Facebook Page
Twitter
RSS
+ Reply to Thread
Page 1 of 2 12 LastLast
Results 1 to 10 of 13
  1. MHB Master
    MHB Site Helper
    mathmari's Avatar
    Status
    Offline
    Join Date
    Apr 2013
    Posts
    3,815
    Thanks
    3,029 times
    Thanked
    995 times
    Awards
    MHB Model User Award (2017)  

MHB Chat Room Award (2015)  

MHB Model User Award (2015)  

MHB LaTeX Award (2015)
    #1
    Hey!!

    Find the solution of the problem $$u_t(x, t)-u_{xx}(x, t)=0, 0<x<1, t>0 \tag {*} \\ u(0, t)=0, t>0 \\ u_x(1,t)+u_t(1,t)=0, t>0$$

    I have done the following:

    We are looking for solutions of the form $$u(x, t)=X(x) \cdot T(t)$$

    $$u(0, t)=X(0) \cdot T(t)=0 \Rightarrow X(0)=0 \\ X'(1) \cdot T(t)+X(1) \cdot T'(t)=0 \Rightarrow \frac{X'(1)}{X(1)}=-\frac{T'(t)}{T(t)}$$

    $$(*) \Rightarrow X(x) \cdot T'(t)-X''(x) \cdot T(t)=0 \\ \Rightarrow \frac{X(x) \cdot T'(t)}{X(x) \cdot T(t)}-\frac{X''(x) \cdot T(t)}{X(x) \cdot T(t)}=0 \\ \Rightarrow \frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}=-\lambda$$

    So, we get the following two problems:
    $$\left.\begin{matrix}
    X''(x)+\lambda X(x)=0, 0<x<1\\
    X(0)=0 \\
    \frac{X'(1)}{X(1)}=\lambda \Rightarrow X'(1)-\lambda X(1)=0
    \end{matrix}\right\}(1)
    $$

    $$\left.\begin{matrix}
    T'(t)+\lambda T(t)=0, t>0
    \end{matrix}\right\}(2)$$

    For the problem $(1)$ we do the following:

    The characteristic polynomial is $d^2+\lambda=0$.
    • $\lambda <0$ :

      $X(x)=c_1 \sinh (\sqrt{-\lambda} x)+c_2 \cosh (\sqrt{-\lambda}x)$

      Using the initial values we get that $X(x)=0$, trivial solution.
    • $\lambda=0$ :

      $X(x)=c_1 x+c_2$

      Using the initial values we get that $X(x)=0$, trivial solution.
    • $\lambda >0$ :

      $X(x)=c_1 cos (\sqrt{\lambda}x)+c_2 \sin (\sqrt{\lambda}x)$

      $X(0)=0 \Rightarrow c_1=0 \Rightarrow X(x)=c_2=\sin (\sqrt{\lambda}x)$

      $X'(1)-\lambda X(1)=0 \Rightarrow \tan (\sqrt{\lambda})=\frac{1}{\sqrt{\lambda}}$


    That means that the eigenvalue problem $(1)$ has only positive eigenvalues $0<\lambda_1 < \lambda_2 < \dots < \lambda_k < \dots $ that are the positive roots of the equation $\tan \sqrt{x}=\frac{1}{\sqrt{x}}$.

    Is this correct??

    Why can we say that the number of the eigenvalues is countable ??

    How can we show that $$\lim_{k \rightarrow +\infty} \frac{\sqrt{\lambda_k}}{k \pi}=1$$ ??

  2. # ADS
    Circuit advertisement
    Join Date
    Always
    Posts
    Many
     

  3. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    MHB Coder
    Klaas van Aarsen's Avatar
    Status
    Offline
    Join Date
    Mar 2012
    Location
    Leiden
    Posts
    8,403
    Thanks
    5,435 times
    Thanked
    14,614 times
    Thank/Post
    1.739
    Awards
    MHB Pre-University Math Award (2018)  

MHB Model Helper Award (2017)  

MHB Best Ideas (2017)  

MHB Analysis Award (2017)  

MHB Calculus Award (2017)
    #2
    Hi!!!

    Quote Originally Posted by mathmari View Post
    Is this correct??
    Yep.

    Quote Quote:
    Why can we say that the number of the eigenvalues is countable ??
    Take a look at of $\tan y$ and $\frac 1 y$.

    The tangent has a period of $\pi$ and in each period it has exactly 1 intersection with $\frac 1 y$.
    Since there are a countable number of periods of the tangent, that means that the number of solutions is also countable.


    Quote Quote:
    How can we show that $$\lim_{k \rightarrow +\infty} \frac{\sqrt{\lambda_k}}{k \pi}=1$$ ??
    In the same graph, you can see that each successive intersection of $\tan y$ and $\frac 1 y$ is closer and closer to where $\tan y$ intersects the axis. That is at $k\pi$.
    So those intersections, that correspond to $\sqrt {\lambda_k}$, approach $k\pi$.
    Thus:
    $$\lim_{k\to \infty} \frac{\sqrt {\lambda_k}}{k\pi} = 1$$

  4. MHB Master
    MHB Site Helper
    mathmari's Avatar
    Status
    Offline
    Join Date
    Apr 2013
    Posts
    3,815
    Thanks
    3,029 times
    Thanked
    995 times
    Awards
    MHB Model User Award (2017)  

MHB Chat Room Award (2015)  

MHB Model User Award (2015)  

MHB LaTeX Award (2015)
    #3 Thread Author
    I understand!!


    Quote Originally Posted by I like Serena View Post
    Take a look at of $\tan y$ and $\frac 1 y$.

    The tangent has a period of $\pi$ and in each period it has exactly 1 intersection with $\frac 1 y$.
    Since there are a countable number of periods of the tangent, that means that the number of solutions is also countable.
    When we have an other problem, and we find that the eigenvalues are the positive roots of the equation $\tan (\sqrt{\lambda } \pi)=\frac{1}{\sqrt{\lambda}}$, is the argument the same??



    I have also an other question...

    Quote Originally Posted by mathmari View Post
    $$u(0, t)=X(0) \cdot T(t)=0 \Rightarrow X(0)=0 \\ X'(1) \cdot T(t)+X(1) \cdot T'(t)=0 \Rightarrow \frac{X'(1)}{X(1)}=-\frac{T'(t)}{T(t)}$$
    To divide with $X(1)$ are we sure that it isn't zero??

  5. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    MHB Coder
    Klaas van Aarsen's Avatar
    Status
    Offline
    Join Date
    Mar 2012
    Location
    Leiden
    Posts
    8,403
    Thanks
    5,435 times
    Thanked
    14,614 times
    Thank/Post
    1.739
    Awards
    MHB Pre-University Math Award (2018)  

MHB Model Helper Award (2017)  

MHB Best Ideas (2017)  

MHB Analysis Award (2017)  

MHB Calculus Award (2017)
    #4
    Quote Originally Posted by mathmari View Post
    I understand!!
    Good!


    Quote Quote:
    When we have an other problem, and we find that the eigenvalues are the positive roots of the equation $\tan (\sqrt{\lambda } \pi)=\frac{1}{\sqrt{\lambda}}$, is the argument the same??
    Yes.
    In this case the limit would be $ \displaystyle \lim_{k\to\infty} \frac{\sqrt{\lambda_k}}{k} = 1$


    Quote Quote:
    I have also an other question...

    To divide with $X(1)$ are we sure that it isn't zero??
    No, we can't be sure.

  6. MHB Master
    MHB Site Helper
    mathmari's Avatar
    Status
    Offline
    Join Date
    Apr 2013
    Posts
    3,815
    Thanks
    3,029 times
    Thanked
    995 times
    Awards
    MHB Model User Award (2017)  

MHB Chat Room Award (2015)  

MHB Model User Award (2015)  

MHB LaTeX Award (2015)
    #5 Thread Author
    Quote Originally Posted by I like Serena View Post
    No, we can't be sure.
    So, can't we do that or do we have to suppose that $X(1) \neq 0$ ??

  7. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    MHB Coder
    Klaas van Aarsen's Avatar
    Status
    Offline
    Join Date
    Mar 2012
    Location
    Leiden
    Posts
    8,403
    Thanks
    5,435 times
    Thanked
    14,614 times
    Thank/Post
    1.739
    Awards
    MHB Pre-University Math Award (2018)  

MHB Model Helper Award (2017)  

MHB Best Ideas (2017)  

MHB Analysis Award (2017)  

MHB Calculus Award (2017)
    #6
    Quote Originally Posted by mathmari View Post
    So, can't we do that or do we have to suppose that $X(1) \neq 0$ ??
    Easiest is to avoid dividing by $X(1)$.
    Otherwise we should distinguish two cases. One where $X(1) \neq 0$ and one where $X(1) = 0$.

  8. MHB Master
    MHB Site Helper
    mathmari's Avatar
    Status
    Offline
    Join Date
    Apr 2013
    Posts
    3,815
    Thanks
    3,029 times
    Thanked
    995 times
    Awards
    MHB Model User Award (2017)  

MHB Chat Room Award (2015)  

MHB Model User Award (2015)  

MHB LaTeX Award (2015)
    #7 Thread Author
    Quote Originally Posted by I like Serena View Post
    Easiest is to avoid dividing by $X(1)$.
    Otherwise we should distinguish two cases. One where $X(1) \neq 0$ and one where $X(1) = 0$.
    How could we avoid dividing by $X(1)$ ??

  9. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    MHB Coder
    Klaas van Aarsen's Avatar
    Status
    Offline
    Join Date
    Mar 2012
    Location
    Leiden
    Posts
    8,403
    Thanks
    5,435 times
    Thanked
    14,614 times
    Thank/Post
    1.739
    Awards
    MHB Pre-University Math Award (2018)  

MHB Model Helper Award (2017)  

MHB Best Ideas (2017)  

MHB Analysis Award (2017)  

MHB Calculus Award (2017)
    #8
    Quote Originally Posted by mathmari View Post
    How could we avoid dividing by $X(1)$ ??
    When do you need to?

  10. MHB Master
    MHB Site Helper
    mathmari's Avatar
    Status
    Offline
    Join Date
    Apr 2013
    Posts
    3,815
    Thanks
    3,029 times
    Thanked
    995 times
    Awards
    MHB Model User Award (2017)  

MHB Chat Room Award (2015)  

MHB Model User Award (2015)  

MHB LaTeX Award (2015)
    #9 Thread Author
    Quote Originally Posted by I like Serena View Post
    When do you need to?
    We have the problem $$u_t(x, t)-u_{xx}(x, t)=0, 0<x<1, t>0 \tag {*} \\ u(0, t)=0, t>0 \\ u_x(1,t)+u_t(1,t)=0, t>0$$

    We are looking for solutions of the form $$u(x, t)=X(x) \cdot T(t)$$

    $$u(0, t)=X(0) \cdot T(t)=0 \Rightarrow X(0)=0 \\ u_x(1,t)+u_t(1,t)=0 \Rightarrow X'(1) \cdot T(t)+X(1) \cdot T'(t)=0 \Rightarrow \frac{X'(1)}{X(1)}=-\frac{T'(t)}{T(t)}$$

    $$(*) \Rightarrow X(x) \cdot T'(t)-X''(x) \cdot T(t)=0 \\ \Rightarrow \frac{X(x) \cdot T'(t)}{X(x) \cdot T(t)}-\frac{X''(x) \cdot T(t)}{X(x) \cdot T(t)}=0 \\ \Rightarrow \frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}=-\lambda$$

    So, we get the following two problems:
    $$\left.\begin{matrix}
    X''(x)+\lambda X(x)=0, 0<x<1\\
    X(0)=0 \\
    \frac{X'(1)}{X(1)}=\lambda \Rightarrow X'(1)-\lambda X(1)=0
    \end{matrix}\right\}(1)
    $$

    $$\left.\begin{matrix}
    T'(t)+\lambda T(t)=0, t>0
    \end{matrix}\right\}(2)$$


    If we don't divide by $X(1)$ how else could we use the condition $u_x(1,t)+u_t(1,t)=0 \Rightarrow X'(1) \cdot T(t)+X(1) \cdot T'(t)=0$ ??

  11. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    MHB Coder
    Klaas van Aarsen's Avatar
    Status
    Offline
    Join Date
    Mar 2012
    Location
    Leiden
    Posts
    8,403
    Thanks
    5,435 times
    Thanked
    14,614 times
    Thank/Post
    1.739
    Awards
    MHB Pre-University Math Award (2018)  

MHB Model Helper Award (2017)  

MHB Best Ideas (2017)  

MHB Analysis Award (2017)  

MHB Calculus Award (2017)
    #10
    Quote Originally Posted by mathmari View Post
    If we don't divide by $X(1)$ how else could we use the condition $u_x(1,t)+u_t(1,t)=0 \Rightarrow X'(1) \cdot T(t)+X(1) \cdot T'(t)=0$ ??
    Substitute $T'(t)=-\lambda T(t)$?

    Afterwards, we can divide by $T(t)$.
    If we assume that the solution for $u$ is non-trivial, $T(t)$ must be non-zero for at least one value of $t$.

Similar Threads

  1. Eigenvalues
    By Yankel in forum Linear and Abstract Algebra
    Replies: 7
    Last Post: January 12th, 2015, 02:32
  2. Eigenvalues
    By mathmari in forum Linear and Abstract Algebra
    Replies: 2
    Last Post: December 30th, 2014, 04:30
  3. Properties of probabilities
    By Kudasai in forum Advanced Probability and Statistics
    Replies: 0
    Last Post: November 7th, 2013, 14:49
  4. Topological properties
    By Siron in forum Topology and Advanced Geometry
    Replies: 2
    Last Post: January 5th, 2013, 11:51
  5. [SOLVED] Eigenvalues
    By Sudharaka in forum Pre-Algebra and Algebra
    Replies: 2
    Last Post: May 27th, 2012, 09:02

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