Reference request: Long-term behaviour of the heat equation for bounded initial dataReference request: Optimal $L^p$-decay for nonhomogenous heat equation in $mathbb R^d$Bounded input Bounded output stability for heat equationMixed norm estimate for the heat equation$L^infty$ estimate on heat equation with a lower order termHeat equation with source term in $L^1$Rate convergence of the heat equation as diffusion tends to zeroBounded solution for parabolic equationLong time existence for heat flow in Corlette-Donaldson Theorem
Reference request: Long-term behaviour of the heat equation for bounded initial data
Reference request: Optimal $L^p$-decay for nonhomogenous heat equation in $mathbb R^d$Bounded input Bounded output stability for heat equationMixed norm estimate for the heat equation$L^infty$ estimate on heat equation with a lower order termHeat equation with source term in $L^1$Rate convergence of the heat equation as diffusion tends to zeroBounded solution for parabolic equationLong time existence for heat flow in Corlette-Donaldson Theorem
$begingroup$
Let us consider the heat equation
beginalign*
fracpartialpartial tu(t,x) & = Delta u(t,x), \
u(0,x) & = f(x)
endalign*
on the whole space $mathbbR^d$. If $f in L^p := L^p(mathbbR^d)$ for $p in [1,infty)$ or if $f in C_0(mathbbR^d)$, the long term behaviour of the solution $u(t) := u(t,cdot)$ is well-known:
If $f in L^p$ and $p in (1,infty)$, then $u(t)$ converges to $0$ with respect to the $L^p$-norm and also with respect to the $L^infty$-norm as $ tto infty$.
If $f in L^1$, then $|u(t)|_L^1 to |int_mathbbR^d f(x) , dx|$ as $t to infty$, and we do not have convergence of $u(t)$ with respect to the $L^1$-norm, in general. However, we still have $|u(t)|_infty to 0$.
If $f in C_0(mathbbR^d)$, then also $|u(t)|_infty to 0$ as $t to infty$.
Now, I am interested in the long-term behaviour of $u(t)$ if $f in C_b(mathbbR^d)$, i.e. if $f$ is a bounded continuous function. (Note: In this case, it is maybe less clear what we mean by a solution of the heat equation, but we can still define a "solution" $u(t)$ as the convolution of $f$ with the heat kernel).
Uniform convergence of $u(t)$ as $t to infty$ is probably too much to expect for many functions $f$, but I would be interested in locally uniform convergence:
Question. Is there a characterization of those $f in C_b(mathbbR^d)$ for which $u(t)$ converges uniformly on compact sets to a constant function as $t to infty$?
I would expect that this has been studied somewhere in the literature, so my question is mainly a reference request.
Disclaimer. I discussed this question with a few experts on the matter, but they did not know the solution nor did they know a reference.
reference-request differential-operators parabolic-pde heat-equation
$endgroup$
add a comment
|
$begingroup$
Let us consider the heat equation
beginalign*
fracpartialpartial tu(t,x) & = Delta u(t,x), \
u(0,x) & = f(x)
endalign*
on the whole space $mathbbR^d$. If $f in L^p := L^p(mathbbR^d)$ for $p in [1,infty)$ or if $f in C_0(mathbbR^d)$, the long term behaviour of the solution $u(t) := u(t,cdot)$ is well-known:
If $f in L^p$ and $p in (1,infty)$, then $u(t)$ converges to $0$ with respect to the $L^p$-norm and also with respect to the $L^infty$-norm as $ tto infty$.
If $f in L^1$, then $|u(t)|_L^1 to |int_mathbbR^d f(x) , dx|$ as $t to infty$, and we do not have convergence of $u(t)$ with respect to the $L^1$-norm, in general. However, we still have $|u(t)|_infty to 0$.
If $f in C_0(mathbbR^d)$, then also $|u(t)|_infty to 0$ as $t to infty$.
Now, I am interested in the long-term behaviour of $u(t)$ if $f in C_b(mathbbR^d)$, i.e. if $f$ is a bounded continuous function. (Note: In this case, it is maybe less clear what we mean by a solution of the heat equation, but we can still define a "solution" $u(t)$ as the convolution of $f$ with the heat kernel).
Uniform convergence of $u(t)$ as $t to infty$ is probably too much to expect for many functions $f$, but I would be interested in locally uniform convergence:
Question. Is there a characterization of those $f in C_b(mathbbR^d)$ for which $u(t)$ converges uniformly on compact sets to a constant function as $t to infty$?
I would expect that this has been studied somewhere in the literature, so my question is mainly a reference request.
Disclaimer. I discussed this question with a few experts on the matter, but they did not know the solution nor did they know a reference.
reference-request differential-operators parabolic-pde heat-equation
$endgroup$
1
$begingroup$
Your claim for the case $fin L^1$ is incorrect. The $L^1$-norm is not constant, unless $f$was signed. Only the integral $m$ of $u(cdot,t)$ is constant. In general, $tmapsto|u(t)|_1$ is non-increasing and converges towards $|m|$. the solution is $L^1$-asymptotic to $mK_t$ where $K_t$ is the heat kernel.
$endgroup$
– Denis Serre
10 hours ago
$begingroup$
Besides $|u(t)|_inftyle(2pi t)^-d/2|f|_1$ so that $u(cdot,t)$ converges uniformly to zero.
$endgroup$
– Denis Serre
10 hours ago
$begingroup$
I don't have a reference, but my inclination would be to look at $partial_t u(t)$, which you can compute from the convolution formula by differentiating under the integral sign, and try to show it goes to 0 sufficiently fast (uniformly on compact sets) to force convergence of $u$.
$endgroup$
– Nate Eldredge
8 hours ago
$begingroup$
@DenisSerre: You're of course right (I have a tendency of always thinking about the case $f ge 0$ tacitly...). Corrected. Concerning the uniform convergence to $0$: yes; moreover, uniform convergence to $0$ also holds for $f in L^p$, $p in (1,infty)$, since the operators of the heat semigroup map $L^p$ continuously into $C_b(mathbbR^d)$. I have also added this information to the post.
$endgroup$
– Jochen Glueck
7 hours ago
$begingroup$
@NateEldredge: Thank you for your comment. I'll have to think about it in more detail.
$endgroup$
– Jochen Glueck
3 hours ago
add a comment
|
$begingroup$
Let us consider the heat equation
beginalign*
fracpartialpartial tu(t,x) & = Delta u(t,x), \
u(0,x) & = f(x)
endalign*
on the whole space $mathbbR^d$. If $f in L^p := L^p(mathbbR^d)$ for $p in [1,infty)$ or if $f in C_0(mathbbR^d)$, the long term behaviour of the solution $u(t) := u(t,cdot)$ is well-known:
If $f in L^p$ and $p in (1,infty)$, then $u(t)$ converges to $0$ with respect to the $L^p$-norm and also with respect to the $L^infty$-norm as $ tto infty$.
If $f in L^1$, then $|u(t)|_L^1 to |int_mathbbR^d f(x) , dx|$ as $t to infty$, and we do not have convergence of $u(t)$ with respect to the $L^1$-norm, in general. However, we still have $|u(t)|_infty to 0$.
If $f in C_0(mathbbR^d)$, then also $|u(t)|_infty to 0$ as $t to infty$.
Now, I am interested in the long-term behaviour of $u(t)$ if $f in C_b(mathbbR^d)$, i.e. if $f$ is a bounded continuous function. (Note: In this case, it is maybe less clear what we mean by a solution of the heat equation, but we can still define a "solution" $u(t)$ as the convolution of $f$ with the heat kernel).
Uniform convergence of $u(t)$ as $t to infty$ is probably too much to expect for many functions $f$, but I would be interested in locally uniform convergence:
Question. Is there a characterization of those $f in C_b(mathbbR^d)$ for which $u(t)$ converges uniformly on compact sets to a constant function as $t to infty$?
I would expect that this has been studied somewhere in the literature, so my question is mainly a reference request.
Disclaimer. I discussed this question with a few experts on the matter, but they did not know the solution nor did they know a reference.
reference-request differential-operators parabolic-pde heat-equation
$endgroup$
Let us consider the heat equation
beginalign*
fracpartialpartial tu(t,x) & = Delta u(t,x), \
u(0,x) & = f(x)
endalign*
on the whole space $mathbbR^d$. If $f in L^p := L^p(mathbbR^d)$ for $p in [1,infty)$ or if $f in C_0(mathbbR^d)$, the long term behaviour of the solution $u(t) := u(t,cdot)$ is well-known:
If $f in L^p$ and $p in (1,infty)$, then $u(t)$ converges to $0$ with respect to the $L^p$-norm and also with respect to the $L^infty$-norm as $ tto infty$.
If $f in L^1$, then $|u(t)|_L^1 to |int_mathbbR^d f(x) , dx|$ as $t to infty$, and we do not have convergence of $u(t)$ with respect to the $L^1$-norm, in general. However, we still have $|u(t)|_infty to 0$.
If $f in C_0(mathbbR^d)$, then also $|u(t)|_infty to 0$ as $t to infty$.
Now, I am interested in the long-term behaviour of $u(t)$ if $f in C_b(mathbbR^d)$, i.e. if $f$ is a bounded continuous function. (Note: In this case, it is maybe less clear what we mean by a solution of the heat equation, but we can still define a "solution" $u(t)$ as the convolution of $f$ with the heat kernel).
Uniform convergence of $u(t)$ as $t to infty$ is probably too much to expect for many functions $f$, but I would be interested in locally uniform convergence:
Question. Is there a characterization of those $f in C_b(mathbbR^d)$ for which $u(t)$ converges uniformly on compact sets to a constant function as $t to infty$?
I would expect that this has been studied somewhere in the literature, so my question is mainly a reference request.
Disclaimer. I discussed this question with a few experts on the matter, but they did not know the solution nor did they know a reference.
reference-request differential-operators parabolic-pde heat-equation
reference-request differential-operators parabolic-pde heat-equation
edited 7 hours ago
Jochen Glueck
asked 10 hours ago
Jochen GlueckJochen Glueck
3,8871 gold badge15 silver badges31 bronze badges
3,8871 gold badge15 silver badges31 bronze badges
1
$begingroup$
Your claim for the case $fin L^1$ is incorrect. The $L^1$-norm is not constant, unless $f$was signed. Only the integral $m$ of $u(cdot,t)$ is constant. In general, $tmapsto|u(t)|_1$ is non-increasing and converges towards $|m|$. the solution is $L^1$-asymptotic to $mK_t$ where $K_t$ is the heat kernel.
$endgroup$
– Denis Serre
10 hours ago
$begingroup$
Besides $|u(t)|_inftyle(2pi t)^-d/2|f|_1$ so that $u(cdot,t)$ converges uniformly to zero.
$endgroup$
– Denis Serre
10 hours ago
$begingroup$
I don't have a reference, but my inclination would be to look at $partial_t u(t)$, which you can compute from the convolution formula by differentiating under the integral sign, and try to show it goes to 0 sufficiently fast (uniformly on compact sets) to force convergence of $u$.
$endgroup$
– Nate Eldredge
8 hours ago
$begingroup$
@DenisSerre: You're of course right (I have a tendency of always thinking about the case $f ge 0$ tacitly...). Corrected. Concerning the uniform convergence to $0$: yes; moreover, uniform convergence to $0$ also holds for $f in L^p$, $p in (1,infty)$, since the operators of the heat semigroup map $L^p$ continuously into $C_b(mathbbR^d)$. I have also added this information to the post.
$endgroup$
– Jochen Glueck
7 hours ago
$begingroup$
@NateEldredge: Thank you for your comment. I'll have to think about it in more detail.
$endgroup$
– Jochen Glueck
3 hours ago
add a comment
|
1
$begingroup$
Your claim for the case $fin L^1$ is incorrect. The $L^1$-norm is not constant, unless $f$was signed. Only the integral $m$ of $u(cdot,t)$ is constant. In general, $tmapsto|u(t)|_1$ is non-increasing and converges towards $|m|$. the solution is $L^1$-asymptotic to $mK_t$ where $K_t$ is the heat kernel.
$endgroup$
– Denis Serre
10 hours ago
$begingroup$
Besides $|u(t)|_inftyle(2pi t)^-d/2|f|_1$ so that $u(cdot,t)$ converges uniformly to zero.
$endgroup$
– Denis Serre
10 hours ago
$begingroup$
I don't have a reference, but my inclination would be to look at $partial_t u(t)$, which you can compute from the convolution formula by differentiating under the integral sign, and try to show it goes to 0 sufficiently fast (uniformly on compact sets) to force convergence of $u$.
$endgroup$
– Nate Eldredge
8 hours ago
$begingroup$
@DenisSerre: You're of course right (I have a tendency of always thinking about the case $f ge 0$ tacitly...). Corrected. Concerning the uniform convergence to $0$: yes; moreover, uniform convergence to $0$ also holds for $f in L^p$, $p in (1,infty)$, since the operators of the heat semigroup map $L^p$ continuously into $C_b(mathbbR^d)$. I have also added this information to the post.
$endgroup$
– Jochen Glueck
7 hours ago
$begingroup$
@NateEldredge: Thank you for your comment. I'll have to think about it in more detail.
$endgroup$
– Jochen Glueck
3 hours ago
1
1
$begingroup$
Your claim for the case $fin L^1$ is incorrect. The $L^1$-norm is not constant, unless $f$was signed. Only the integral $m$ of $u(cdot,t)$ is constant. In general, $tmapsto|u(t)|_1$ is non-increasing and converges towards $|m|$. the solution is $L^1$-asymptotic to $mK_t$ where $K_t$ is the heat kernel.
$endgroup$
– Denis Serre
10 hours ago
$begingroup$
Your claim for the case $fin L^1$ is incorrect. The $L^1$-norm is not constant, unless $f$was signed. Only the integral $m$ of $u(cdot,t)$ is constant. In general, $tmapsto|u(t)|_1$ is non-increasing and converges towards $|m|$. the solution is $L^1$-asymptotic to $mK_t$ where $K_t$ is the heat kernel.
$endgroup$
– Denis Serre
10 hours ago
$begingroup$
Besides $|u(t)|_inftyle(2pi t)^-d/2|f|_1$ so that $u(cdot,t)$ converges uniformly to zero.
$endgroup$
– Denis Serre
10 hours ago
$begingroup$
Besides $|u(t)|_inftyle(2pi t)^-d/2|f|_1$ so that $u(cdot,t)$ converges uniformly to zero.
$endgroup$
– Denis Serre
10 hours ago
$begingroup$
I don't have a reference, but my inclination would be to look at $partial_t u(t)$, which you can compute from the convolution formula by differentiating under the integral sign, and try to show it goes to 0 sufficiently fast (uniformly on compact sets) to force convergence of $u$.
$endgroup$
– Nate Eldredge
8 hours ago
$begingroup$
I don't have a reference, but my inclination would be to look at $partial_t u(t)$, which you can compute from the convolution formula by differentiating under the integral sign, and try to show it goes to 0 sufficiently fast (uniformly on compact sets) to force convergence of $u$.
$endgroup$
– Nate Eldredge
8 hours ago
$begingroup$
@DenisSerre: You're of course right (I have a tendency of always thinking about the case $f ge 0$ tacitly...). Corrected. Concerning the uniform convergence to $0$: yes; moreover, uniform convergence to $0$ also holds for $f in L^p$, $p in (1,infty)$, since the operators of the heat semigroup map $L^p$ continuously into $C_b(mathbbR^d)$. I have also added this information to the post.
$endgroup$
– Jochen Glueck
7 hours ago
$begingroup$
@DenisSerre: You're of course right (I have a tendency of always thinking about the case $f ge 0$ tacitly...). Corrected. Concerning the uniform convergence to $0$: yes; moreover, uniform convergence to $0$ also holds for $f in L^p$, $p in (1,infty)$, since the operators of the heat semigroup map $L^p$ continuously into $C_b(mathbbR^d)$. I have also added this information to the post.
$endgroup$
– Jochen Glueck
7 hours ago
$begingroup$
@NateEldredge: Thank you for your comment. I'll have to think about it in more detail.
$endgroup$
– Jochen Glueck
3 hours ago
$begingroup$
@NateEldredge: Thank you for your comment. I'll have to think about it in more detail.
$endgroup$
– Jochen Glueck
3 hours ago
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
First of all, with initial data in $C_b$, classical solution exist, so there is no need for quotation marks. It is easy to see that the convolution of the initial data $f(x)$ with the Gauss–Weierstrass kernel $p_t(x)$ defines the unique bounded classical solution $u(t, x)$.
It is easy to see that $u(t,x) - u(t,y)$ converges to zero as $t to infty$: the functions $p_t(cdot - x) - p_t(cdot - y)$ converge in $L^1$ to zero as $t to infty$. It is only slightly more difficult to see that this convergence is locally uniform with respect to $x$ and $y$. Therefore, your question is equivalent to the following simpler one: for what initial data $f(x)$, the limit of $u(t, 0)$ exists as $t to infty$.
I doubt there is a simpler "if and only if" characterization. One can easily provide sufficient conditions, for example it is sufficient to assume that the mean value of $f$ over the ball $B(0, R)$ has a limit as $R to infty$ (by the same argument that one uses when proving that Cesàro convergence implies Abel convergence). And one can equally easily construct counterexamples.
$endgroup$
$begingroup$
+1 The observation that the solution converges uniformly on compact subsets if and only if it converges at the point $0$ is really interesting. Of course, I would still hope for a characterization of those $f$, maybe as members of some kind of nice function space. (But I agree with you that this hope is somewhat bold.) Concerning the quotation marks around "solution": I only used them because $u(t)$ does not converge to $f$ with respect to the $|cdot|_infty$-norm as $t downarrow 0$, so this notion of solution is slightly weaker that what one has, for instance, for $C_0$-semigroups.
$endgroup$
– Jochen Glueck
3 hours ago
add a comment
|
$begingroup$
I doubt that there is a characterization, but one thing I can say is that the complement of the set of such $f$ contains a dense open set.
Let $U(f)$ be the solution with initial condition $u(0,x) = f(x)$, and
$$G = f in C_b: lim_t to infty U(f)(t,cdot) textconverges uniformly on compact sets$$
Take some $f_0 in C_b$ such that $U(f_0)(t,0)$ does not converge as $t to infty$. Let $delta = limsup_t to infty U(f_0)(t,0) - liminf_t to infty U(f_0)(t,0) > 0$. Then for any $f in G$ and any $c ne 0$, if $|g- (f + c f_0)|_infty < |c|delta/2$
then $$limsup_t infty U(g)(t,0) - liminf_t to infty U(g)(t,0) ge |c|delta - 2 |g-(f+c f_0)|_infty > 0 $$
$endgroup$
$begingroup$
+1 Thank you for your answer! Alternatively, one can also argue that $G$ is a proper subspace of $C_b$ that is closed with respect to $|cdot|_infty$, so its complement is an open and dense set.
$endgroup$
– Jochen Glueck
7 hours ago
add a comment
|
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2 Answers
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2 Answers
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votes
$begingroup$
First of all, with initial data in $C_b$, classical solution exist, so there is no need for quotation marks. It is easy to see that the convolution of the initial data $f(x)$ with the Gauss–Weierstrass kernel $p_t(x)$ defines the unique bounded classical solution $u(t, x)$.
It is easy to see that $u(t,x) - u(t,y)$ converges to zero as $t to infty$: the functions $p_t(cdot - x) - p_t(cdot - y)$ converge in $L^1$ to zero as $t to infty$. It is only slightly more difficult to see that this convergence is locally uniform with respect to $x$ and $y$. Therefore, your question is equivalent to the following simpler one: for what initial data $f(x)$, the limit of $u(t, 0)$ exists as $t to infty$.
I doubt there is a simpler "if and only if" characterization. One can easily provide sufficient conditions, for example it is sufficient to assume that the mean value of $f$ over the ball $B(0, R)$ has a limit as $R to infty$ (by the same argument that one uses when proving that Cesàro convergence implies Abel convergence). And one can equally easily construct counterexamples.
$endgroup$
$begingroup$
+1 The observation that the solution converges uniformly on compact subsets if and only if it converges at the point $0$ is really interesting. Of course, I would still hope for a characterization of those $f$, maybe as members of some kind of nice function space. (But I agree with you that this hope is somewhat bold.) Concerning the quotation marks around "solution": I only used them because $u(t)$ does not converge to $f$ with respect to the $|cdot|_infty$-norm as $t downarrow 0$, so this notion of solution is slightly weaker that what one has, for instance, for $C_0$-semigroups.
$endgroup$
– Jochen Glueck
3 hours ago
add a comment
|
$begingroup$
First of all, with initial data in $C_b$, classical solution exist, so there is no need for quotation marks. It is easy to see that the convolution of the initial data $f(x)$ with the Gauss–Weierstrass kernel $p_t(x)$ defines the unique bounded classical solution $u(t, x)$.
It is easy to see that $u(t,x) - u(t,y)$ converges to zero as $t to infty$: the functions $p_t(cdot - x) - p_t(cdot - y)$ converge in $L^1$ to zero as $t to infty$. It is only slightly more difficult to see that this convergence is locally uniform with respect to $x$ and $y$. Therefore, your question is equivalent to the following simpler one: for what initial data $f(x)$, the limit of $u(t, 0)$ exists as $t to infty$.
I doubt there is a simpler "if and only if" characterization. One can easily provide sufficient conditions, for example it is sufficient to assume that the mean value of $f$ over the ball $B(0, R)$ has a limit as $R to infty$ (by the same argument that one uses when proving that Cesàro convergence implies Abel convergence). And one can equally easily construct counterexamples.
$endgroup$
$begingroup$
+1 The observation that the solution converges uniformly on compact subsets if and only if it converges at the point $0$ is really interesting. Of course, I would still hope for a characterization of those $f$, maybe as members of some kind of nice function space. (But I agree with you that this hope is somewhat bold.) Concerning the quotation marks around "solution": I only used them because $u(t)$ does not converge to $f$ with respect to the $|cdot|_infty$-norm as $t downarrow 0$, so this notion of solution is slightly weaker that what one has, for instance, for $C_0$-semigroups.
$endgroup$
– Jochen Glueck
3 hours ago
add a comment
|
$begingroup$
First of all, with initial data in $C_b$, classical solution exist, so there is no need for quotation marks. It is easy to see that the convolution of the initial data $f(x)$ with the Gauss–Weierstrass kernel $p_t(x)$ defines the unique bounded classical solution $u(t, x)$.
It is easy to see that $u(t,x) - u(t,y)$ converges to zero as $t to infty$: the functions $p_t(cdot - x) - p_t(cdot - y)$ converge in $L^1$ to zero as $t to infty$. It is only slightly more difficult to see that this convergence is locally uniform with respect to $x$ and $y$. Therefore, your question is equivalent to the following simpler one: for what initial data $f(x)$, the limit of $u(t, 0)$ exists as $t to infty$.
I doubt there is a simpler "if and only if" characterization. One can easily provide sufficient conditions, for example it is sufficient to assume that the mean value of $f$ over the ball $B(0, R)$ has a limit as $R to infty$ (by the same argument that one uses when proving that Cesàro convergence implies Abel convergence). And one can equally easily construct counterexamples.
$endgroup$
First of all, with initial data in $C_b$, classical solution exist, so there is no need for quotation marks. It is easy to see that the convolution of the initial data $f(x)$ with the Gauss–Weierstrass kernel $p_t(x)$ defines the unique bounded classical solution $u(t, x)$.
It is easy to see that $u(t,x) - u(t,y)$ converges to zero as $t to infty$: the functions $p_t(cdot - x) - p_t(cdot - y)$ converge in $L^1$ to zero as $t to infty$. It is only slightly more difficult to see that this convergence is locally uniform with respect to $x$ and $y$. Therefore, your question is equivalent to the following simpler one: for what initial data $f(x)$, the limit of $u(t, 0)$ exists as $t to infty$.
I doubt there is a simpler "if and only if" characterization. One can easily provide sufficient conditions, for example it is sufficient to assume that the mean value of $f$ over the ball $B(0, R)$ has a limit as $R to infty$ (by the same argument that one uses when proving that Cesàro convergence implies Abel convergence). And one can equally easily construct counterexamples.
answered 4 hours ago
Mateusz KwaśnickiMateusz Kwaśnicki
6,0401 gold badge7 silver badges22 bronze badges
6,0401 gold badge7 silver badges22 bronze badges
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+1 The observation that the solution converges uniformly on compact subsets if and only if it converges at the point $0$ is really interesting. Of course, I would still hope for a characterization of those $f$, maybe as members of some kind of nice function space. (But I agree with you that this hope is somewhat bold.) Concerning the quotation marks around "solution": I only used them because $u(t)$ does not converge to $f$ with respect to the $|cdot|_infty$-norm as $t downarrow 0$, so this notion of solution is slightly weaker that what one has, for instance, for $C_0$-semigroups.
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– Jochen Glueck
3 hours ago
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+1 The observation that the solution converges uniformly on compact subsets if and only if it converges at the point $0$ is really interesting. Of course, I would still hope for a characterization of those $f$, maybe as members of some kind of nice function space. (But I agree with you that this hope is somewhat bold.) Concerning the quotation marks around "solution": I only used them because $u(t)$ does not converge to $f$ with respect to the $|cdot|_infty$-norm as $t downarrow 0$, so this notion of solution is slightly weaker that what one has, for instance, for $C_0$-semigroups.
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– Jochen Glueck
3 hours ago
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+1 The observation that the solution converges uniformly on compact subsets if and only if it converges at the point $0$ is really interesting. Of course, I would still hope for a characterization of those $f$, maybe as members of some kind of nice function space. (But I agree with you that this hope is somewhat bold.) Concerning the quotation marks around "solution": I only used them because $u(t)$ does not converge to $f$ with respect to the $|cdot|_infty$-norm as $t downarrow 0$, so this notion of solution is slightly weaker that what one has, for instance, for $C_0$-semigroups.
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– Jochen Glueck
3 hours ago
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+1 The observation that the solution converges uniformly on compact subsets if and only if it converges at the point $0$ is really interesting. Of course, I would still hope for a characterization of those $f$, maybe as members of some kind of nice function space. (But I agree with you that this hope is somewhat bold.) Concerning the quotation marks around "solution": I only used them because $u(t)$ does not converge to $f$ with respect to the $|cdot|_infty$-norm as $t downarrow 0$, so this notion of solution is slightly weaker that what one has, for instance, for $C_0$-semigroups.
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– Jochen Glueck
3 hours ago
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I doubt that there is a characterization, but one thing I can say is that the complement of the set of such $f$ contains a dense open set.
Let $U(f)$ be the solution with initial condition $u(0,x) = f(x)$, and
$$G = f in C_b: lim_t to infty U(f)(t,cdot) textconverges uniformly on compact sets$$
Take some $f_0 in C_b$ such that $U(f_0)(t,0)$ does not converge as $t to infty$. Let $delta = limsup_t to infty U(f_0)(t,0) - liminf_t to infty U(f_0)(t,0) > 0$. Then for any $f in G$ and any $c ne 0$, if $|g- (f + c f_0)|_infty < |c|delta/2$
then $$limsup_t infty U(g)(t,0) - liminf_t to infty U(g)(t,0) ge |c|delta - 2 |g-(f+c f_0)|_infty > 0 $$
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+1 Thank you for your answer! Alternatively, one can also argue that $G$ is a proper subspace of $C_b$ that is closed with respect to $|cdot|_infty$, so its complement is an open and dense set.
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– Jochen Glueck
7 hours ago
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I doubt that there is a characterization, but one thing I can say is that the complement of the set of such $f$ contains a dense open set.
Let $U(f)$ be the solution with initial condition $u(0,x) = f(x)$, and
$$G = f in C_b: lim_t to infty U(f)(t,cdot) textconverges uniformly on compact sets$$
Take some $f_0 in C_b$ such that $U(f_0)(t,0)$ does not converge as $t to infty$. Let $delta = limsup_t to infty U(f_0)(t,0) - liminf_t to infty U(f_0)(t,0) > 0$. Then for any $f in G$ and any $c ne 0$, if $|g- (f + c f_0)|_infty < |c|delta/2$
then $$limsup_t infty U(g)(t,0) - liminf_t to infty U(g)(t,0) ge |c|delta - 2 |g-(f+c f_0)|_infty > 0 $$
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+1 Thank you for your answer! Alternatively, one can also argue that $G$ is a proper subspace of $C_b$ that is closed with respect to $|cdot|_infty$, so its complement is an open and dense set.
$endgroup$
– Jochen Glueck
7 hours ago
add a comment
|
$begingroup$
I doubt that there is a characterization, but one thing I can say is that the complement of the set of such $f$ contains a dense open set.
Let $U(f)$ be the solution with initial condition $u(0,x) = f(x)$, and
$$G = f in C_b: lim_t to infty U(f)(t,cdot) textconverges uniformly on compact sets$$
Take some $f_0 in C_b$ such that $U(f_0)(t,0)$ does not converge as $t to infty$. Let $delta = limsup_t to infty U(f_0)(t,0) - liminf_t to infty U(f_0)(t,0) > 0$. Then for any $f in G$ and any $c ne 0$, if $|g- (f + c f_0)|_infty < |c|delta/2$
then $$limsup_t infty U(g)(t,0) - liminf_t to infty U(g)(t,0) ge |c|delta - 2 |g-(f+c f_0)|_infty > 0 $$
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I doubt that there is a characterization, but one thing I can say is that the complement of the set of such $f$ contains a dense open set.
Let $U(f)$ be the solution with initial condition $u(0,x) = f(x)$, and
$$G = f in C_b: lim_t to infty U(f)(t,cdot) textconverges uniformly on compact sets$$
Take some $f_0 in C_b$ such that $U(f_0)(t,0)$ does not converge as $t to infty$. Let $delta = limsup_t to infty U(f_0)(t,0) - liminf_t to infty U(f_0)(t,0) > 0$. Then for any $f in G$ and any $c ne 0$, if $|g- (f + c f_0)|_infty < |c|delta/2$
then $$limsup_t infty U(g)(t,0) - liminf_t to infty U(g)(t,0) ge |c|delta - 2 |g-(f+c f_0)|_infty > 0 $$
answered 8 hours ago
Robert IsraelRobert Israel
45.4k54 silver badges129 bronze badges
45.4k54 silver badges129 bronze badges
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+1 Thank you for your answer! Alternatively, one can also argue that $G$ is a proper subspace of $C_b$ that is closed with respect to $|cdot|_infty$, so its complement is an open and dense set.
$endgroup$
– Jochen Glueck
7 hours ago
add a comment
|
$begingroup$
+1 Thank you for your answer! Alternatively, one can also argue that $G$ is a proper subspace of $C_b$ that is closed with respect to $|cdot|_infty$, so its complement is an open and dense set.
$endgroup$
– Jochen Glueck
7 hours ago
$begingroup$
+1 Thank you for your answer! Alternatively, one can also argue that $G$ is a proper subspace of $C_b$ that is closed with respect to $|cdot|_infty$, so its complement is an open and dense set.
$endgroup$
– Jochen Glueck
7 hours ago
$begingroup$
+1 Thank you for your answer! Alternatively, one can also argue that $G$ is a proper subspace of $C_b$ that is closed with respect to $|cdot|_infty$, so its complement is an open and dense set.
$endgroup$
– Jochen Glueck
7 hours ago
add a comment
|
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Your claim for the case $fin L^1$ is incorrect. The $L^1$-norm is not constant, unless $f$was signed. Only the integral $m$ of $u(cdot,t)$ is constant. In general, $tmapsto|u(t)|_1$ is non-increasing and converges towards $|m|$. the solution is $L^1$-asymptotic to $mK_t$ where $K_t$ is the heat kernel.
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– Denis Serre
10 hours ago
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Besides $|u(t)|_inftyle(2pi t)^-d/2|f|_1$ so that $u(cdot,t)$ converges uniformly to zero.
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– Denis Serre
10 hours ago
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I don't have a reference, but my inclination would be to look at $partial_t u(t)$, which you can compute from the convolution formula by differentiating under the integral sign, and try to show it goes to 0 sufficiently fast (uniformly on compact sets) to force convergence of $u$.
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– Nate Eldredge
8 hours ago
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@DenisSerre: You're of course right (I have a tendency of always thinking about the case $f ge 0$ tacitly...). Corrected. Concerning the uniform convergence to $0$: yes; moreover, uniform convergence to $0$ also holds for $f in L^p$, $p in (1,infty)$, since the operators of the heat semigroup map $L^p$ continuously into $C_b(mathbbR^d)$. I have also added this information to the post.
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– Jochen Glueck
7 hours ago
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@NateEldredge: Thank you for your comment. I'll have to think about it in more detail.
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– Jochen Glueck
3 hours ago