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Very simple: second order derivative in wave equation

  1. Dec 2, 2014 #1
    In the equation regarding an array of masses connected by springs in wikipedia the step from
    $$\frac {u(x+2h,t)-2u(x+h,t)+u(x,t)} { h^2}$$

    To
    $$\frac {\partial ^2 u(x,t)}{\partial x^2}$$

    By making ##h \to 0## is making me wonder how is it rigorously demonstrated. I mean:
    $$\frac {\partial ^2 u(x,t)}{\partial x^2} = \lim_{h\to 0} \frac {u_x(x+h,t)-u_x(x,t)} { h}$$
    But we have
    $$\lim_{h\to 0}\frac {u(x+2h,t)-2u(x+h,t)+u(x,t)} { h^2}=\lim_{h\to 0}\frac{\frac {u(x+2h)-u(x+h)}{h}-\frac {u(x+h)-u(x)}{h}}{h}$$
    How do we demonstrate that these two expressions are equal?
     
  2. jcsd
  3. Dec 3, 2014 #2
    By the way, I am using $$u_x(x,t) = \frac {\partial u} {\partial x}$$

    I keep looking at it and it has to do with continuity, differentiability and properties of limits... but I cannot figure it out. Once again I would like to know the mathematical justification that states that doing the following is safe (I've get rid of the time dependency for simplicity):

    $$\lim_{h\to 0}\frac{\frac {u(x+2h)-u(x+h)}{h}-\frac {u(x+h)-u(x)}{h}}{h} =
    \lim_{h\to 0}\frac{\lim_{d\to 0}\frac {u(x+h+d)-u(x+h)}{d}-\lim_{d\to 0}\frac {u(x+d)-u(x)}{d}}{h}$$
     
  4. Dec 7, 2014 #3

    Erland

    User Avatar
    Science Advisor

    So, we assume that ##u(x)## is a twice continuously differentiable function in an interval ##I## which contains ##a##, and we want to prove that

    ##\lim_{h\to 0}\frac {u(a+2h)-2u(a+h)+u(a)}{h^2}=u''(a).##

    Fix ##h## and put ##\phi(x)=u(x+h)-u(x)## for all ##x## such that both ##x## and ##x+h## lie in ##I##. Assuming that ##a+2h\in I##, we obtain, by the Mean Value Theorem:

    ##u(a+2h)-2u(a+h)+u(a)=\phi(a+h)-\phi(a)=\phi'(a+\theta h)h=(u'(a+\theta h +h) - u'(a+\theta h))h,##

    for some ##\theta \in (0,1)##.

    Then, you can use the Mean Value Theorem again and use that ##u''## is continuous at ##a## to obtain the desired result. I leave this as an exercise.
     
    Last edited: Dec 7, 2014
  5. Dec 8, 2014 #4
    I see. Then, using the Mean Value Theorem again

    ##(a+2h)-2u(a+h)+u(a)=(u'(a+\theta h +h) - u'(a+\theta h))h = u''(a+\theta h + \lambda h) h^2##

    for some ##\lambda \in (0,1)##, and using that ##u''(x)## is continuous,

    ##\lim_{h\to 0} \frac {u(a+2h)-2u(a+h)+u(a)}{h^2}=\lim_{h\to 0} u''(a+(\theta + \lambda) h) =u''(a) .##


    Thank you very much!
     
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