Facebook Page
Twitter
RSS
+ Reply to Thread
Page 2 of 2 FirstFirst 12
Results 11 to 18 of 18
  1. MHB Master
    MHB Site Helper
    MHB Math Helper
    evinda's Avatar
    Status
    Offline
    Join Date
    Apr 2013
    Posts
    3,300
    Thanks
    3,255 times
    Thanked
    992 times
    Awards
    MHB Model User Award (2016)  

MHB Model User Award (2014)
    #11 Thread Author
    Quote Originally Posted by I like Serena View Post
    Not for any $\epsilon>0$...
    For any given $x$ in the inside of the ball there exists an $\epsilon>0$ such that for each $\xi$ on the boundary with $|\xi - x_0|\le\delta$ we have $|\xi - x|>\epsilon$.

    So for every given $x$ in the inside of the ball $\sup_{|\xi-x_0| \leq \delta} |P(x,\xi)|$ is bounded.
    Ah I see...

    Can you also explain to me how we get the inequality for $I_2$ ?

  2. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    MHB Coder
    I like Serena's Avatar
    Status
    Online
    Join Date
    Mar 2012
    Location
    Netherlands
    Posts
    6,636
    Thanks
    4,423 times
    Thanked
    12,388 times
    Thank/Post
    1.867
    Awards
    MHB Model Helper Award (2016)  

MHB Best Ideas (2016)  

MHB LaTeX Award (2016)  

MHB Calculus Award (2014)  

MHB Discrete Mathematics Award (Jul-Dec 2013)
    #12
    Quote Originally Posted by evinda View Post
    $P(x, \xi)=\frac{R^2-|x|^2}{w_n R |x-\xi|^n}$

    $I_2=\int_{|\xi-x_0| > \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS$.

    $\forall \epsilon >0 \exists \delta>0$ so that if $|\xi-x_0| \leq \delta$ then $|I_1|< \frac{\epsilon}{2}$.

    $I_2 \leq 2M \frac{R^2-|x|^2}{w_n R \delta^n} \int_{|\xi-x_0|=R} dS=2M \frac{R^2-|x|^2}{\delta^n} R^{n-2} \leq \frac{\epsilon}{2}$ if $x$ is near to $ x_0 $.
    Quote Originally Posted by evinda View Post
    Ah I see...

    Can you also explain to me how we get the inequality for $I_2$ ?
    We can write:
    $$|I_2|=\left|\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS\right| \\
    \le \sup_{\xi \in \partial B_R,\ |\xi-x_0| > \delta} |P(x, \xi)| \cdot \sup_{\xi \in \partial B_R,\ |\xi-x_0| > \delta} |\phi(\xi)-\phi(x_0)| \cdot \left|\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta}dS\right|
    $$
    How might we continue?

  3. MHB Master
    MHB Site Helper
    MHB Math Helper
    evinda's Avatar
    Status
    Offline
    Join Date
    Apr 2013
    Posts
    3,300
    Thanks
    3,255 times
    Thanked
    992 times
    Awards
    MHB Model User Award (2016)  

MHB Model User Award (2014)
    #13 Thread Author
    How can we get a relation for the integral $\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta}dS$ ?

    Also we have that $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$. Does this imply that $\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi)ds=1$ ?

  4. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    MHB Coder
    I like Serena's Avatar
    Status
    Online
    Join Date
    Mar 2012
    Location
    Netherlands
    Posts
    6,636
    Thanks
    4,423 times
    Thanked
    12,388 times
    Thank/Post
    1.867
    Awards
    MHB Model Helper Award (2016)  

MHB Best Ideas (2016)  

MHB LaTeX Award (2016)  

MHB Calculus Award (2014)  

MHB Discrete Mathematics Award (Jul-Dec 2013)
    #14
    Quote Originally Posted by evinda View Post
    How can we get a relation for the integral $\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta}dS$ ?

    Also we have that $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$. Does this imply that $\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi)ds=1$ ?
    Don't we have that:
    $$\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta}dS \le \int_{\xi \in \partial B_R}dS = w_n R^{n-1}$$



    Btw, I don't understand yet why it's supposed to hold that $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$.
    Do you?
    Is it somewhere in your notes?
    Anyway, if it holds, then, since $P(x,\xi) > 0$, we have:
    $$\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi)ds \le \int_{\partial{B_R}} P(x,\xi) ds_{\xi} = 1$$
    It seems we don't need it though.

  5. MHB Master
    MHB Site Helper
    MHB Math Helper
    evinda's Avatar
    Status
    Offline
    Join Date
    Apr 2013
    Posts
    3,300
    Thanks
    3,255 times
    Thanked
    992 times
    Awards
    MHB Model User Award (2016)  

MHB Model User Award (2014)
    #15 Thread Author
    Quote Originally Posted by I like Serena View Post
    Don't we have that:
    $$\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta}dS \le \int_{\xi \in \partial B_R}dS = w_n R^{n-1}$$
    Oh yes, right.

    Quote Originally Posted by I like Serena View Post
    Btw, I don't understand yet why it's supposed to hold that $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$.
    Do you?
    Is it somewhere in your notes?
    We have that if $u \in C^2(B_R(0)) \cap C^1(\overline{B_R}(0))$ is harmonic in $B_R(0)$, then the following formula holds:

    $u(x)=\frac{R^2-|x|^2}{w_n R} \int_{\partial{B_R}} \frac{u}{|x-\xi|^n} ds_{\xi}$.

    For $u=1$ we get that $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$.

    Quote Originally Posted by I like Serena View Post
    Anyway, if it holds, then, since $P(x,\xi) > 0$, we have:
    $$\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi)ds \le \int_{\partial{B_R}} P(x,\xi) ds_{\xi} = 1$$
    It seems we don't need it though.
    So then we have $\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi) |\phi(\xi)-\phi(x_0)| dS \leq \sup_{\xi \in \partial{B_R}, |\xi-x_0|> \delta} |\phi(\xi)-\phi(x_0)| \int_{|\xi-x_0|>\delta, \xi \in \partial{B_R}} P(x, \xi) dS \leq 2M \int_{\partial{B_R}} P(x,\xi) dS=2M $


    So we have to pick $4M$ as $\epsilon$ ?

  6. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    MHB Coder
    I like Serena's Avatar
    Status
    Online
    Join Date
    Mar 2012
    Location
    Netherlands
    Posts
    6,636
    Thanks
    4,423 times
    Thanked
    12,388 times
    Thank/Post
    1.867
    Awards
    MHB Model Helper Award (2016)  

MHB Best Ideas (2016)  

MHB LaTeX Award (2016)  

MHB Calculus Award (2014)  

MHB Discrete Mathematics Award (Jul-Dec 2013)
    #16
    Quote Originally Posted by evinda View Post
    We have that if $u \in C^2(B_R(0)) \cap C^1(\overline{B_R}(0))$ is harmonic in $B_R(0)$, then the following formula holds:

    $u(x)=\frac{R^2-|x|^2}{w_n R} \int_{\partial{B_R}} \frac{u}{|x-\xi|^n} ds_{\xi}$.

    For $u=1$ we get that $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$.
    Ok!

    Quote Quote:
    So then we have $\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi) |\phi(\xi)-\phi(x_0)| dS \leq \sup_{\xi \in \partial{B_R}, |\xi-x_0|> \delta} |\phi(\xi)-\phi(x_0)| \int_{|\xi-x_0|>\delta, \xi \in \partial{B_R}} P(x, \xi) dS \leq 2M \int_{\partial{B_R}} P(x,\xi) dS=2M $

    So we have to pick $4M$ as $\epsilon$ ?
    I don't think so. We cannot make $4M$ as small as we want to.

    Shouldn't we pick $\xi$ sufficiently close to $x_0$ for the proof?
    That is, $|\xi-x_0|<\delta$?

  7. MHB Master
    MHB Site Helper
    MHB Math Helper
    evinda's Avatar
    Status
    Offline
    Join Date
    Apr 2013
    Posts
    3,300
    Thanks
    3,255 times
    Thanked
    992 times
    Awards
    MHB Model User Award (2016)  

MHB Model User Award (2014)
    #17 Thread Author
    Quote Originally Posted by I like Serena View Post
    Ok!



    I don't think so. We cannot make $4M$ as small as we want to.

    Shouldn't we pick $\xi$ sufficiently close to $x_0$ for the proof?
    That is, $|\xi-x_0|<\delta$?
    We pick $|x-x_0|$ to be sufficiently small.

    I think that I understood the inequality of my notes.

    Since $\phi$ is bounded, there is a $M>0$ such that $|\phi| \leq M$.

    So $\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi) |\phi(\xi)-\phi(x_0)| ds \leq 2M \int_{|\xi-x_0|>\delta, \xi \in \partial{B_R}} \frac{R^2-|x|^2}{w_n R |x-\xi|^n} ds < 2M \frac{R^2-|x|^2}{R} \frac{1}{\delta^n} R^{n-1}, \text{ since } |\xi-x_0|> \delta$.

    Since $|x-x_0|$ is sufficiently small , $|x| \to |x_0|=|R|$ and so $\frac{R^2-|x|^2}{R} \frac{1}{\delta^n} R^{n-1} \to 0$ and so it is $\leq \frac{\epsilon}{2}$.

  8. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    MHB Coder
    I like Serena's Avatar
    Status
    Online
    Join Date
    Mar 2012
    Location
    Netherlands
    Posts
    6,636
    Thanks
    4,423 times
    Thanked
    12,388 times
    Thank/Post
    1.867
    Awards
    MHB Model Helper Award (2016)  

MHB Best Ideas (2016)  

MHB LaTeX Award (2016)  

MHB Calculus Award (2014)  

MHB Discrete Mathematics Award (Jul-Dec 2013)
    #18
    Quote Originally Posted by evinda View Post
    We pick $|x-x_0|$ to be sufficiently small.
    Oh yes.

    Quote Quote:
    I think that I understood the inequality of my notes.

    Since $\phi$ is bounded, there is a $M>0$ such that $|\phi| \leq M$.

    So $\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi) |\phi(\xi)-\phi(x_0)| ds \leq 2M \int_{|\xi-x_0|>\delta, \xi \in \partial{B_R}} \frac{R^2-|x|^2}{w_n R |x-\xi|^n} ds < 2M \frac{R^2-|x|^2}{R} \frac{1}{\delta^n} R^{n-1}, \text{ since } |\xi-x_0|> \delta$.

    Since $|x-x_0|$ is sufficiently small , $|x| \to |x_0|=|R|$ and so $\frac{R^2-|x|^2}{R} \frac{1}{\delta^n} R^{n-1} \to 0$ and so it is $\leq \frac{\epsilon}{2}$.
    Good!

Similar Threads

  1. Uniqueness
    By mathmari in forum Differential Equations
    Replies: 1
    Last Post: December 6th, 2015, 15:26
  2. Maximum principle-Uniqueness of solution
    By mathmari in forum Differential Equations
    Replies: 1
    Last Post: May 23rd, 2015, 05:51
  3. Uniqueness
    By onie mti in forum Calculus
    Replies: 0
    Last Post: May 25th, 2014, 06:45
  4. Uniqueness
    By onie mti in forum Differential Equations
    Replies: 0
    Last Post: May 23rd, 2014, 05:37
  5. local uniqueness
    By onie mti in forum Differential Equations
    Replies: 1
    Last Post: April 9th, 2014, 07:35

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards