I was given the following exercise and wasn't really able to make heads-or-tails out of it. It goes like so:
Let $f: \mathbb{R}^3 \rightarrow \mathbb{R}$ be twice differentiable continuously, and statisfy the following:
\begin{align*}
\quad & \forall (x,y,z) \in \mathbb{R}^3 \\
(*) \quad & \frac{\partial f}{\partial y}(x,y,z) = \frac{\partial^2 f}{\partial x^2}(x,y,z)
\end{align*}
Then define:
\begin{gather*}
g: \mathbb{R}^2 \rightarrow \mathbb{R} \\
(x,y) \rightarrow \int_0^y f(x,y-z,z)dz
\end{gather*}
Prove for all $(x,y) \in \mathbb{R}^2$
\begin{gather*}
\frac{\partial g}{\partial y}(x,y) = \frac{\partial^2 g}{\partial x^2}(x,y) + f(x,0,y)
\end{gather*}
I do know this question must involve Leibniz's Rule of Integration Under The Integral Sign, yet I am not sure how this is applicable here.
The only sensible step to take is to fix $g$ for both variables and derive using Leibniz's rule (as we know $g$ is continuously differentiable twice in $\mathbb{R}^3$).
So for the second variable:
\begin{gather*}
\frac{\partial g}{\partial y} = f(y,y,z)\cdot\frac{d}{dy}y + \int_0^y\frac{\partial f}{\partial y}(x,y-z,z)dz
\end{gather*}
For the first variable it seemed to me that I should derive twice using Leibniz's rule so I can apply $(*)$, and I got:
\begin{gather*}
\frac{\partial^2 g}{\partial x^2} = \int_0^y\frac{\partial^2 f}{\partial x^2}(x,y-z,z)dz
\end{gather*}
We can easily see that by applying $(*)$ we get close to the required term, yet it is not quite there.
I assume my evaluation of $\dfrac{\partial g}{\partial y}$ is incorrect.
So first and foremost, how would one comprehensively justify using Leibniz's rule here in the preceding manner – it is not your usual case, there are two parameters in the integral represented by $g$. Is it even justifiable here?
And secondly, did I evaluate $\dfrac{\partial g}{\partial y}$ wrongly, or have I just missed something and the evaluation is incomplete?
Any hint or explanation would be extremely appreciated.
Thank you so much, and have a great day!
Best Answer
The way I learned differentiation under the integral sign is as follows: $$ \partial_x \int_{a(x)}^{b(x)} F(x,y) dx = F(b(x),y)b'(x) - F(a(x),y) a'(x) + \int_{a(x)}^{b(x)} \partial_x F(x,y) dx$$ So applying this to $g$: \begin{align*} \partial_y g(x,y) &= \partial_y \int_0^y f(x,y-z,z)dz \\ &= \frac{dy}{dy} f(x,y-y,y) - \frac{d0}{dy} f(x,y-0,z) + \int_0^y \partial_y f(x,y-z,z) dz \\ &= f(x,0,y) + \int_0^y \partial_x^2 f(x,y-z,z) dz \\ \end{align*} With respect to your comment on why it's not $f(x,y,z)$, you can see above that $f(x,y-z,z)\big|_{z=y} = f(x,0,y)$.
Now we show that the integral in the last line is equal to $\partial_x^2 g$. Note that because the bounds of integration don't depend on $x$ we have \begin{align*} \partial_x g(x,y) &= \partial_x \int_0^y f(x,y-z,z)dz \\ &= \frac{dy}{dx} f(x,y-y,y) - \frac{d0}{dx} f(x,y-0,z) + \int_0^y \partial_x f(x,y-z,z)dz \\ &= \int_0^y \partial_x f(x,y-z,z)dz \end{align*} Doing the same calculation with the second $\partial_x$ shows that it goes through too, so we have \begin{align*} \partial_y g(x,y) &= f(x,0,y) + \int_0^y \partial_x^2 f(x,y-z,z) dz \\ &= f(x,0,y) + \partial_x^2 \int_0^y f(x,y-z,z) dz \\ &= f(x,0,y) + \partial_x^2 g(x,y) \end{align*}
I hope that helps.