discontinuity of $G$ by a step function, noticed that there was a contradiction and consequently stated that "so the function must be continuous". But isn't it a bit restrictive, to just assume a jump discontinuity?
2) Can this be generalized for a generic $n$-order equation?
My questions
To resume, I would like to know:
1) What is the correct way to proceed to demonstrate that $G$ and its derivatives up to the $n-2$-th are continuous in $x_0$ and its $(n-1)$-th derivative has a jump discontinuity
2) As a side question: if the $(n-1)$-th derivative of the function is discontinuous in $x_0$ how can $G^{n}$ exist in $x_0$?
I'm a physicist not a mathematician, so please be kind =)
ordinary-differential-equations greens-function
asked Jan 13 at 2:00
LuthienLuthien
1448
$endgroup$
add a comment |
$begingroup$
I am studying Green functions and I want to find the Green function $G(x,x_0)$ of the linear operator
begin{equation}
mathcal{L}=left(frac{d^n}{dx^n}+a_1(x)frac{d^{n-1}}{dx^{n-1}}+dots+a_n(x) right)
end{equation}
This means solving the following problem
begin{equation}
frac{d^n G}{dx^n}+a_1(x)frac{d^{n-1}G}{dx^{n-1}}+dots+a_n(x)G=delta(x-x_0)
tag{1}
end{equation}
I understand that I have to impose continuity in $x_0$ for all the derivatives, up to $n-2$, and impose the jump discontinuity in $x_0$ for the $(n-1)$-th derivative. I don't exactly get why this is the case. I have searched in some lecture notes and found the following clues:
First approach
They proceed by integrating equation $(1)$, and they say "we obtain$G^{(n-1)}=H(x-x_0)$ + some continuous functions. The $(n-1)$-th derivative is not continuous, but suffers a discontinuous jump there. Integrating again shows that $G^{(n−2)}$ is continuous".
My problem here is the following:
how can I say that the result of integrating equation $(1)$ gets me "some continuous functions" if I don't know anything about the continuity of $G(x, x_0)$ and its derivatives?
Second approach
I found it for a $2^{nd}$ order ODE but I assume it can be generalized. It proceeds by saying "Suppose first that $G(x,x_0)$ was discontinuous at $x=x_0$, with the discontinuity modelled by a step function. Then $G' propto delta(x-x_0)$ and consequently $G''propto delta(x−x_0)$. However, the form of the ODE shows that $mathcal{L}G$ involves no generalized functions beyond $delta(x-x_0)$, thus $G(x, ξ)$ must be continuous." He THEN proceeds in integrating the ODE in a small neighborhood of $x_0$, and he asserts that "Also, since $G$ is continuous, $G'$ must be bounded so the term $G'$ also cannot contribute as the integration region shrinks to zero size". He then concludes that it's only the second derivative that contributes to the magnitude 1 jump.
My problems here:
1) we started by modeling the discontinuity of $G$ by a step function, noticed that there was a contradiction and consequently stated that "so the function must be continuous". But isn't it a bit restrictive, to just assume a jump discontinuity?
2) Can this be generalized for a generic $n$-order equation?
My questions
To resume, I would like to know:
1) What is the correct way to proceed to demonstrate that $G$ and its derivatives up to the $n-2$-th are continuous in $x_0$ and its $(n-1)$-th derivative has a jump discontinuity
2) As a side question: if the $(n-1)$-th derivative of the function is discontinuous in $x_0$ how can $G^{n}$ exist in $x_0$?
I'm a physicist not a mathematician, so please be kind =)
ordinary-differential-equations greens-function
asked Jan 13 at 2:00
LuthienLuthien
1448
$endgroup$
I am studying Green functions and I want to find the Green function $G(x,x_0)$ of the linear operator
begin{equation}
mathcal{L}=left(frac{d^n}{dx^n}+a_1(x)frac{d^{n-1}}{dx^{n-1}}+dots+a_n(x) right)
end{equation}
This means solving the following problem
begin{equation}
frac{d^n G}{dx^n}+a_1(x)frac{d^{n-1}G}{dx^{n-1}}+dots+a_n(x)G=delta(x-x_0)
tag{1}
end{equation}
I understand that I have to impose continuity in $x_0$ for all the derivatives, up to $n-2$, and impose the jump discontinuity in $x_0$ for the $(n-1)$-th derivative. I don't exactly get why this is the case. I have searched in some lecture notes and found the following clues:
First approach
They proceed by integrating equation $(1)$, and they say "we obtain$G^{(n-1)}=H(x-x_0)$ + some continuous functions. The $(n-1)$-th derivative is not continuous, but suffers a discontinuous jump there. Integrating again shows that $G^{(n−2)}$ is continuous".
My problem here is the following:
how can I say that the result of integrating equation $(1)$ gets me "some continuous functions" if I don't know anything about the continuity of $G(x, x_0)$ and its derivatives?
Second approach
I found it for a $2^{nd}$ order ODE but I assume it can be generalized. It proceeds by saying "Suppose first that $G(x,x_0)$ was discontinuous at $x=x_0$, with the discontinuity modelled by a step function. Then $G' propto delta(x-x_0)$ and consequently $G''propto delta(x−x_0)$. However, the form of the ODE shows that $mathcal{L}G$ involves no generalized functions beyond $delta(x-x_0)$, thus $G(x, ξ)$ must be continuous." He THEN proceeds in integrating the ODE in a small neighborhood of $x_0$, and he asserts that "Also, since $G$ is continuous, $G'$ must be bounded so the term $G'$ also cannot contribute as the integration region shrinks to zero size". He then concludes that it's only the second derivative that contributes to the magnitude 1 jump.
My problems here:
1) we started by modeling the discontinuity of $G$ by a step function, noticed that there was a contradiction and consequently stated that "so the function must be continuous". But isn't it a bit restrictive, to just assume a jump discontinuity?
2) Can this be generalized for a generic $n$-order equation?
My questions
To resume, I would like to know:
1) What is the correct way to proceed to demonstrate that $G$ and its derivatives up to the $n-2$-th are continuous in $x_0$ and its $(n-1)$-th derivative has a jump discontinuity
2) As a side question: if the $(n-1)$-th derivative of the function is discontinuous in $x_0$ how can $G^{n}$ exist in $x_0$?
I'm a physicist not a mathematician, so please be kind =)
ordinary-differential-equations greens-function
ordinary-differential-equations greens-function
asked Jan 13 at 2:00
LuthienLuthien
1448
asked Jan 13 at 2:00
LuthienLuthien
1448
edited Jan 13 at 2:12
Luthien
asked Jan 13 at 2:00
LuthienLuthien
1448
asked Jan 13 at 2:00
LuthienLuthien
1448
asked Jan 13 at 2:00
LuthienLuthien
1448
1448
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
1) One can start other way round. Namely to present a Green's function with required properties. Let for simplicity sake the equation's coefficients be constant. Let $v$ be the solution of the Cauchy problem ${cal L}v=0$, $v(x_0)=v'(x_0)=ldots=v^{(n-2)}(x_0)=0$, $v^{(n-1)}(x_0)=1$ and put $G(x,x_0)=v(x)$ for $x>x_0$ and $G(x,x_0)=0$ for $x<x_0$. It's straightforward to check that $G(x,x_0)$ is a Green's function for ${cal L}$.
Btw in the PDE theory such functions are called also fundamental solutions and the term Green's function is usually reserved for fundamental solutions with some homogeneous boundary conditions (e.g. zero Dirichlet condition).
By construction function $G(x,x_0)$ has continuous derivatives up to order $n-2$ and the derivative of order $n-1$ has a jump discontinuity in $x_0$. Now if $tilde G(x,x_0)$ is another Green's function then their difference $G-tilde G$ satisfy homogeneous equation ${cal L}u=0$ and therefore is smooth. In particular it has a continuous derivative of order $n-1$. So derivative $tilde G^{(n-1)}(x,x_0)$ also has a jump discontinuity...
2) It doesn't exist, no problem here. The Dirac function $delta(x-x_0)$ also "does not exist" in $x_0$. The equation with the delta function in the RHS is understood in the sense of distributions.
ADDED:
I don't see the need to follow their argument to the point.
To see that the required parts are "some continuous functions" one can start other way round without assuming any smoothness of $G$ but keeping in mind the theorems about smoothness of ODE solutions (with regular enough coefficients). Fix $x_0$ and define function $v$ as $v(x)=(x-x_0)^{n-1}/(n-1)!$ for $xge x_0$ and $v(x)=0$ for $x< x_0$. Since $partial_x^n v(x)=delta(x-x_0)$ for function $u(x)=G(x,x_0)-v(x)$ it follows that
$$
cal Lu=-(L-partial_x^n)v=f(x),
$$
where the rhs $f$ is bounded in some segment containing $x_0$, say $I=[x_0-1,x_0+1]$. In the (linear) ODE theory there are theorems stating that smoothness of the solution's senior derivative $partial_x^nu$ has the same smoothness as the rhs. In particular if $fin L_infty(I)$ then $partial_x^nuin L_infty(I)$. And this means that $partial_x^{n-1}u$ is continuous (even Lipschitz continuous). Returning to the Green function one has
$$
partial_x^{n-1}G(x,x_0)=partial_x^{n-1}u(x)+partial_x^{n-1}v(x)=
partial_x^{n-1}u(x)+H(x-x_0)
$$
which is a sum of a continuous function and the step function.
answered Jan 13 at 3:12
AndrewAndrew
9,55211946
$endgroup$
$begingroup$
Thank you for your answer! Can I also ask you what do you think of the two approaches I mentioned in my question? Are they "wrong" in some part?
$endgroup$
– Luthien
Jan 15 at 15:36
1
$begingroup$
@Luthien I'e edited the answer rewriting the arguments from your points 1) and 2) in somewhat other form. But essentially it's all the same.
$endgroup$
– Andrew
Jan 16 at 12:59
$begingroup$
Thank you, great!
$endgroup$
– Luthien
Jan 16 at 13:40
add a comment |
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1 Answer
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$begingroup$
1) One can start other way round. Namely to present a Green's function with required properties. Let for simplicity sake the equation's coefficients be constant. Let $v$ be the solution of the Cauchy problem ${cal L}v=0$, $v(x_0)=v'(x_0)=ldots=v^{(n-2)}(x_0)=0$, $v^{(n-1)}(x_0)=1$ and put $G(x,x_0)=v(x)$ for $x>x_0$ and $G(x,x_0)=0$ for $x<x_0$. It's straightforward to check that $G(x,x_0)$ is a Green's function for ${cal L}$.
Btw in the PDE theory such functions are called also fundamental solutions and the term Green's function is usually reserved for fundamental solutions with some homogeneous boundary conditions (e.g. zero Dirichlet condition).
By construction function $G(x,x_0)$ has continuous derivatives up to order $n-2$ and the derivative of order $n-1$ has a jump discontinuity in $x_0$. Now if $tilde G(x,x_0)$ is another Green's function then their difference $G-tilde G$ satisfy homogeneous equation ${cal L}u=0$ and therefore is smooth. In particular it has a continuous derivative of order $n-1$. So derivative $tilde G^{(n-1)}(x,x_0)$ also has a jump discontinuity...
2) It doesn't exist, no problem here. The Dirac function $delta(x-x_0)$ also "does not exist" in $x_0$. The equation with the delta function in the RHS is understood in the sense of distributions.
ADDED:
I don't see the need to follow their argument to the point.
To see that the required parts are "some continuous functions" one can start other way round without assuming any smoothness of $G$ but keeping in mind the theorems about smoothness of ODE solutions (with regular enough coefficients). Fix $x_0$ and define function $v$ as $v(x)=(x-x_0)^{n-1}/(n-1)!$ for $xge x_0$ and $v(x)=0$ for $x< x_0$. Since $partial_x^n v(x)=delta(x-x_0)$ for function $u(x)=G(x,x_0)-v(x)$ it follows that
$$
cal Lu=-(L-partial_x^n)v=f(x),
$$
where the rhs $f$ is bounded in some segment containing $x_0$, say $I=[x_0-1,x_0+1]$. In the (linear) ODE theory there are theorems stating that smoothness of the solution's senior derivative $partial_x^nu$ has the same smoothness as the rhs. In particular if $fin L_infty(I)$ then $partial_x^nuin L_infty(I)$. And this means that $partial_x^{n-1}u$ is continuous (even Lipschitz continuous). Returning to the Green function one has
$$
partial_x^{n-1}G(x,x_0)=partial_x^{n-1}u(x)+partial_x^{n-1}v(x)=
partial_x^{n-1}u(x)+H(x-x_0)
$$
which is a sum of a continuous function and the step function.
answered Jan 13 at 3:12
AndrewAndrew
9,55211946
$endgroup$
$begingroup$
Thank you for your answer! Can I also ask you what do you think of the two approaches I mentioned in my question? Are they "wrong" in some part?
$endgroup$
– Luthien
Jan 15 at 15:36
1
$begingroup$
@Luthien I'e edited the answer rewriting the arguments from your points 1) and 2) in somewhat other form. But essentially it's all the same.
$endgroup$
– Andrew
Jan 16 at 12:59
$begingroup$
Thank you, great!
$endgroup$
– Luthien
Jan 16 at 13:40
add a comment |
$begingroup$
1) One can start other way round. Namely to present a Green's function with required properties. Let for simplicity sake the equation's coefficients be constant. Let $v$ be the solution of the Cauchy problem ${cal L}v=0$, $v(x_0)=v'(x_0)=ldots=v^{(n-2)}(x_0)=0$, $v^{(n-1)}(x_0)=1$ and put $G(x,x_0)=v(x)$ for $x>x_0$ and $G(x,x_0)=0$ for $x<x_0$. It's straightforward to check that $G(x,x_0)$ is a Green's function for ${cal L}$.
Btw in the PDE theory such functions are called also fundamental solutions and the term Green's function is usually reserved for fundamental solutions with some homogeneous boundary conditions (e.g. zero Dirichlet condition).
By construction function $G(x,x_0)$ has continuous derivatives up to order $n-2$ and the derivative of order $n-1$ has a jump discontinuity in $x_0$. Now if $tilde G(x,x_0)$ is another Green's function then their difference $G-tilde G$ satisfy homogeneous equation ${cal L}u=0$ and therefore is smooth. In particular it has a continuous derivative of order $n-1$. So derivative $tilde G^{(n-1)}(x,x_0)$ also has a jump discontinuity...
2) It doesn't exist, no problem here. The Dirac function $delta(x-x_0)$ also "does not exist" in $x_0$. The equation with the delta function in the RHS is understood in the sense of distributions.
ADDED:
I don't see the need to follow their argument to the point.
To see that the required parts are "some continuous functions" one can start other way round without assuming any smoothness of $G$ but keeping in mind the theorems about smoothness of ODE solutions (with regular enough coefficients). Fix $x_0$ and define function $v$ as $v(x)=(x-x_0)^{n-1}/(n-1)!$ for $xge x_0$ and $v(x)=0$ for $x< x_0$. Since $partial_x^n v(x)=delta(x-x_0)$ for function $u(x)=G(x,x_0)-v(x)$ it follows that
$$
cal Lu=-(L-partial_x^n)v=f(x),
$$
where the rhs $f$ is bounded in some segment containing $x_0$, say $I=[x_0-1,x_0+1]$. In the (linear) ODE theory there are theorems stating that smoothness of the solution's senior derivative $partial_x^nu$ has the same smoothness as the rhs. In particular if $fin L_infty(I)$ then $partial_x^nuin L_infty(I)$. And this means that $partial_x^{n-1}u$ is continuous (even Lipschitz continuous). Returning to the Green function one has
$$
partial_x^{n-1}G(x,x_0)=partial_x^{n-1}u(x)+partial_x^{n-1}v(x)=
partial_x^{n-1}u(x)+H(x-x_0)
$$
which is a sum of a continuous function and the step function.
answered Jan 13 at 3:12
AndrewAndrew
9,55211946
$endgroup$
$begingroup$
Thank you for your answer! Can I also ask you what do you think of the two approaches I mentioned in my question? Are they "wrong" in some part?
$endgroup$
– Luthien
Jan 15 at 15:36
1
$begingroup$
@Luthien I'e edited the answer rewriting the arguments from your points 1) and 2) in somewhat other form. But essentially it's all the same.
$endgroup$
– Andrew
Jan 16 at 12:59
$begingroup$
Thank you, great!
$endgroup$
– Luthien
Jan 16 at 13:40
add a comment |
$begingroup$
1) One can start other way round. Namely to present a Green's function with required properties. Let for simplicity sake the equation's coefficients be constant. Let $v$ be the solution of the Cauchy problem ${cal L}v=0$, $v(x_0)=v'(x_0)=ldots=v^{(n-2)}(x_0)=0$, $v^{(n-1)}(x_0)=1$ and put $G(x,x_0)=v(x)$ for $x>x_0$ and $G(x,x_0)=0$ for $x<x_0$. It's straightforward to check that $G(x,x_0)$ is a Green's function for ${cal L}$.
Btw in the PDE theory such functions are called also fundamental solutions and the term Green's function is usually reserved for fundamental solutions with some homogeneous boundary conditions (e.g. zero Dirichlet condition).
By construction function $G(x,x_0)$ has continuous derivatives up to order $n-2$ and the derivative of order $n-1$ has a jump discontinuity in $x_0$. Now if $tilde G(x,x_0)$ is another Green's function then their difference $G-tilde G$ satisfy homogeneous equation ${cal L}u=0$ and therefore is smooth. In particular it has a continuous derivative of order $n-1$. So derivative $tilde G^{(n-1)}(x,x_0)$ also has a jump discontinuity...
2) It doesn't exist, no problem here. The Dirac function $delta(x-x_0)$ also "does not exist" in $x_0$. The equation with the delta function in the RHS is understood in the sense of distributions.
ADDED:
I don't see the need to follow their argument to the point.
To see that the required parts are "some continuous functions" one can start other way round without assuming any smoothness of $G$ but keeping in mind the theorems about smoothness of ODE solutions (with regular enough coefficients). Fix $x_0$ and define function $v$ as $v(x)=(x-x_0)^{n-1}/(n-1)!$ for $xge x_0$ and $v(x)=0$ for $x< x_0$. Since $partial_x^n v(x)=delta(x-x_0)$ for function $u(x)=G(x,x_0)-v(x)$ it follows that
$$
cal Lu=-(L-partial_x^n)v=f(x),
$$
where the rhs $f$ is bounded in some segment containing $x_0$, say $I=[x_0-1,x_0+1]$. In the (linear) ODE theory there are theorems stating that smoothness of the solution's senior derivative $partial_x^nu$ has the same smoothness as the rhs. In particular if $fin L_infty(I)$ then $partial_x^nuin L_infty(I)$. And this means that $partial_x^{n-1}u$ is continuous (even Lipschitz continuous). Returning to the Green function one has
$$
partial_x^{n-1}G(x,x_0)=partial_x^{n-1}u(x)+partial_x^{n-1}v(x)=
partial_x^{n-1}u(x)+H(x-x_0)
$$
which is a sum of a continuous function and the step function.
answered Jan 13 at 3:12
AndrewAndrew
9,55211946
$endgroup$
1) One can start other way round. Namely to present a Green's function with required properties. Let for simplicity sake the equation's coefficients be constant. Let $v$ be the solution of the Cauchy problem ${cal L}v=0$, $v(x_0)=v'(x_0)=ldots=v^{(n-2)}(x_0)=0$, $v^{(n-1)}(x_0)=1$ and put $G(x,x_0)=v(x)$ for $x>x_0$ and $G(x,x_0)=0$ for $x<x_0$. It's straightforward to check that $G(x,x_0)$ is a Green's function for ${cal L}$.
Btw in the PDE theory such functions are called also fundamental solutions and the term Green's function is usually reserved for fundamental solutions with some homogeneous boundary conditions (e.g. zero Dirichlet condition).
By construction function $G(x,x_0)$ has continuous derivatives up to order $n-2$ and the derivative of order $n-1$ has a jump discontinuity in $x_0$. Now if $tilde G(x,x_0)$ is another Green's function then their difference $G-tilde G$ satisfy homogeneous equation ${cal L}u=0$ and therefore is smooth. In particular it has a continuous derivative of order $n-1$. So derivative $tilde G^{(n-1)}(x,x_0)$ also has a jump discontinuity...
2) It doesn't exist, no problem here. The Dirac function $delta(x-x_0)$ also "does not exist" in $x_0$. The equation with the delta function in the RHS is understood in the sense of distributions.
ADDED:
I don't see the need to follow their argument to the point.
To see that the required parts are "some continuous functions" one can start other way round without assuming any smoothness of $G$ but keeping in mind the theorems about smoothness of ODE solutions (with regular enough coefficients). Fix $x_0$ and define function $v$ as $v(x)=(x-x_0)^{n-1}/(n-1)!$ for $xge x_0$ and $v(x)=0$ for $x< x_0$. Since $partial_x^n v(x)=delta(x-x_0)$ for function $u(x)=G(x,x_0)-v(x)$ it follows that
$$
cal Lu=-(L-partial_x^n)v=f(x),
$$
where the rhs $f$ is bounded in some segment containing $x_0$, say $I=[x_0-1,x_0+1]$. In the (linear) ODE theory there are theorems stating that smoothness of the solution's senior derivative $partial_x^nu$ has the same smoothness as the rhs. In particular if $fin L_infty(I)$ then $partial_x^nuin L_infty(I)$. And this means that $partial_x^{n-1}u$ is continuous (even Lipschitz continuous). Returning to the Green function one has
$$
partial_x^{n-1}G(x,x_0)=partial_x^{n-1}u(x)+partial_x^{n-1}v(x)=
partial_x^{n-1}u(x)+H(x-x_0)
$$
which is a sum of a continuous function and the step function.
answered Jan 13 at 3:12
AndrewAndrew
9,55211946
edited Jan 16 at 12:57
answered Jan 13 at 3:12
AndrewAndrew
9,55211946
answered Jan 13 at 3:12
AndrewAndrew
9,55211946
answered Jan 13 at 3:12
AndrewAndrew
9,55211946
9,55211946
$begingroup$
Thank you for your answer! Can I also ask you what do you think of the two approaches I mentioned in my question? Are they "wrong" in some part?
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– Luthien
Jan 15 at 15:36
1
$begingroup$
@Luthien I'e edited the answer rewriting the arguments from your points 1) and 2) in somewhat other form. But essentially it's all the same.
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– Andrew
Jan 16 at 12:59
$begingroup$
Thank you, great!
$endgroup$
– Luthien
Jan 16 at 13:40
add a comment |
$begingroup$
Thank you for your answer! Can I also ask you what do you think of the two approaches I mentioned in my question? Are they "wrong" in some part?
$endgroup$
– Luthien
Jan 15 at 15:36
1
$begingroup$
@Luthien I'e edited the answer rewriting the arguments from your points 1) and 2) in somewhat other form. But essentially it's all the same.
$endgroup$
– Andrew
Jan 16 at 12:59
$begingroup$
Thank you, great!
$endgroup$
– Luthien
Jan 16 at 13:40
$begingroup$
Thank you for your answer! Can I also ask you what do you think of the two approaches I mentioned in my question? Are they "wrong" in some part?
$endgroup$
– Luthien
Jan 15 at 15:36
$begingroup$
Thank you for your answer! Can I also ask you what do you think of the two approaches I mentioned in my question? Are they "wrong" in some part?
$endgroup$
– Luthien
Jan 15 at 15:36
1
1
$begingroup$
@Luthien I'e edited the answer rewriting the arguments from your points 1) and 2) in somewhat other form. But essentially it's all the same.
$endgroup$
– Andrew
Jan 16 at 12:59
$begingroup$
@Luthien I'e edited the answer rewriting the arguments from your points 1) and 2) in somewhat other form. But essentially it's all the same.
$endgroup$
– Andrew
Jan 16 at 12:59
$begingroup$
Thank you, great!
$endgroup$
– Luthien
Jan 16 at 13:40
$begingroup$
Thank you, great!
$endgroup$
– Luthien
Jan 16 at 13:40
add a comment |
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