building an orthogonal grid on 2d unbounded domain by solving laplace's equation using FEM

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I am trying to build an orthogonal grid on an unbounded two dimensional domain by solving Laplace equation with FEM.

As an example, consider the domain $D$ defined as

$x \in [0,1], y\in[f(x),y_0]$

An orthogonal grid on such a domain can be built by defining a set of curvilinear coordinates $(\tilde{x},\tilde{y})$ and solving

$\Delta \tilde{x}=\Delta \tilde{y}=0$

with opportune boundary conditions (e.g. orthogonal BC on all the sides). If I consider the trivial case $f(x)=0$, then I get simply $\tilde{x}=x,\tilde{y}=y$.

Now let $y_0$ tend to infinite. In this case, I can still define a bounded domain for my calculations by introducing a compactification of the unbounded domain. Consider as an example the transformation:

$(x,y)\longrightarrow(x,\frac{1}{y+1})$.

which will map the unbounded domain D into a domain $D^\prime$ defined a

$x \in [0,1], y^{\prime}\in[0,\frac{1}{y_0+1}]$.

In this bounded domain I can solve the problems

$\Delta_{x,y^{\prime}} \tilde{x}=\Delta_{x,y^{\prime}} \tilde{y}=0$

where $\Delta_{x,y^{\prime}}$ is the cartesian gradient $\Delta$ expressed in the compact coordinates. The problem is: what kind of BC shall I consider here?

If I set Dirichlet boundary condition on the "infinite" boundary, in fact, I get an exponential decay of the solution $\tilde{y}$. In my case, however, the solution should tend asymptotically as

$\lim\limits_{y\rightarrow\infty}\tilde{y}=y=\infty$

while setting a Dirichlet BC I get

$\lim\limits_{y\rightarrow\infty}\tilde{y}=C$

where C is a constant. How can I deal with this specific case? Is there any "compactification" possible for the solution $\tilde{y}$ itself?