How does this limit turn into partial derivatives?

Question

$$ \lim_{a\to0} \frac{1}{a} \left[ \left(\frac{y(x+a,t) - y(x,t)}{a}\right)^3 - \left(\frac{y(x,t) - y(x-a,t)}{a}\right)^3 \right]= \frac{\partial }{\partial x}\left[ \left( \frac{\partial y}{\partial x}\right )^3\right] $$ Where $y(x,t)$ is a function of $x$ and $t$. I encountered this equation when seeing a derivation of a nonlinear wave equation for a string, however I'm not seeing the jump from the LHS to the RHS. If this equation is true, how is it so? If you can point me to a topic/resource where similar limits are carried out, that would be greatly appreciated.

Answer 1

I would do it this way. First I will use notations that are comfortable to me. For all points $(x,t) \in \Bbb{R}^2$, the partial derivative of $y$ with respect to the first variable at $(x,t)$ is defined as

$$D_1y(x,t) := \lim_{z \to 0}\frac{y(x+z,t) - y(x,t)}{z}$$

For all $a \neq 0$, let $\phi(k) = y(x+k,t)$. Then if we do some factorization, we see that

\begin{align}&\quad \ \frac{1}{a}\bigg[ \bigg( \frac{y(x+a,t)-y(x,t)}{a} \bigg)^3 - \bigg(\frac{y(x,t)-y(x-a,t)}{a} \bigg)^3 \bigg]\\ &=\frac{1}{a}\bigg[ \bigg(\underbrace{\frac{\phi(a)-\phi(0)}{a}}_{\alpha}\bigg)^3-\bigg(\underbrace{\frac{\phi(0)-\phi(-a)}{a}}_{\beta}\bigg)^3\bigg]\\[0.5em] &= \underbrace{\frac{\phi(a)-2\phi(0)+\phi(-a)}{a^2}}_{\gamma}(\alpha^2+\alpha\beta+\beta^2)\end{align}

Suppose $D_1(D_1y)= D_{1,1}y$ exists everywhere. Then $$\frac{\partial}{\partial x}\bigg[ \bigg(\frac{\partial y}{\partial x} \bigg)^3\bigg] = D_1[(D_1y)^3]=3[D_1y]^2 D_{1,1}y$$ We would argue that as $a \to 0$, $\alpha, \beta \to D_1y$ and $\gamma \to D_{1,1}y$. As the "sublimits" converge, we can perform arithmetic on them and show that the original limit is indeed $3[D_1y]^2 D_{1,1}y= D_1[(D_1y)^3]$.

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$\alpha, \beta \to D_1y$:

It is easy to see that $\phi'(k) = D_1y(x+k,t)$. If you cannot see that, let's go to the definition. Write $x_1 = x+k$.

\begin{align} \phi'(k)&:= \lim_{w \to 0} \frac{y(x+(w+k),t)-y(x+k,t)}{w}\tag{1}\\ &= a\lim_{w \to 0} \frac{y(x_1+w,t)-y(x_1,t)}{wa}\\ \phi'(k)&= D_1y(x_1,t) = D_1y(x+k,t)\tag{2}\end{align}

Therefore, by the mean value theorem, $\phi(a)-\phi(0)= a\phi'(c_1)$ for some $c_1 \in (0,a)$ and $\phi(0)-\phi(-a)= a\phi'(c_2)$ for some $c_2 \in (-a,0)$. Therefore, $$\alpha = \frac{\phi(a)-\phi(0)}{a}= \frac{\phi'(c_1)a}{a}=D_1y(x+c_1,t) $$ By continuity of $D_1y$, we see that as $a \to 0$ and $c_1, c_2 \in (-a,a)$, $\alpha = D_1y(x+c_1,t) \to D_1y(x,t)$. Similarly, $\beta = D_1y(x+c_2,t) \to D_1y(x,t)$.

(Actually $\alpha \to D_1y(x,t)$ directly follows from $(1)$ and $(2)$ by putting $k =0$, and $\alpha \to D_1y(x,t)$ follows from $(1)$ by letting $w' = -w$ and the result $(2)$.)

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$\gamma \to D_{1,1}y$:

We use the fact about second derivative that if $\phi''(k)$ exists, then

$$\lim_{h \to 0} \frac{\phi(k+h)-2\phi(k)+\phi(k-h)}{h^2}$$

exists, and converges to $\phi''(k)$. It is usually proven with L'Hospital's rule. As before, we perform derivative on $\phi'(k)$ again, $$\phi''(k)=\frac{d}{dk} D_1y(x+k,t)= D_1[D_1y(x+k,t)] = D_{1,1}y(x+k,t)$$ Therefore, $$\lim_{a \to 0} \frac{\phi(0+a)-2\phi(0)+\phi(0-a)}{a^2} = D_{1,1}y(x+0,t) = D_{1,1}(x,t)$$ and we are done.

Answer 2

By taking into account the definition of a single variable derivative, given by $$\lim_{h \to 0} \, \frac{f(x+h) - f(x)}{h} = \frac{d f(x)}{dx}.$$

Now, a partial derivative is applied when taking a derivative with respect to one variable of a multi variable function, \begin{align} \lim_{h \to 0+} \frac{f(x + h, t) - f(x, t)}{h} &= \frac{\partial f(x,t)}{\partial x} \\ \lim_{h \to 0-} \frac{f(x,t) - f(x-h,t)}{h} &= \frac{\partial f(x,t)}{\partial x}. \end{align}

Applying this to the question at hand leads to \begin{align} \lim_{a \to 0} \frac{1}{a} \, \left[ \left(\frac{f(x + a, t) - f(x, t)}{h}\right)^{3} - \left(\frac{f(x,t) - f(x-a,t)}{h} \right)^{3} \right] = \frac{\partial}{\partial x} \, \left[ \left(\frac{\partial f(x,t)}{\partial x}\right)^{3} \right] \end{align}