Let $S^+$ be the upper hemisphere of the standard sphere in $\mathbb R^3$ and $b$ -- the boundary of $S^+$ (the equator). 
Let $b'$ be a small isometric deformation of $b$ (a nearby curve of the same length). Is there a smooth (or of regularity higher than $C^2$) isometric immersion of $S^+$ which bounds $b'$?


I'm also interested in the infinitesimal version, i.e. if $X$ is a vector field along $b$ that does not stretch $b$, can $X$ be extended to a vector field along $S^+$ that is the derivative of a family of isometric immersions of $S^+$?


In other words: can an open set in the space of configurations of $b$ coincide with the subspace of configurations of $b$ induced from isometric immersions of $S^+$? 



Bending the hemisphere
The answer to the infinitesimal version is 'no', which makes it very unlikely that the answer to the isometric deformation version is 'yes'. Here is how one can see this:


One can parametrize the upper hemisphere by the unit disk $x^2+y^2\le 1$ conformally by the well-known formula 
$$
F(x,y) = \left(\frac{2x}{1+x^2+y^2},\frac{2y}{1+x^2+y^2},\frac{1-x^2-y^2}{1+x^2+y^2}\right).
$$
Let $V(x,y)$ be a vector field along the image of $F$, which can be expressed uniquely in the form
$$
V(x,y) = a(x,y)\,F\_x(x,y) + b(x,y)\,F\_y(x,y) + c(x,y)\,F(x,y)
$$
for some functions $a$, $b$, and $c$ on the unit disk. Then the condition that $V$ determine an infinitesimal isometric deformation, i.e., $\mathrm{d}F \cdot \mathrm{d}V = 0$, is easily seen to be the system of $3$ equations
$$
a\_x-b\_y = 0,\qquad a\_y+b\_x = 0,\qquad c = \frac{2(xa+yb)}{1+x^2+y^2} - a\_x\,.
$$
Thus, $a+ib$ must be a holomorphic function of $z = x+iy$, and $c$ is determined in explicitly in terms of $a$ and $b$. In particular, 
$$
a + ib = \sum\_{n=0}^{\infty} c\_n\,z^n
$$
for some complex coefficients $c\_n$, $n\ge 0$.


Now, restrict everything to the boundary of the disk, set
$$
E\_0(\theta) = F(\cos\theta,\sin\theta),\qquad
E\_1(\theta) = F\_x(\cos\theta,\sin\theta),\qquad
E\_2(\theta) = F\_y(\cos\theta,\sin\theta),
$$
so that $E\_0$, $E\_1$, and $E\_2$ are an orthonormal frame field along the boundary curve (i.e., the equator). Let $W(\theta)$ be a vector field along the boundary. It can be written uniquely in the form
$$
W(\theta) = f\_0(\theta)\,E\_0(\theta)+f\_1(\theta)\,E\_1(\theta)+f\_2(\theta)\,E\_2(\theta)
$$
for some $2\pi$-periodic functions of $\theta$. The condition that $W$ furnish an infinitesimal isometric deformation of the boundary curve, i.e., $\mathrm{d}E\_0\cdot\mathrm{d}W = 0$, is easily seen to be the differential equation
$$
f\_0(\theta) = \frac{\mathrm{d}}{\mathrm{d}\theta}\bigl(\sin\theta\,f\_1(\theta)-\cos\theta\,f\_2(\theta)\bigr)
$$
Notice that $f\_1$ and $f\_2$ can be arbitrary $2\pi$-periodic functions of $\theta$. However, if $W$ is to be the boundary value of an isometric deformation vector field $V$ as above, we will have to have
$$
f\_1 + i f\_2 = a(\cos\theta,\sin\theta) + ib(\cos\theta,\sin\theta) 
= \sum\_{n=0}^{\infty} c\_n\,\mathrm{e}^{in\theta},
$$ 
i.e., the 'negative' Fourier coefficients of $f\_1+if\_2$ must all vanish. 


Thus, the generic infinitesimal isometric deformation of the boundary cannot be tracked by an infinitesimal isometric deformation of the hemisphere.


