Mixing
Problem 7 - Chapter 6 - Evans' PDE (Second Edition)
14
$begingroup$
Problem 7 in §6.6 states as follows:
Let $uin H^1(mathbb{R}^n)$ have compact support and be a weak solution of the semilinear PDE
$$-Delta u+c(u)=f,,text{ in } mathbb{R}^n,$$
where $fin L^2(mathbb{R}^n)$ and $c:mathbb{R}tomathbb{R}$ is smooth, with $c(0)=0$ and $c'geqslant 0.$ Prove $uin H^2(mathbb{R}^n)$.
I am trying to prove $c(u(x))in L^2(mathbb{R}^n)$ or $c'(u(x))in L^{infty}(mathbb{R}^n)$ but both failed. I know in fact $u$ can be in $H_0^1(mathbb{R}^n)$, but I don't know whether this is useful here. And $c(u(x))$ has compact support since $u$ has compact support. However, again, I can't figure out that $uin L^{infty}(mathbb{R}^n)$. And I really don't know how to use the condition $c'geqslant 0$.
In conclusion, I've no idea about this problem.
Anyone could help me? Any advice will be appreciated.
pde
edited Feb 1 at 0:34
Pedro
10.9k23475
asked Dec 26 '11 at 7:22
Y.ZY.Z
664515
$endgroup$
add a comment |
14
$begingroup$
Problem 7 in §6.6 states as follows:
Let $uin H^1(mathbb{R}^n)$ have compact support and be a weak solution of the semilinear PDE
$$-Delta u+c(u)=f,,text{ in } mathbb{R}^n,$$
where $fin L^2(mathbb{R}^n)$ and $c:mathbb{R}tomathbb{R}$ is smooth, with $c(0)=0$ and $c'geqslant 0.$ Prove $uin H^2(mathbb{R}^n)$.
I am trying to prove $c(u(x))in L^2(mathbb{R}^n)$ or $c'(u(x))in L^{infty}(mathbb{R}^n)$ but both failed. I know in fact $u$ can be in $H_0^1(mathbb{R}^n)$, but I don't know whether this is useful here. And $c(u(x))$ has compact support since $u$ has compact support. However, again, I can't figure out that $uin L^{infty}(mathbb{R}^n)$. And I really don't know how to use the condition $c'geqslant 0$.
In conclusion, I've no idea about this problem.
Anyone could help me? Any advice will be appreciated.
pde
edited Feb 1 at 0:34
Pedro
10.9k23475
asked Dec 26 '11 at 7:22
Y.ZY.Z
664515
$endgroup$
$begingroup$
Have you tried a Fourier transform approach? In Fourier space the problem boils down to prove $mathscr{F}[c(u)] in L^2(mathbb{R}^n)$, which looks more manageable. What do you think?
$endgroup$
– Giuseppe Negro
Dec 26 '11 at 11:01
$begingroup$
@GiuseppeNegro: Thanks. I haven't tried this approach before. Acturally I'm not familiar with Fourier transform, I mean I know some basic facts about it but I've never done any exercises. Here I still do not know how to deal with the $c(u(x))$ under the integral. I think if I can deal with this integral I can also deal with the former approach. Do you have some more details?
$endgroup$
– Y.Z
Dec 26 '11 at 13:24
$begingroup$
@GiuseppeNegro Could you please add more details about the Fourier transform approach? Thanks!
$endgroup$
– Sherry
Apr 10 '15 at 6:13
$begingroup$
@Sherry: I am afraid I cannot answer right now. I don't remember what was my idea back in '11, and frankly, there is a big possibility that it was just plain wrong.
$endgroup$
– Giuseppe Negro
Apr 10 '15 at 7:20
add a comment |
14
14
14
10
$begingroup$
Problem 7 in §6.6 states as follows:
Let $uin H^1(mathbb{R}^n)$ have compact support and be a weak solution of the semilinear PDE
$$-Delta u+c(u)=f,,text{ in } mathbb{R}^n,$$
where $fin L^2(mathbb{R}^n)$ and $c:mathbb{R}tomathbb{R}$ is smooth, with $c(0)=0$ and $c'geqslant 0.$ Prove $uin H^2(mathbb{R}^n)$.
I am trying to prove $c(u(x))in L^2(mathbb{R}^n)$ or $c'(u(x))in L^{infty}(mathbb{R}^n)$ but both failed. I know in fact $u$ can be in $H_0^1(mathbb{R}^n)$, but I don't know whether this is useful here. And $c(u(x))$ has compact support since $u$ has compact support. However, again, I can't figure out that $uin L^{infty}(mathbb{R}^n)$. And I really don't know how to use the condition $c'geqslant 0$.
In conclusion, I've no idea about this problem.
Anyone could help me? Any advice will be appreciated.
pde
edited Feb 1 at 0:34
Pedro
10.9k23475
asked Dec 26 '11 at 7:22
Y.ZY.Z
664515
$endgroup$
Problem 7 in §6.6 states as follows:
Let $uin H^1(mathbb{R}^n)$ have compact support and be a weak solution of the semilinear PDE
$$-Delta u+c(u)=f,,text{ in } mathbb{R}^n,$$
where $fin L^2(mathbb{R}^n)$ and $c:mathbb{R}tomathbb{R}$ is smooth, with $c(0)=0$ and $c'geqslant 0.$ Prove $uin H^2(mathbb{R}^n)$.
I am trying to prove $c(u(x))in L^2(mathbb{R}^n)$ or $c'(u(x))in L^{infty}(mathbb{R}^n)$ but both failed. I know in fact $u$ can be in $H_0^1(mathbb{R}^n)$, but I don't know whether this is useful here. And $c(u(x))$ has compact support since $u$ has compact support. However, again, I can't figure out that $uin L^{infty}(mathbb{R}^n)$. And I really don't know how to use the condition $c'geqslant 0$.
In conclusion, I've no idea about this problem.
Anyone could help me? Any advice will be appreciated.
pde
pde
edited Feb 1 at 0:34
Pedro
10.9k23475
asked Dec 26 '11 at 7:22
Y.ZY.Z
664515
edited Feb 1 at 0:34
Pedro
10.9k23475
asked Dec 26 '11 at 7:22
Y.ZY.Z
664515
edited Feb 1 at 0:34
Pedro
10.9k23475
edited Feb 1 at 0:34
Pedro
10.9k23475
edited Feb 1 at 0:34
Pedro
10.9k23475
10.9k23475
asked Dec 26 '11 at 7:22
Y.ZY.Z
664515
asked Dec 26 '11 at 7:22
Y.ZY.Z
664515
asked Dec 26 '11 at 7:22
Y.ZY.Z
664515
664515
$begingroup$
Have you tried a Fourier transform approach? In Fourier space the problem boils down to prove $mathscr{F}[c(u)] in L^2(mathbb{R}^n)$, which looks more manageable. What do you think?
$endgroup$
– Giuseppe Negro
Dec 26 '11 at 11:01
$begingroup$
@GiuseppeNegro: Thanks. I haven't tried this approach before. Acturally I'm not familiar with Fourier transform, I mean I know some basic facts about it but I've never done any exercises. Here I still do not know how to deal with the $c(u(x))$ under the integral. I think if I can deal with this integral I can also deal with the former approach. Do you have some more details?
$endgroup$
– Y.Z
Dec 26 '11 at 13:24
$begingroup$
@GiuseppeNegro Could you please add more details about the Fourier transform approach? Thanks!
$endgroup$
– Sherry
Apr 10 '15 at 6:13
$begingroup$
@Sherry: I am afraid I cannot answer right now. I don't remember what was my idea back in '11, and frankly, there is a big possibility that it was just plain wrong.
$endgroup$
– Giuseppe Negro
Apr 10 '15 at 7:20
add a comment |
$begingroup$
Have you tried a Fourier transform approach? In Fourier space the problem boils down to prove $mathscr{F}[c(u)] in L^2(mathbb{R}^n)$, which looks more manageable. What do you think?
$endgroup$
– Giuseppe Negro
Dec 26 '11 at 11:01
$begingroup$
@GiuseppeNegro: Thanks. I haven't tried this approach before. Acturally I'm not familiar with Fourier transform, I mean I know some basic facts about it but I've never done any exercises. Here I still do not know how to deal with the $c(u(x))$ under the integral. I think if I can deal with this integral I can also deal with the former approach. Do you have some more details?
$endgroup$
– Y.Z
Dec 26 '11 at 13:24
$begingroup$
@GiuseppeNegro Could you please add more details about the Fourier transform approach? Thanks!
$endgroup$
– Sherry
Apr 10 '15 at 6:13
$begingroup$
@Sherry: I am afraid I cannot answer right now. I don't remember what was my idea back in '11, and frankly, there is a big possibility that it was just plain wrong.
$endgroup$
– Giuseppe Negro
Apr 10 '15 at 7:20
$begingroup$
Have you tried a Fourier transform approach? In Fourier space the problem boils down to prove $mathscr{F}[c(u)] in L^2(mathbb{R}^n)$, which looks more manageable. What do you think?
$endgroup$
– Giuseppe Negro
Dec 26 '11 at 11:01
$begingroup$
Have you tried a Fourier transform approach? In Fourier space the problem boils down to prove $mathscr{F}[c(u)] in L^2(mathbb{R}^n)$, which looks more manageable. What do you think?
$endgroup$
– Giuseppe Negro
Dec 26 '11 at 11:01
$begingroup$
@GiuseppeNegro: Thanks. I haven't tried this approach before. Acturally I'm not familiar with Fourier transform, I mean I know some basic facts about it but I've never done any exercises. Here I still do not know how to deal with the $c(u(x))$ under the integral. I think if I can deal with this integral I can also deal with the former approach. Do you have some more details?
$endgroup$
– Y.Z
Dec 26 '11 at 13:24
$begingroup$
@GiuseppeNegro: Thanks. I haven't tried this approach before. Acturally I'm not familiar with Fourier transform, I mean I know some basic facts about it but I've never done any exercises. Here I still do not know how to deal with the $c(u(x))$ under the integral. I think if I can deal with this integral I can also deal with the former approach. Do you have some more details?
$endgroup$
– Y.Z
Dec 26 '11 at 13:24
$begingroup$
@GiuseppeNegro Could you please add more details about the Fourier transform approach? Thanks!
$endgroup$
– Sherry
Apr 10 '15 at 6:13
$begingroup$
@GiuseppeNegro Could you please add more details about the Fourier transform approach? Thanks!
$endgroup$
– Sherry
Apr 10 '15 at 6:13
$begingroup$
@Sherry: I am afraid I cannot answer right now. I don't remember what was my idea back in '11, and frankly, there is a big possibility that it was just plain wrong.
$endgroup$
– Giuseppe Negro
Apr 10 '15 at 7:20
$begingroup$
@Sherry: I am afraid I cannot answer right now. I don't remember what was my idea back in '11, and frankly, there is a big possibility that it was just plain wrong.
$endgroup$
– Giuseppe Negro
Apr 10 '15 at 7:20
add a comment |
2 Answers
2
active
oldest
votes
9
+50
$begingroup$
Here's my try.
The hint in Evans suggests to mimic the proof of the $H^2$ interior regularity theorem.
This seems a very good hint, so let's try to follow it.
- (Weak form of the problem + choice of a test function) Take a test function $vin H_0^1(mathbb{R}^N)$, multiply both sides of equation $-Delta u+c(u)=f$ by $v$ and integrate by parts to get:
$$tag{1} int_{mathbb{R}^N} langle nabla u,nabla vrangle = int_{mathbb{R}^N} Big( f-c(u)Big) v$$
(recall that $text{supt }u$ is compact, hence the integrals are actually over a suitably large ball).
Now, choose $|h|>0$ small, an index $nin {1,ldots ,N}$ and set $v=-D_n^{-h}[D_n^h u]$ in (1), where:
$$D_k^hu(x):= frac{u(x+hmathbf{e}^n)-u(x)}{h}; $$
(we use the "double" difference quotient for we want estimetes for the second derivatives of $u$); for sake of simplicity, we rewrite the resulting expression as:
$$I=J; .$$ - (On rewriting $I$) Recalling that:
$$int_{mathbb{R}^N} phi D_n^{-h}psi = int_{mathbb{R}^N} psi D_n^{h}phiquad text{and}quad D_n^hphi_{x_i}=(D_n^hphi)_{x_i}; ,$$
we get:
$$begin{split} I &= -int_{mathbb{R}^N} langle nabla u, nabla (D_n^{-h} D_n^h u)rangle \
&= -sum_{i=1}^N int_{mathbb{R}^N} u_{x_i} (D_n^{-h}D_n^h u)_{x_i}\
&= -sum_{i=1}^N int_{mathbb{R}^N} u_{x_i} D_n^{-h}(D_n^h u)_{x_i}\
&= sum_{i=1}^N int_{mathbb{R}^N} D_n^hu_{x_i} (D_n^h u)_{x_i}\
&= int_{mathbb{R}^N} langle D_n^h nabla u, D_n^h nabla urangle ; ,end{split}$$
i.e.:
$$tag{I} I=lVert D_n^h nabla urVert_2^2; .$$ - (Estimate for $J$) We have:
$$|J|=left| int_{mathbb{R}^N} -f D_n^{-h}D_n^h u +c(u) D_n^{-h}D_n^h u right| leq int_{mathbb{R}^N} Big( |f|+|c(u)|Big) |D_n^{-h}D_n^h u|; ;$$
using FTIC, $c(0)=0$ and $c^prime in L^infty$ we find:
$$|c(u(x))|=left| int_0^{u(x)} c^prime (t) text{d} tright| leq lVert c^primerVert_infty |u(x)|$$
hence plugging the latter inequality in the former we obtain:
$$tag{2} |J|leq int_{mathbb{R}^N} Big( |f|+lVert c^primerVert_infty |u|Big) |D_n^{-h}D_n^h u|; .$$
Now, from Thm 3(i), §5.8.2 we get:
$$tag{3} int_{mathbb{R}^N} |D_n^{-h}D_n^h u|^2leq C_1int_{mathbb{R}^N} |nabla D_n^h u|^2 leq C_2int_{mathbb{R}^N} |nabla u|^2; ,$$
thus from (2)-(3), Cauchy inequality and an elementary inequality we infer:
$$begin{split} |J| &leq int_{mathbb{R}^N} Big( |f|+lVert c^primerVert_infty |u|Big) |D_n^{-h}D_n^h u| &qquad text{[by (2)]}\
&leq frac{1}{2} int_{mathbb{R}^N} Big( |f|+lVert c^primerVert_infty |u|Big)^2 + frac{1}{2} int_{mathbb{R}^N} |D_n^{-h}D_n^h u|^2 &qquad text{[by Cauchy's]}\
&leq int_{mathbb{R}^N}Big(|f|^2+lVert c^prime rVert_infty^2 |u|^2Big) +frac{C_2}{2} int_{mathbb{R}^N} |nabla u|^2 &qquad text{[by (3)]}; ,end{split}$$
hence:
$$tag{J} |J|leq C_3 left( lVert frVert_2^2+lVert urVert_2^2 +lVert nabla urVert_2^2right)$$
for some suitable constant $C_3geq 0$. - (Conclusion) Therefore from $I=J$, (I) and (J):
$$lVert D_n^h u_{x_i}rVert_2^2 leq lVert D_n^h nabla urVert_2^2leq C_3 left( lVert frVert_2^2+lVert urVert_2^2 +lVert nabla urVert_2^2right); ,$$
for $|h|>0$ small and each $iin {1,ldots, N}$. Thm 3(ii), §5.8.2 yields $u_{x_i}in H^1(mathbb{R}^N)$ and $lVert nabla u_{x_i}rVert_2^2 leq C_3 left( lVert frVert_2^2+lVert urVert_2^2 +lVert nabla urVert_2^2right)$, therefore $uin H^2(mathbb{R}^N)$.
What do you think?
answered Dec 26 '11 at 20:06
PacciuPacciu
3,99611030
$endgroup$
$begingroup$
Thanks for your details. But I still do not know why the inequality above (2) holds. Why $c'in L^{infty}$(this is my question, may seems obvious for you). I know $c(u(x))$ has compact support, say, $B_R(0)$, but for $c(t),t=u(x)$, I'm not sure $t$ is in a ball.
$endgroup$
– Y.Z
Dec 27 '11 at 1:39
$begingroup$
Sorry, I misread your post so $c^prime in L^infty$ appeared as an assumption to my eyes... There should be a way to overcome this difficulty, but I have to think about it.
$endgroup$
– Pacciu
Dec 27 '11 at 1:50
add a comment |
3
$begingroup$
First step:
Since $u$ has compact support and $c(0)=0$, and $c(x)$ is continuous, $c(u(x))$ is continuous and has compact support and so it is in $L^2(R^n)$.
Second step:
Take Fourier transform of $$-Delta u+c(u(x))=f(x)$$
Then you can get $$|xi|^2hat u(xi)+widehat{c(u(x))}=hat f(xi)$$
It follows from the first step that $widehat {c(u(x))} in L^2$. Thus $|xi|^2hat u(xi) in L^2$
We also know $hat u in L^2$
So $(1+|xi|^2)hat u(xi) in L^2$
It follows that $u in H^2$.
answered Apr 17 '15 at 0:16
SherrySherry
1,687625
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
9
+50
$begingroup$
Here's my try.
The hint in Evans suggests to mimic the proof of the $H^2$ interior regularity theorem.
This seems a very good hint, so let's try to follow it.
- (Weak form of the problem + choice of a test function) Take a test function $vin H_0^1(mathbb{R}^N)$, multiply both sides of equation $-Delta u+c(u)=f$ by $v$ and integrate by parts to get:
$$tag{1} int_{mathbb{R}^N} langle nabla u,nabla vrangle = int_{mathbb{R}^N} Big( f-c(u)Big) v$$
(recall that $text{supt }u$ is compact, hence the integrals are actually over a suitably large ball).
Now, choose $|h|>0$ small, an index $nin {1,ldots ,N}$ and set $v=-D_n^{-h}[D_n^h u]$ in (1), where:
$$D_k^hu(x):= frac{u(x+hmathbf{e}^n)-u(x)}{h}; $$
(we use the "double" difference quotient for we want estimetes for the second derivatives of $u$); for sake of simplicity, we rewrite the resulting expression as:
$$I=J; .$$ - (On rewriting $I$) Recalling that:
$$int_{mathbb{R}^N} phi D_n^{-h}psi = int_{mathbb{R}^N} psi D_n^{h}phiquad text{and}quad D_n^hphi_{x_i}=(D_n^hphi)_{x_i}; ,$$
we get:
$$begin{split} I &= -int_{mathbb{R}^N} langle nabla u, nabla (D_n^{-h} D_n^h u)rangle \
&= -sum_{i=1}^N int_{mathbb{R}^N} u_{x_i} (D_n^{-h}D_n^h u)_{x_i}\
&= -sum_{i=1}^N int_{mathbb{R}^N} u_{x_i} D_n^{-h}(D_n^h u)_{x_i}\
&= sum_{i=1}^N int_{mathbb{R}^N} D_n^hu_{x_i} (D_n^h u)_{x_i}\
&= int_{mathbb{R}^N} langle D_n^h nabla u, D_n^h nabla urangle ; ,end{split}$$
i.e.:
$$tag{I} I=lVert D_n^h nabla urVert_2^2; .$$ - (Estimate for $J$) We have:
$$|J|=left| int_{mathbb{R}^N} -f D_n^{-h}D_n^h u +c(u) D_n^{-h}D_n^h u right| leq int_{mathbb{R}^N} Big( |f|+|c(u)|Big) |D_n^{-h}D_n^h u|; ;$$
using FTIC, $c(0)=0$ and $c^prime in L^infty$ we find:
$$|c(u(x))|=left| int_0^{u(x)} c^prime (t) text{d} tright| leq lVert c^primerVert_infty |u(x)|$$
hence plugging the latter inequality in the former we obtain:
$$tag{2} |J|leq int_{mathbb{R}^N} Big( |f|+lVert c^primerVert_infty |u|Big) |D_n^{-h}D_n^h u|; .$$
Now, from Thm 3(i), §5.8.2 we get:
$$tag{3} int_{mathbb{R}^N} |D_n^{-h}D_n^h u|^2leq C_1int_{mathbb{R}^N} |nabla D_n^h u|^2 leq C_2int_{mathbb{R}^N} |nabla u|^2; ,$$
thus from (2)-(3), Cauchy inequality and an elementary inequality we infer:
$$begin{split} |J| &leq int_{mathbb{R}^N} Big( |f|+lVert c^primerVert_infty |u|Big) |D_n^{-h}D_n^h u| &qquad text{[by (2)]}\
&leq frac{1}{2} int_{mathbb{R}^N} Big( |f|+lVert c^primerVert_infty |u|Big)^2 + frac{1}{2} int_{mathbb{R}^N} |D_n^{-h}D_n^h u|^2 &qquad text{[by Cauchy's]}\
&leq int_{mathbb{R}^N}Big(|f|^2+lVert c^prime rVert_infty^2 |u|^2Big) +frac{C_2}{2} int_{mathbb{R}^N} |nabla u|^2 &qquad text{[by (3)]}; ,end{split}$$
hence:
$$tag{J} |J|leq C_3 left( lVert frVert_2^2+lVert urVert_2^2 +lVert nabla urVert_2^2right)$$
for some suitable constant $C_3geq 0$. - (Conclusion) Therefore from $I=J$, (I) and (J):
$$lVert D_n^h u_{x_i}rVert_2^2 leq lVert D_n^h nabla urVert_2^2leq C_3 left( lVert frVert_2^2+lVert urVert_2^2 +lVert nabla urVert_2^2right); ,$$
for $|h|>0$ small and each $iin {1,ldots, N}$. Thm 3(ii), §5.8.2 yields $u_{x_i}in H^1(mathbb{R}^N)$ and $lVert nabla u_{x_i}rVert_2^2 leq C_3 left( lVert frVert_2^2+lVert urVert_2^2 +lVert nabla urVert_2^2right)$, therefore $uin H^2(mathbb{R}^N)$.
What do you think?
answered Dec 26 '11 at 20:06
PacciuPacciu
3,99611030
$endgroup$
$begingroup$
Thanks for your details. But I still do not know why the inequality above (2) holds. Why $c'in L^{infty}$(this is my question, may seems obvious for you). I know $c(u(x))$ has compact support, say, $B_R(0)$, but for $c(t),t=u(x)$, I'm not sure $t$ is in a ball.
$endgroup$
– Y.Z
Dec 27 '11 at 1:39
$begingroup$
Sorry, I misread your post so $c^prime in L^infty$ appeared as an assumption to my eyes... There should be a way to overcome this difficulty, but I have to think about it.
$endgroup$
– Pacciu
Dec 27 '11 at 1:50
add a comment |
9
+50
$begingroup$
Here's my try.
The hint in Evans suggests to mimic the proof of the $H^2$ interior regularity theorem.
This seems a very good hint, so let's try to follow it.
- (Weak form of the problem + choice of a test function) Take a test function $vin H_0^1(mathbb{R}^N)$, multiply both sides of equation $-Delta u+c(u)=f$ by $v$ and integrate by parts to get:
$$tag{1} int_{mathbb{R}^N} langle nabla u,nabla vrangle = int_{mathbb{R}^N} Big( f-c(u)Big) v$$
(recall that $text{supt }u$ is compact, hence the integrals are actually over a suitably large ball).
Now, choose $|h|>0$ small, an index $nin {1,ldots ,N}$ and set $v=-D_n^{-h}[D_n^h u]$ in (1), where:
$$D_k^hu(x):= frac{u(x+hmathbf{e}^n)-u(x)}{h}; $$
(we use the "double" difference quotient for we want estimetes for the second derivatives of $u$); for sake of simplicity, we rewrite the resulting expression as:
$$I=J; .$$ - (On rewriting $I$) Recalling that:
$$int_{mathbb{R}^N} phi D_n^{-h}psi = int_{mathbb{R}^N} psi D_n^{h}phiquad text{and}quad D_n^hphi_{x_i}=(D_n^hphi)_{x_i}; ,$$
we get:
$$begin{split} I &= -int_{mathbb{R}^N} langle nabla u, nabla (D_n^{-h} D_n^h u)rangle \
&= -sum_{i=1}^N int_{mathbb{R}^N} u_{x_i} (D_n^{-h}D_n^h u)_{x_i}\
&= -sum_{i=1}^N int_{mathbb{R}^N} u_{x_i} D_n^{-h}(D_n^h u)_{x_i}\
&= sum_{i=1}^N int_{mathbb{R}^N} D_n^hu_{x_i} (D_n^h u)_{x_i}\
&= int_{mathbb{R}^N} langle D_n^h nabla u, D_n^h nabla urangle ; ,end{split}$$
i.e.:
$$tag{I} I=lVert D_n^h nabla urVert_2^2; .$$ - (Estimate for $J$) We have:
$$|J|=left| int_{mathbb{R}^N} -f D_n^{-h}D_n^h u +c(u) D_n^{-h}D_n^h u right| leq int_{mathbb{R}^N} Big( |f|+|c(u)|Big) |D_n^{-h}D_n^h u|; ;$$
using FTIC, $c(0)=0$ and $c^prime in L^infty$ we find:
$$|c(u(x))|=left| int_0^{u(x)} c^prime (t) text{d} tright| leq lVert c^primerVert_infty |u(x)|$$
hence plugging the latter inequality in the former we obtain:
$$tag{2} |J|leq int_{mathbb{R}^N} Big( |f|+lVert c^primerVert_infty |u|Big) |D_n^{-h}D_n^h u|; .$$
Now, from Thm 3(i), §5.8.2 we get:
$$tag{3} int_{mathbb{R}^N} |D_n^{-h}D_n^h u|^2leq C_1int_{mathbb{R}^N} |nabla D_n^h u|^2 leq C_2int_{mathbb{R}^N} |nabla u|^2; ,$$
thus from (2)-(3), Cauchy inequality and an elementary inequality we infer:
$$begin{split} |J| &leq int_{mathbb{R}^N} Big( |f|+lVert c^primerVert_infty |u|Big) |D_n^{-h}D_n^h u| &qquad text{[by (2)]}\
&leq frac{1}{2} int_{mathbb{R}^N} Big( |f|+lVert c^primerVert_infty |u|Big)^2 + frac{1}{2} int_{mathbb{R}^N} |D_n^{-h}D_n^h u|^2 &qquad text{[by Cauchy's]}\
&leq int_{mathbb{R}^N}Big(|f|^2+lVert c^prime rVert_infty^2 |u|^2Big) +frac{C_2}{2} int_{mathbb{R}^N} |nabla u|^2 &qquad text{[by (3)]}; ,end{split}$$
hence:
$$tag{J} |J|leq C_3 left( lVert frVert_2^2+lVert urVert_2^2 +lVert nabla urVert_2^2right)$$
for some suitable constant $C_3geq 0$. - (Conclusion) Therefore from $I=J$, (I) and (J):
$$lVert D_n^h u_{x_i}rVert_2^2 leq lVert D_n^h nabla urVert_2^2leq C_3 left( lVert frVert_2^2+lVert urVert_2^2 +lVert nabla urVert_2^2right); ,$$
for $|h|>0$ small and each $iin {1,ldots, N}$. Thm 3(ii), §5.8.2 yields $u_{x_i}in H^1(mathbb{R}^N)$ and $lVert nabla u_{x_i}rVert_2^2 leq C_3 left( lVert frVert_2^2+lVert urVert_2^2 +lVert nabla urVert_2^2right)$, therefore $uin H^2(mathbb{R}^N)$.
What do you think?
answered Dec 26 '11 at 20:06
PacciuPacciu
3,99611030
$endgroup$
$begingroup$
Thanks for your details. But I still do not know why the inequality above (2) holds. Why $c'in L^{infty}$(this is my question, may seems obvious for you). I know $c(u(x))$ has compact support, say, $B_R(0)$, but for $c(t),t=u(x)$, I'm not sure $t$ is in a ball.
$endgroup$
– Y.Z
Dec 27 '11 at 1:39
$begingroup$
Sorry, I misread your post so $c^prime in L^infty$ appeared as an assumption to my eyes... There should be a way to overcome this difficulty, but I have to think about it.
$endgroup$
– Pacciu
Dec 27 '11 at 1:50
add a comment |
9
+50
9
+50
9
+50
$begingroup$
Here's my try.
The hint in Evans suggests to mimic the proof of the $H^2$ interior regularity theorem.
This seems a very good hint, so let's try to follow it.
- (Weak form of the problem + choice of a test function) Take a test function $vin H_0^1(mathbb{R}^N)$, multiply both sides of equation $-Delta u+c(u)=f$ by $v$ and integrate by parts to get:
$$tag{1} int_{mathbb{R}^N} langle nabla u,nabla vrangle = int_{mathbb{R}^N} Big( f-c(u)Big) v$$
(recall that $text{supt }u$ is compact, hence the integrals are actually over a suitably large ball).
Now, choose $|h|>0$ small, an index $nin {1,ldots ,N}$ and set $v=-D_n^{-h}[D_n^h u]$ in (1), where:
$$D_k^hu(x):= frac{u(x+hmathbf{e}^n)-u(x)}{h}; $$
(we use the "double" difference quotient for we want estimetes for the second derivatives of $u$); for sake of simplicity, we rewrite the resulting expression as:
$$I=J; .$$ - (On rewriting $I$) Recalling that:
$$int_{mathbb{R}^N} phi D_n^{-h}psi = int_{mathbb{R}^N} psi D_n^{h}phiquad text{and}quad D_n^hphi_{x_i}=(D_n^hphi)_{x_i}; ,$$
we get:
$$begin{split} I &= -int_{mathbb{R}^N} langle nabla u, nabla (D_n^{-h} D_n^h u)rangle \
&= -sum_{i=1}^N int_{mathbb{R}^N} u_{x_i} (D_n^{-h}D_n^h u)_{x_i}\
&= -sum_{i=1}^N int_{mathbb{R}^N} u_{x_i} D_n^{-h}(D_n^h u)_{x_i}\
&= sum_{i=1}^N int_{mathbb{R}^N} D_n^hu_{x_i} (D_n^h u)_{x_i}\
&= int_{mathbb{R}^N} langle D_n^h nabla u, D_n^h nabla urangle ; ,end{split}$$
i.e.:
$$tag{I} I=lVert D_n^h nabla urVert_2^2; .$$ - (Estimate for $J$) We have:
$$|J|=left| int_{mathbb{R}^N} -f D_n^{-h}D_n^h u +c(u) D_n^{-h}D_n^h u right| leq int_{mathbb{R}^N} Big( |f|+|c(u)|Big) |D_n^{-h}D_n^h u|; ;$$
using FTIC, $c(0)=0$ and $c^prime in L^infty$ we find:
$$|c(u(x))|=left| int_0^{u(x)} c^prime (t) text{d} tright| leq lVert c^primerVert_infty |u(x)|$$
hence plugging the latter inequality in the former we obtain:
$$tag{2} |J|leq int_{mathbb{R}^N} Big( |f|+lVert c^primerVert_infty |u|Big) |D_n^{-h}D_n^h u|; .$$
Now, from Thm 3(i), §5.8.2 we get:
$$tag{3} int_{mathbb{R}^N} |D_n^{-h}D_n^h u|^2leq C_1int_{mathbb{R}^N} |nabla D_n^h u|^2 leq C_2int_{mathbb{R}^N} |nabla u|^2; ,$$
thus from (2)-(3), Cauchy inequality and an elementary inequality we infer:
$$begin{split} |J| &leq int_{mathbb{R}^N} Big( |f|+lVert c^primerVert_infty |u|Big) |D_n^{-h}D_n^h u| &qquad text{[by (2)]}\
&leq frac{1}{2} int_{mathbb{R}^N} Big( |f|+lVert c^primerVert_infty |u|Big)^2 + frac{1}{2} int_{mathbb{R}^N} |D_n^{-h}D_n^h u|^2 &qquad text{[by Cauchy's]}\
&leq int_{mathbb{R}^N}Big(|f|^2+lVert c^prime rVert_infty^2 |u|^2Big) +frac{C_2}{2} int_{mathbb{R}^N} |nabla u|^2 &qquad text{[by (3)]}; ,end{split}$$
hence:
$$tag{J} |J|leq C_3 left( lVert frVert_2^2+lVert urVert_2^2 +lVert nabla urVert_2^2right)$$
for some suitable constant $C_3geq 0$. - (Conclusion) Therefore from $I=J$, (I) and (J):
$$lVert D_n^h u_{x_i}rVert_2^2 leq lVert D_n^h nabla urVert_2^2leq C_3 left( lVert frVert_2^2+lVert urVert_2^2 +lVert nabla urVert_2^2right); ,$$
for $|h|>0$ small and each $iin {1,ldots, N}$. Thm 3(ii), §5.8.2 yields $u_{x_i}in H^1(mathbb{R}^N)$ and $lVert nabla u_{x_i}rVert_2^2 leq C_3 left( lVert frVert_2^2+lVert urVert_2^2 +lVert nabla urVert_2^2right)$, therefore $uin H^2(mathbb{R}^N)$.
What do you think?
answered Dec 26 '11 at 20:06
PacciuPacciu
3,99611030
$endgroup$
Here's my try.
The hint in Evans suggests to mimic the proof of the $H^2$ interior regularity theorem.
This seems a very good hint, so let's try to follow it.
- (Weak form of the problem + choice of a test function) Take a test function $vin H_0^1(mathbb{R}^N)$, multiply both sides of equation $-Delta u+c(u)=f$ by $v$ and integrate by parts to get:
$$tag{1} int_{mathbb{R}^N} langle nabla u,nabla vrangle = int_{mathbb{R}^N} Big( f-c(u)Big) v$$
(recall that $text{supt }u$ is compact, hence the integrals are actually over a suitably large ball).
Now, choose $|h|>0$ small, an index $nin {1,ldots ,N}$ and set $v=-D_n^{-h}[D_n^h u]$ in (1), where:
$$D_k^hu(x):= frac{u(x+hmathbf{e}^n)-u(x)}{h}; $$
(we use the "double" difference quotient for we want estimetes for the second derivatives of $u$); for sake of simplicity, we rewrite the resulting expression as:
$$I=J; .$$ - (On rewriting $I$) Recalling that:
$$int_{mathbb{R}^N} phi D_n^{-h}psi = int_{mathbb{R}^N} psi D_n^{h}phiquad text{and}quad D_n^hphi_{x_i}=(D_n^hphi)_{x_i}; ,$$
we get:
$$begin{split} I &= -int_{mathbb{R}^N} langle nabla u, nabla (D_n^{-h} D_n^h u)rangle \
&= -sum_{i=1}^N int_{mathbb{R}^N} u_{x_i} (D_n^{-h}D_n^h u)_{x_i}\
&= -sum_{i=1}^N int_{mathbb{R}^N} u_{x_i} D_n^{-h}(D_n^h u)_{x_i}\
&= sum_{i=1}^N int_{mathbb{R}^N} D_n^hu_{x_i} (D_n^h u)_{x_i}\
&= int_{mathbb{R}^N} langle D_n^h nabla u, D_n^h nabla urangle ; ,end{split}$$
i.e.:
$$tag{I} I=lVert D_n^h nabla urVert_2^2; .$$ - (Estimate for $J$) We have:
$$|J|=left| int_{mathbb{R}^N} -f D_n^{-h}D_n^h u +c(u) D_n^{-h}D_n^h u right| leq int_{mathbb{R}^N} Big( |f|+|c(u)|Big) |D_n^{-h}D_n^h u|; ;$$
using FTIC, $c(0)=0$ and $c^prime in L^infty$ we find:
$$|c(u(x))|=left| int_0^{u(x)} c^prime (t) text{d} tright| leq lVert c^primerVert_infty |u(x)|$$
hence plugging the latter inequality in the former we obtain:
$$tag{2} |J|leq int_{mathbb{R}^N} Big( |f|+lVert c^primerVert_infty |u|Big) |D_n^{-h}D_n^h u|; .$$
Now, from Thm 3(i), §5.8.2 we get:
$$tag{3} int_{mathbb{R}^N} |D_n^{-h}D_n^h u|^2leq C_1int_{mathbb{R}^N} |nabla D_n^h u|^2 leq C_2int_{mathbb{R}^N} |nabla u|^2; ,$$
thus from (2)-(3), Cauchy inequality and an elementary inequality we infer:
$$begin{split} |J| &leq int_{mathbb{R}^N} Big( |f|+lVert c^primerVert_infty |u|Big) |D_n^{-h}D_n^h u| &qquad text{[by (2)]}\
&leq frac{1}{2} int_{mathbb{R}^N} Big( |f|+lVert c^primerVert_infty |u|Big)^2 + frac{1}{2} int_{mathbb{R}^N} |D_n^{-h}D_n^h u|^2 &qquad text{[by Cauchy's]}\
&leq int_{mathbb{R}^N}Big(|f|^2+lVert c^prime rVert_infty^2 |u|^2Big) +frac{C_2}{2} int_{mathbb{R}^N} |nabla u|^2 &qquad text{[by (3)]}; ,end{split}$$
hence:
$$tag{J} |J|leq C_3 left( lVert frVert_2^2+lVert urVert_2^2 +lVert nabla urVert_2^2right)$$
for some suitable constant $C_3geq 0$. - (Conclusion) Therefore from $I=J$, (I) and (J):
$$lVert D_n^h u_{x_i}rVert_2^2 leq lVert D_n^h nabla urVert_2^2leq C_3 left( lVert frVert_2^2+lVert urVert_2^2 +lVert nabla urVert_2^2right); ,$$
for $|h|>0$ small and each $iin {1,ldots, N}$. Thm 3(ii), §5.8.2 yields $u_{x_i}in H^1(mathbb{R}^N)$ and $lVert nabla u_{x_i}rVert_2^2 leq C_3 left( lVert frVert_2^2+lVert urVert_2^2 +lVert nabla urVert_2^2right)$, therefore $uin H^2(mathbb{R}^N)$.
What do you think?
answered Dec 26 '11 at 20:06
PacciuPacciu
3,99611030
answered Dec 26 '11 at 20:06
PacciuPacciu
3,99611030
answered Dec 26 '11 at 20:06
PacciuPacciu
3,99611030
answered Dec 26 '11 at 20:06
PacciuPacciu
3,99611030
3,99611030
$begingroup$
Thanks for your details. But I still do not know why the inequality above (2) holds. Why $c'in L^{infty}$(this is my question, may seems obvious for you). I know $c(u(x))$ has compact support, say, $B_R(0)$, but for $c(t),t=u(x)$, I'm not sure $t$ is in a ball.
$endgroup$
– Y.Z
Dec 27 '11 at 1:39
$begingroup$
Sorry, I misread your post so $c^prime in L^infty$ appeared as an assumption to my eyes... There should be a way to overcome this difficulty, but I have to think about it.
$endgroup$
– Pacciu
Dec 27 '11 at 1:50
add a comment |
$begingroup$
Thanks for your details. But I still do not know why the inequality above (2) holds. Why $c'in L^{infty}$(this is my question, may seems obvious for you). I know $c(u(x))$ has compact support, say, $B_R(0)$, but for $c(t),t=u(x)$, I'm not sure $t$ is in a ball.
$endgroup$
– Y.Z
Dec 27 '11 at 1:39
$begingroup$
Sorry, I misread your post so $c^prime in L^infty$ appeared as an assumption to my eyes... There should be a way to overcome this difficulty, but I have to think about it.
$endgroup$
– Pacciu
Dec 27 '11 at 1:50
$begingroup$
Thanks for your details. But I still do not know why the inequality above (2) holds. Why $c'in L^{infty}$(this is my question, may seems obvious for you). I know $c(u(x))$ has compact support, say, $B_R(0)$, but for $c(t),t=u(x)$, I'm not sure $t$ is in a ball.
$endgroup$
– Y.Z
Dec 27 '11 at 1:39
$begingroup$
Thanks for your details. But I still do not know why the inequality above (2) holds. Why $c'in L^{infty}$(this is my question, may seems obvious for you). I know $c(u(x))$ has compact support, say, $B_R(0)$, but for $c(t),t=u(x)$, I'm not sure $t$ is in a ball.
$endgroup$
– Y.Z
Dec 27 '11 at 1:39
$begingroup$
Sorry, I misread your post so $c^prime in L^infty$ appeared as an assumption to my eyes... There should be a way to overcome this difficulty, but I have to think about it.
$endgroup$
– Pacciu
Dec 27 '11 at 1:50
$begingroup$
Sorry, I misread your post so $c^prime in L^infty$ appeared as an assumption to my eyes... There should be a way to overcome this difficulty, but I have to think about it.
$endgroup$
– Pacciu
Dec 27 '11 at 1:50
add a comment |
3
$begingroup$
First step:
Since $u$ has compact support and $c(0)=0$, and $c(x)$ is continuous, $c(u(x))$ is continuous and has compact support and so it is in $L^2(R^n)$.
Second step:
Take Fourier transform of $$-Delta u+c(u(x))=f(x)$$
Then you can get $$|xi|^2hat u(xi)+widehat{c(u(x))}=hat f(xi)$$
It follows from the first step that $widehat {c(u(x))} in L^2$. Thus $|xi|^2hat u(xi) in L^2$
We also know $hat u in L^2$
So $(1+|xi|^2)hat u(xi) in L^2$
It follows that $u in H^2$.
answered Apr 17 '15 at 0:16
SherrySherry
1,687625
$endgroup$
add a comment |
3
$begingroup$
First step:
Since $u$ has compact support and $c(0)=0$, and $c(x)$ is continuous, $c(u(x))$ is continuous and has compact support and so it is in $L^2(R^n)$.
Second step:
Take Fourier transform of $$-Delta u+c(u(x))=f(x)$$
Then you can get $$|xi|^2hat u(xi)+widehat{c(u(x))}=hat f(xi)$$
It follows from the first step that $widehat {c(u(x))} in L^2$. Thus $|xi|^2hat u(xi) in L^2$
We also know $hat u in L^2$
So $(1+|xi|^2)hat u(xi) in L^2$
It follows that $u in H^2$.
answered Apr 17 '15 at 0:16
SherrySherry
1,687625
$endgroup$
add a comment |
3
3
3
$begingroup$
First step:
Since $u$ has compact support and $c(0)=0$, and $c(x)$ is continuous, $c(u(x))$ is continuous and has compact support and so it is in $L^2(R^n)$.
Second step:
Take Fourier transform of $$-Delta u+c(u(x))=f(x)$$
Then you can get $$|xi|^2hat u(xi)+widehat{c(u(x))}=hat f(xi)$$
It follows from the first step that $widehat {c(u(x))} in L^2$. Thus $|xi|^2hat u(xi) in L^2$
We also know $hat u in L^2$
So $(1+|xi|^2)hat u(xi) in L^2$
It follows that $u in H^2$.
answered Apr 17 '15 at 0:16
SherrySherry
1,687625
$endgroup$
First step:
Since $u$ has compact support and $c(0)=0$, and $c(x)$ is continuous, $c(u(x))$ is continuous and has compact support and so it is in $L^2(R^n)$.
Second step:
Take Fourier transform of $$-Delta u+c(u(x))=f(x)$$
Then you can get $$|xi|^2hat u(xi)+widehat{c(u(x))}=hat f(xi)$$
It follows from the first step that $widehat {c(u(x))} in L^2$. Thus $|xi|^2hat u(xi) in L^2$
We also know $hat u in L^2$
So $(1+|xi|^2)hat u(xi) in L^2$
It follows that $u in H^2$.
answered Apr 17 '15 at 0:16
SherrySherry
1,687625
answered Apr 17 '15 at 0:16
SherrySherry
1,687625
answered Apr 17 '15 at 0:16
SherrySherry
1,687625
answered Apr 17 '15 at 0:16
SherrySherry
1,687625
1,687625
add a comment |
add a comment |
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Have you tried a Fourier transform approach? In Fourier space the problem boils down to prove $mathscr{F}[c(u)] in L^2(mathbb{R}^n)$, which looks more manageable. What do you think?
$endgroup$
– Giuseppe Negro
Dec 26 '11 at 11:01
$begingroup$
@GiuseppeNegro: Thanks. I haven't tried this approach before. Acturally I'm not familiar with Fourier transform, I mean I know some basic facts about it but I've never done any exercises. Here I still do not know how to deal with the $c(u(x))$ under the integral. I think if I can deal with this integral I can also deal with the former approach. Do you have some more details?
$endgroup$
– Y.Z
Dec 26 '11 at 13:24
$begingroup$
@GiuseppeNegro Could you please add more details about the Fourier transform approach? Thanks!
$endgroup$
– Sherry
Apr 10 '15 at 6:13
$begingroup$
@Sherry: I am afraid I cannot answer right now. I don't remember what was my idea back in '11, and frankly, there is a big possibility that it was just plain wrong.
$endgroup$
– Giuseppe Negro
Apr 10 '15 at 7:20