TITLE: Solutions for $-\Delta v=\lambda v$ on $\Omega=[0,\pi]^3$
QUESTION [1 upvotes]: In Strauss's Partial Differential Equations, the eigenvalue problem
$$-\Delta v=\lambda v,\qquad     v\big|_{\partial \Omega}=0$$
is solved by separating the $x,y,z$ variables: $v=X(x)Y(y)Z(z)$,
$$ \frac{X''}{X}+\frac{Y''}{Y}+\frac{Z''}{Z}=-\lambda$$
The separated BCs are $$X(0)=X(\pi)=Y(0)=Y(\pi)=Z(0)=Z(\pi)=0$$
Here are my questions:

How do we deduce that the solutions are
$$v(x,y,z)=\sin lx\sin my \sin nz$$
where $$l^2+m^2+n^2=\lambda\quad (1\leq m,l,n<\infty)$$
Why the shape of $\Omega=[a,b]\times[c,d]\times[e,f]$ is needed for this kind of method?

REPLY [2 votes]: From $$\frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}+\frac{Z''(z)}{Z(z)}=-\lambda,$$ we get that exists $(c_1,c_2,c_3)$ such that
$$\frac{X''(x)}{X(x)} =c_1, \quad\frac{Y''(y)}{Y(y)}=c_2, \quad\frac{Z''(z)}{Z(z)}=c_3$$ and $c_1+c_2+c_3=-\lambda$. You will get the result by solving those differential equations.
We have to define $X$, $Y$ and $Z$ and these functions have to be defined on intervals, which explains the  shape of $\Omega$.


TITLE: Chain rules and calculus
QUESTION [0 upvotes]: I've forgotten simple chain rules of calculus.
Let $G(x,y) = x - \frac{F(x,y)}{1+F(x,y)}$ where $F(x,y) = h(x,y) \frac{x}{1-x}$. $x$ and $y$ are independent. What is the second derivative of $G$ wrt $h$?

REPLY [0 votes]: Let $p=h(x,y)$ and $q=F(x,y).$ Then we have that $$q=\frac{px}{1-x}.$$ Thus, we have that $$r=G(x,y)=x-\frac{q}{1+q}=x-\frac{\frac{px}{1-x}}{1+\frac{px}{1-x}}=x-\frac{px}{1-x+px},$$ which is equal to $$x-\frac{1}{1+\frac{1-x}{px}},$$ provided $px\ne 0.$
Therefore, to find the second derivative of $r$ with respect to $p,$ we consider $x$ a function of $p$ and $y,$ with these last variables now independent. Then $r$ depends on $p,\,y.$ Hence we have $$r'_p=x'_p-\frac{\left(\frac{1-x}{px}\right)'_p}{\left(1+\frac{1-x}{px}\right)^2},$$ or $$x'_p-\frac{\frac{1}{p^2}-\frac{1}{xp^2}-\frac{x'_p}{px^2}}{\left(1+\frac{1-x}{px}\right)^2}.$$ Simplify this and go for the second derivative.
If $px=0,$ then we have that $r=x,$ and therefore we have $r'_p=x'_p$ for this case, so that $r''_p=x''_p.$