# Very simple: second order derivative in wave equation

1. Dec 2, 2014

### mr_sparxx

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. Dec 3, 2014

### mr_sparxx

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):

$$\lim_{h\to 0}\frac{\frac {u(x+2h)-u(x+h)}{h}-\frac {u(x+h)-u(x)}{h}}{h} = \lim_{h\to 0}\frac{\lim_{d\to 0}\frac {u(x+h+d)-u(x+h)}{d}-\lim_{d\to 0}\frac {u(x+d)-u(x)}{d}}{h}$$

3. Dec 7, 2014

### Erland

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
4. Dec 8, 2014

### mr_sparxx

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!