Very simple: second order derivative in wave equation

[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?

 

[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):

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

 

[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.

 

[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!

 

[blue_raver22]

I understand your curiosity about the rigor of mathematical expressions. In this case, the demonstration of the equality between the two expressions can be done using the definition of the second derivative and the limit definition of the derivative.

First, let's rewrite the first expression in terms of the second derivative:
$$\frac {u(x+2h,t)-2u(x+h,t)+u(x,t)} { h^2} = \frac {u(x+2h,t)-u(x+h,t)}{h^2} - \frac {u(x+h,t)-u(x,t)}{h^2}$$
Now, we can use the definition of the second derivative:
$$\frac {u(x+2h,t)-u(x+h,t)}{h^2} = \frac{\partial^2 u}{\partial x^2} (x+h,t)$$
$$\frac {u(x+h,t)-u(x,t)}{h^2} = \frac{\partial^2 u}{\partial x^2} (x,t)$$
Substituting these back into the original expression, we get:
$$\frac {u(x+2h,t)-2u(x+h,t)+u(x,t)} { h^2} = \frac{\partial^2 u}{\partial x^2} (x+h,t) - \frac{\partial^2 u}{\partial x^2} (x,t)$$
Now, using the limit definition of the derivative, we can rewrite this as:
$$\lim_{h\to 0} \frac {u(x+2h,t)-2u(x+h,t)+u(x,t)} { h^2} = \lim_{h\to 0} \frac{\frac{\partial u}{\partial x} (x+h,t) - \frac{\partial u}{\partial x} (x,t)}{h}$$
Which is the same as the second expression you provided. Therefore, it can be demonstrated that the two expressions are equal.

I hope this helps to clarify the rigor behind the step in the wave equation. Keep questioning and seeking understanding in your scientific pursuits.