Solution of ∇²𝑇 = 0 in a square domain (0 < x < 1 and 0 < y < 1) with boundary conditions: T(x, 0) = x; T(0, y) = y; T(x, 1) = 1 + x; T(1,y) = 1 + y is

The correct option is T(x, y) = x + y.

To verify why this option is correct, we need to check if the proposed temperature distribution $T(x,y)=x+y$ satisfies both Laplace's equation in the given square domain and all the specified boundary conditions.

Step 1: Check Laplace's Equation
Laplace's equation in two dimensions is given by:
${\nabla }^{2}T=\frac{{\partial }^{2}T\partial x^2}{}+\frac{{\partial }^{2}T\partial y^2}{}=0$

Let us find the first and second partial derivatives of $T(x,y)=x+y$:
First partial derivative with respect to $x$:
$\frac{\partial T}{\partial x}=\frac{\partial }{\partial x}(x+y)=1$
Second partial derivative with respect to $x$:
$\frac{{\partial }^{2}T}{\partial x^2}=\frac{\partial }{\partial x}\left(1\right)=0$

First partial derivative with respect to $y$:
$\frac{\partial T}{\partial y}=\frac{\partial }{\partial y}(x+y)=1$
Second partial derivative with respect to $y$:
$\frac{{\partial }^{2}T}{\partial y^2}=\frac{\partial }{\partial y}\left(1\right)=0$

Adding the second derivatives:
${\nabla }^{2}T=0+0=0$
Thus, Laplace's equation is satisfied.

Step 2: Verify the Boundary Conditions
Now we substitute the boundaries into our solution $T(x,y)=x+y$:
1. At the bottom boundary where $y=0$:
$T(x,0)=x+0=x$ (Matches the boundary condition)

2. At the left boundary where $x=0$:
$T(0,y)=0+y=y$ (Matches the boundary condition)

3. At the top boundary where $y=1$:
$T(x,1)=x+1=1+x$ (Matches the boundary condition)

4. At the right boundary where $x=1$:
$T(1,y)=1+y$ (Matches the boundary condition)

Since the solution $T(x,y)=x+y$ satisfies Laplace's equation and all boundary conditions, by the uniqueness theorem of differential equations, it is the unique and correct solution.