Results 1 to 4 of 4

Math Help - Autonomous PDE admits a solution of exponential order?

  1. #1
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2

    Autonomous PDE admits a solution of exponential order?

    Dear MHF members,

    consider the PDE IVP
    \dfrac{\partial\Phi}{\partial x}+\dfrac{\partial\Phi}{\partial t}=0 for (x,t)\in\Omega:=\{(x,t)\in\mathbb{R}^{2}:\ x\geq t\geq0\}
    under the initial condition
    \Phi(x,0)=\Phi_{0}(x) for x\geq0, where \Phi_{0}\in\mathrm{C}^{1} is a function of exponential order,
    i.e., there exists M,a>0 such that |\Phi_{0}(x)|\leq M\mathrm{e}^{ax} for all x\geq0.
    Prove that for each fixed x\geq0, the solution \Phi(x,\cdot) is of exponential order, i.e.,
    there exists N_{x},b_{x}>0 such that |\Phi(x,t)|\leq N_{x}\mathrm{e}^{b_{x}t} for all t\geq0.

    Note. Do not use the solution \Phi(x,t)=\Phi_{0}(x-t) directly to show it.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    At my house.
    Posts
    527
    Thanks
    7
    Expand by Taylor in {x\geq t>0} to get \Phi(x,t)=\Phi_0(x)+\Phi_x(x,0)x+R(\xi,0)x^2\leq \M{\rm e}^{ax}+\Phi_x(x,0)x+g(x), with the obvious choices. Invoking \Phi_x=-\Phi_t we obtain |\Phi_t|\leq \frac{1}{x}(|\Phi|+g(x)) and the result follows by using Gronwall's inequality.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2
    Quote Originally Posted by Rebesques View Post
    Expand by Taylor in {x\geq t>0} to get \Phi(x,t)=\Phi_0(x)+\Phi_x(x,0)x+R(\xi,0)x^2\leq \M{\rm e}^{ax}+\Phi_x(x,0)x+g(x), with the obvious choices. Invoking \Phi_x=-\Phi_t we obtain |\Phi_t|\leq \frac{1}{x}(|\Phi|+g(x)) and the result follows by using Gronwall's inequality.
    Rebesques thanks for the reply, but it seems that there is a mistake with your Taylor's formula, it should read as follows.
    \Phi(x,t)=\Phi_{0}(x)+\Phi_{\mathrm{t}}(x,0)t+\dis  playstyle\int_{0}^{t}(t-u)\Phi_{\mathrm{tt}}(x,u)\mathrm{d}u.
    However, as the function has first order derivatives, we can not expand it up to its second-order derivative, i.e., we can only use
    \Phi(x,t)=\Phi_{0}(x)+\displaystyle\int_{0}^{t}\Ph  i_{\mathrm{t}}(x,u)\mathrm{d}u
    ..........._ =\Phi_{0}(x)-\displaystyle\int_{0}^{t}\Phi_{\mathrm{x}}(x,u)\ma  thrm{d}u.
    But I dont know if we can move somewhere from here...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2

    Lightbulb Re: Autonomous PDE admits a solution of exponential order?

    If \Phi_{0}\in\mathrm{C}^{\infty} has the series expansion
    \Phi_{0}(x)=\sum_{k=0}^{\infty}a_{k}\frac{x^{k}}{k  !}
    satisfying |a_{k}|\leq Mr^{k} for all k\in\mathbb{N}_{0},
    where M,r>0, then \Phi is of exponential order.
    One can show by direct substitution that
    \Phi(x,t)=\sum_{k=0}^{\infty}a_{k}\frac{(x-t)^{k}}{k!},
    which yields
    |\Phi(x,t)|\leq M\sum_{k=0}^{\infty}r^{k}\frac{x^{k}}{k!}=M\mathrm  {e}^{rx} for all (x,t)\in\Omega.
    Also if \{a_{k}\} is a nonnegative sequence, then we obtain
    \Phi(x,t)\leq\sum_{k=0}^{\infty}a_{k}\frac{x^{k}}{  k!}=\Phi_{0}(t) for all (x,t)\in\Omega,
    which implies \Phi is of exponential order too.
    But I dont know how we can show this only with \Phi_{0}\in\mathrm{C}^{1}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Show that the solution is of exponential order
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: September 24th 2010, 02:14 AM
  2. first order autonomous eq
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: June 1st 2010, 06:24 AM
  3. simple question about autonomous first-order DEs
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: May 12th 2010, 11:39 PM
  4. Replies: 2
    Last Post: February 23rd 2009, 05:54 AM

Search Tags


/mathhelpforum @mathhelpforum