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Here's my justification for setting the nodal values to $g(\mathbf{x})$. There might be some details missing because this is an online answer and not a math paper.

Let's start with the homogeneous variational problem of finding $u \in V$ such that

$$ a(u,v) = \mathcal{B}(v) \quad \forall \quad v \in V $$ where $V$ is the space of $L_2$ functions that vanish at $\Gamma_D$. Let's assume for simplicity the problem is 2D:

Then the Galerkin approximation involves writing both $u$ and $v$ as linear combinations of 32 shape functions $h_j(\mathbf{x})$ nodal values

 $$ u(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot u_j = H \cdot \mathbf{u} $$

$$ v(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot v_j = H \cdot \mathbf{v} $$

for a matrix $H \in \mathbb{R}^{1 \times 32}$ such that the problem can be cast in matrix form as

$$ A \cdot \mathbf{u} = \mathbf{b} $$ where the elements of the matrix $A$ are

$$ a_{i,j} = a(h_i,h_j) $$ and the elements of vector $\mathbf{b}$ are

$$ b_i = \mathcal{B}(h_i) $$

Note that since the $h_j(\mathbf{x})$ form a basis of $V$ they all vanish on $\Gamma_D$ so the solution $u(\mathbf{x})$ also vanishes on $\Gamma_D$. This procedure is right but it is inconvenient because it is not easy to find the right shape functions. We can simplify it by allowing nodes to exist on $\Gamma_D$

and "extending" the expansion of $u$ and $v$ with zeros:

$$ u_h(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot u_j + \sum_{j=33}^{35} h_j(\mathbf{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\mathbf{u} \\ \mathbf{0}\end{bmatrix} $$

$$ v(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot v_j + \sum_{j=33}^{35} h_j(\mathbf{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\mathbf{v} \\ \mathbf{0}\end{bmatrix} $$ for an extended matrix $\tilde{H} \in \mathbb{R}^{1 \times 35}$.

First, we note that these "extended" objects

$$ \tilde{\mathbf{v}} = \begin{bmatrix} \mathbf{v} \\ \mathbf{0} \end{bmatrix} \quad \tilde{{A}} = \begin{bmatrix} {A} & {C} \\ {D} & {E} \\ \end{bmatrix} \quad \tilde{\mathbf{u}} = \begin{bmatrix} \mathbf{u} \\ \mathbf{0} \end{bmatrix} \quad \tilde{\mathbf{b}} = \begin{bmatrix} \mathbf{b} \\ \mathbf{e} \end{bmatrix} $$ still represent the original Galerkin problem. In effect,

$$ \begin{aligned} \tilde{\mathbf{v}}^T \cdot \tilde{{A}} \cdot \tilde{\mathbf{u}} &= \tilde{\mathbf{v}}^T \cdot \tilde{\mathbf{b}} \\ \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} {A} & {C} \\ {D} & {E} \\ \end{bmatrix} \cdot \begin{bmatrix} \mathbf{u} \\ \mathbf{0} \end{bmatrix} & = \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \begin{bmatrix} \mathbf{b} \\ \mathbf{e} \end{bmatrix} \\ \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} {A} \cdot \mathbf{u} + {C} \cdot \mathbf{0} \\ {D} \cdot \mathbf{u} + {E} \cdot \mathbf{0} \\ \end{bmatrix} &= \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \begin{bmatrix} \mathbf{b} \\ \mathbf{e} \end{bmatrix} \\ \mathbf{v}^T \cdot {A} \cdot \mathbf{u} + \mathbf{0}^T \cdot {D} \cdot \mathbf{u} &= \mathbf{v}^T \cdot \mathbf{b} + \mathbf{0}^T \cdot \mathbf{e}\\ \mathbf{v}^T \cdot {A} \cdot \mathbf{u} &= \mathbf{v}^T \cdot \mathbf{b}\\ \end{aligned} $$

Since this identity must hold $\forall \mathbf{v}$, then ${A} \cdot \mathbf{u} - \mathbf{b} = 0$, which is the original homogeneous problem. We now need to prove that if we have

$$ {K} = \begin{bmatrix} {A} & {C} \\ {0} & {I} \\ \end{bmatrix} \quad \mathbf{f} = \begin{bmatrix} \mathbf{b} \\ \mathbf{0} \\ \end{bmatrix} $$ such that ${A} \cdot \mathbf{u} = \mathbf{b}$, where ${I}$ is the identity matrix of size $3 \times 3$, then the vector $\mathbf{\varphi}$ such that ${K} \cdot \mathbf{\varphi} = \mathbf{f}$ is equal to

$$ \mathbf{\varphi} = \begin{bmatrix} \mathbf{u} \\ \mathbf{0} \\ \end{bmatrix} $$

Indeed, let $\mathbf{\varphi} = \begin{bmatrix} \mathbf{\varphi}_1 & \mathbf{\varphi}_2 \end{bmatrix}^T$. Then ${K} \cdot \mathbf{\varphi}$ is

$$ \begin{bmatrix} {A} & {C} \\ {0} & {I} \end{bmatrix} \cdot \begin{bmatrix} \mathbf{\varphi}_1 \\ \mathbf{\varphi}_2 \end{bmatrix} = \begin{bmatrix} {A} \cdot \mathbf{\varphi}_1 + {C} \cdot \mathbf{\varphi}_2\\ {0} \cdot \mathbf{\varphi}_1 + {I} \cdot \mathbf{\varphi}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{b}\\ \mathbf{0} \end{bmatrix} $$

From the second row, $\mathbf{\varphi}_2 = \mathbf{0}$. Replacing this result in the first row, ${A} \cdot \mathbf{\varphi}_1 = \mathbf{b}$. Therefore $\mathbf{\varphi}_1 = {A}^{-1} \cdot \mathbf{b} = \mathbf{u}$.

This part proves that putting a one in the diagonal of the stiffness matrix and a zero in the RHS vector in the rows corresponding to the nodes at $\Gamma_D$ is correct.

Now let's investigate non-homogeneous Dirichlet BCs. The "lifting" procedure from textbooks translates into finding $u_h \in V$ such that

$$ a(u_h,v) = \mathcal{B}(v) - a(u_g,v) \quad \forall \quad v \in V $$

where functions in $V$ vanish at $\Gamma_D$, and $u_g$ is a known function (i.e. a known lifting) that satisfies $u_g(\mathbf{x}) = g(\mathbf{x})$ for $\mathbf{x} \in \Gamma_D$. Note that the right-hand side of the formulation contains known functions only.

Now, let's take back the $u_g$ into the left-hand side (we have already assumed $a$ was bi-linear) then we have to solve

$$ a(u_h+u_g,v) = \mathcal{B}(v) $$

Let's first write the functions $u_h$ and $v$ over the whole set of nodes

$$ u_h(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot u_j + \sum_{j=33}^{35} h_j(\mathbf{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\mathbf{u} \\ \mathbf{0}\end{bmatrix} $$

$$ v(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot v_j + \sum_{j=33}^{35} h_j(\mathbf{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\mathbf{v} \\ \mathbf{0}\end{bmatrix} $$ for another matrix $\tilde{H}$ of size $1 \times 35$.

We then write the non-homogeneous part $u_g$ as

$$ u_h(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot 0 + \sum_{j=33}^{35} h_j(\mathbf{x}) \cdot g(\mathbf{x}_j) = \tilde{H} \cdot \begin{bmatrix}\mathbf{0} \\ \mathbf{g}\end{bmatrix} $$ so the sum $u_h + u_g$ is

$$ u_h(\mathbf{x}) + u_g(\mathbf{x}) = \tilde{H} \cdot \begin{bmatrix}\mathbf{u} \\ \mathbf{g}\end{bmatrix} $$

Now, the stiffness matrix of the first homogeneous problem was $A \in \mathbb{R}^{32 \times 32}$. We then extend it to a new matrix of size $35 \times 35$

$$ \begin{bmatrix} A & C \\ D & E \end{bmatrix} $$ such that the discretized problem is now

$$ \begin{aligned} \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} A & C \\ D & E \end{bmatrix} \cdot \begin{bmatrix} \mathbf{u} \\ \mathbf{g} \end{bmatrix} &= \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} \mathbf{b} \\ \mathbf{e} \end{bmatrix} \\ \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} A \cdot \mathbf{u} + C \cdot \mathbf{g} \\ D \cdot \mathbf{u} + E \cdot \mathbf{g} \end{bmatrix} &= \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} \mathbf{b} \\ \mathbf{e} \end{bmatrix} \\ \mathbf{v}^T \cdot A \cdot \mathbf{u} + \mathbf{v}^T \cdot C \cdot \mathbf{g} &= \mathbf{v}^T \cdot \mathbf{b} \end{aligned} $$ for all $\mathbf{u} \in \mathbb{R}^J$. That is to say,

$$ A \cdot \mathbf{u} + C \cdot \mathbf{g} = \mathbf{b} $$

Now you can prove as an exercise that if we have

$$ K = \begin{bmatrix} A & C \\ 0 & I \\ \end{bmatrix} \quad \mathbf{f} = \begin{bmatrix} \mathbf{b} \\ \mathbf{g} \\ \end{bmatrix} $$ such that $A \cdot \mathbf{u} + C \cdot \mathbf{g} = \mathbf{b}$, then the vector $\mathbf{\varphi}$ such that $K \cdot \mathbf{\varphi} = \mathbf{f}$ is equal to

$$ \mathbf{\varphi} = \begin{bmatrix} \mathbf{u} \\ \mathbf{g} \\ \end{bmatrix} $$

Here's my justification for setting the nodal values to $g(\mathbf{x})$. There might be some details missing because this is an online answer and not a math paper.

Let's start with the homogeneous variational problem of finding $u \in V$ such that

$$ a(u,v) = \mathcal{B}(v) \quad \forall \quad v \in V $$ where $V$ is the space of $L_2$ functions that vanish at $\Gamma_D$. Let's assume for simplicity the problem is 2D:

Then the Galerkin approximation involves writing both $u$ and $v$ as linear combinations of 32 shape functions $h_j(\mathbf{x})$ times the nodal values $u_j$ and $v_j$ $$ u(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot u_j = H \cdot \mathbf{u} $$

$$ v(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot v_j = H \cdot \mathbf{v} $$

for a matrix $H \in \mathbb{R}^{1 \times 32}$ such that the problem can be cast in matrix form as

$$ A \cdot \mathbf{u} = \mathbf{b} $$ where the elements of the matrix $A$ are

$$ a_{i,j} = a(h_i,h_j) $$ and the elements of vector $\mathbf{b}$ are

$$ b_i = \mathcal{B}(h_i) $$

Note that since the $h_j(\mathbf{x})$ form a basis of $V$ they all vanish on $\Gamma_D$ so the solution $u(\mathbf{x})$ also vanishes on $\Gamma_D$. This procedure is right but it is inconvenient because it is not easy to find the right shape functions that vanish on $\Gamma_D$. We can simplify it by allowing nodes to exist on $\Gamma_D$

and "extending" the expansion of $u$ and $v$ with zeros for $j=33,34,35$:

$$ u_h(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot u_j + \sum_{j=33}^{35} h_j(\mathbf{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\mathbf{u} \\ \mathbf{0}\end{bmatrix} $$

$$ v(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot v_j + \sum_{j=33}^{35} h_j(\mathbf{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\mathbf{v} \\ \mathbf{0}\end{bmatrix} $$ for an extended matrix $\tilde{H} \in \mathbb{R}^{1 \times 35}$.

First, we note that these "extended" objects

$$ \tilde{\mathbf{v}} = \begin{bmatrix} \mathbf{v} \\ \mathbf{0} \end{bmatrix} \quad \tilde{{A}} = \begin{bmatrix} {A} & {C} \\ {D} & {E} \\ \end{bmatrix} \quad \tilde{\mathbf{u}} = \begin{bmatrix} \mathbf{u} \\ \mathbf{0} \end{bmatrix} \quad \tilde{\mathbf{b}} = \begin{bmatrix} \mathbf{b} \\ \mathbf{e} \end{bmatrix} $$ still represent the original Galerkin problem. In effect,

$$ \begin{aligned} \tilde{\mathbf{v}}^T \cdot \tilde{{A}} \cdot \tilde{\mathbf{u}} &= \tilde{\mathbf{v}}^T \cdot \tilde{\mathbf{b}} \\ \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} {A} & {C} \\ {D} & {E} \\ \end{bmatrix} \cdot \begin{bmatrix} \mathbf{u} \\ \mathbf{0} \end{bmatrix} & = \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \begin{bmatrix} \mathbf{b} \\ \mathbf{e} \end{bmatrix} \\ \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} {A} \cdot \mathbf{u} + {C} \cdot \mathbf{0} \\ {D} \cdot \mathbf{u} + {E} \cdot \mathbf{0} \\ \end{bmatrix} &= \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \begin{bmatrix} \mathbf{b} \\ \mathbf{e} \end{bmatrix} \\ \mathbf{v}^T \cdot {A} \cdot \mathbf{u} + \mathbf{0}^T \cdot {D} \cdot \mathbf{u} &= \mathbf{v}^T \cdot \mathbf{b} + \mathbf{0}^T \cdot \mathbf{e}\\ \mathbf{v}^T \cdot {A} \cdot \mathbf{u} &= \mathbf{v}^T \cdot \mathbf{b}\\ \end{aligned} $$

Since this identity must hold $\forall \mathbf{v}$, then ${A} \cdot \mathbf{u} - \mathbf{b} = 0$, which is the original homogeneous problem. We now need to prove that if we have

$$ {K} = \begin{bmatrix} {A} & {C} \\ {0} & {I} \\ \end{bmatrix} \quad \mathbf{f} = \begin{bmatrix} \mathbf{b} \\ \mathbf{0} \\ \end{bmatrix} $$ such that ${A} \cdot \mathbf{u} = \mathbf{b}$, where ${I}$ is the identity matrix of size $3 \times 3$, then the vector $\mathbf{\varphi}$ such that ${K} \cdot \mathbf{\varphi} = \mathbf{f}$ is equal to

$$ \mathbf{\varphi} = \begin{bmatrix} \mathbf{u} \\ \mathbf{0} \\ \end{bmatrix} $$

Indeed, let $\mathbf{\varphi} = \begin{bmatrix} \mathbf{\varphi}_1 & \mathbf{\varphi}_2 \end{bmatrix}^T$. Then ${K} \cdot \mathbf{\varphi}$ is

$$ \begin{bmatrix} {A} & {C} \\ {0} & {I} \end{bmatrix} \cdot \begin{bmatrix} \mathbf{\varphi}_1 \\ \mathbf{\varphi}_2 \end{bmatrix} = \begin{bmatrix} {A} \cdot \mathbf{\varphi}_1 + {C} \cdot \mathbf{\varphi}_2\\ {0} \cdot \mathbf{\varphi}_1 + {I} \cdot \mathbf{\varphi}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{b}\\ \mathbf{0} \end{bmatrix} $$

From the second row, $\mathbf{\varphi}_2 = \mathbf{0}$. Replacing this result in the first row, ${A} \cdot \mathbf{\varphi}_1 = \mathbf{b}$. Therefore $\mathbf{\varphi}_1 = {A}^{-1} \cdot \mathbf{b} = \mathbf{u}$.

This part proves that putting a one in the diagonal of the stiffness matrix and a zero in the RHS vector in the rows corresponding to the nodes at $\Gamma_D$ is correct.

Now let's investigate non-homogeneous Dirichlet BCs. The "lifting" procedure from textbooks translates into finding $u_h \in V$ such that

$$ a(u_h,v) = \mathcal{B}(v) - a(u_g,v) \quad \forall \quad v \in V $$

where functions in $V$ vanish at $\Gamma_D$, and $u_g$ is a known function (i.e. a known lifting) that satisfies $u_g(\mathbf{x}) = g(\mathbf{x})$ for $\mathbf{x} \in \Gamma_D$. Note that the right-hand side of the formulation contains known functions only.

Now, let's take back the $u_g$ into the left-hand side (we have already assumed $a$ was bi-linear) then we have to solve

$$ a(u_h+u_g,v) = \mathcal{B}(v) $$

Let's first write the functions $u_h$ and $v$ over the whole set of nodes

$$ u_h(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot u_j + \sum_{j=33}^{35} h_j(\mathbf{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\mathbf{u} \\ \mathbf{0}\end{bmatrix} $$

$$ v(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot v_j + \sum_{j=33}^{35} h_j(\mathbf{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\mathbf{v} \\ \mathbf{0}\end{bmatrix} $$ for the "extended" matrix $\tilde{H}$ of size $1 \times 35$.

We then write the non-homogeneous part $u_g$ as

$$ u_h(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot 0 + \sum_{j=33}^{35} h_j(\mathbf{x}) \cdot g(\mathbf{x}_j) = \tilde{H} \cdot \begin{bmatrix}\mathbf{0} \\ \mathbf{g}\end{bmatrix} $$ so the sum $u_h + u_g$ is

$$ u_h(\mathbf{x}) + u_g(\mathbf{x}) = \tilde{H} \cdot \begin{bmatrix}\mathbf{u} \\ \mathbf{g}\end{bmatrix} $$

Now, the stiffness matrix of the first homogeneous problem was $A \in \mathbb{R}^{32 \times 32}$. We then extend it to a new matrix of size $35 \times 35$

$$ \begin{bmatrix} A & C \\ D & E \end{bmatrix} $$ such that the discretized problem is now

$$ \begin{aligned} \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} A & C \\ D & E \end{bmatrix} \cdot \begin{bmatrix} \mathbf{u} \\ \mathbf{g} \end{bmatrix} &= \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} \mathbf{b} \\ \mathbf{e} \end{bmatrix} \\ \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} A \cdot \mathbf{u} + C \cdot \mathbf{g} \\ D \cdot \mathbf{u} + E \cdot \mathbf{g} \end{bmatrix} &= \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} \mathbf{b} \\ \mathbf{e} \end{bmatrix} \\ \mathbf{v}^T \cdot A \cdot \mathbf{u} + \mathbf{v}^T \cdot C \cdot \mathbf{g} &= \mathbf{v}^T \cdot \mathbf{b} \end{aligned} $$ for all $\mathbf{u} \in \mathbb{R}^J$. That is to say,

$$ A \cdot \mathbf{u} + C \cdot \mathbf{g} = \mathbf{b} $$

Now you can prove as an exercise that if we have

$$ K = \begin{bmatrix} A & C \\ 0 & I \\ \end{bmatrix} \quad \mathbf{f} = \begin{bmatrix} \mathbf{b} \\ \mathbf{g} \\ \end{bmatrix} $$ such that $A \cdot \mathbf{u} + C \cdot \mathbf{g} = \mathbf{b}$, then the vector $\mathbf{\varphi}$ such that $K \cdot \mathbf{\varphi} = \mathbf{f}$ is equal to

$$ \mathbf{\varphi} = \begin{bmatrix} \mathbf{u} \\ \mathbf{g} \\ \end{bmatrix} $$

Here's my justification for setting the nodal values to $g(\vec{x})$. There might be some details missing because this is an online answer and not a math paper.

Then the Galerkin approximation involves writing both $u$ and $v$ as linear combinations of 32 shape functions $h_j(\vec{x})$ nodal values

$$ u(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot u_j = H \cdot \vec{u} $$

$$ v(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot v_j = H \cdot \vec{v} $$

$$ A \cdot \vec{u} = \vec{b} $$ where the elements of the matrix  $A$ are

$$ a_{i,j} = a(h_i,h_j) $$ and the elements of vector $\vec{b}$ are

Note that since the $h_j(\vec{x})$ form a basis of $V$ they all vanish on  $\Gamma_D$ so the solution $u(\vec{x})$ also vanishes on  $\Gamma_D$. This procedure is right but it is inconvenient because it is not easy to find the right shape functions. We can simplify it by allowing nodes to exist on  $\Gamma_D$

$$ u_h(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot u_j + \sum_{j=33}^{35} h_j(\vec{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\vec{u} \\ \vec{0}\end{bmatrix} $$

$$ v(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot v_j + \sum_{j=33}^{35} h_j(\vec{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\vec{v} \\ \vec{0}\end{bmatrix} $$ for an extended matrix  $\tilde{H} \in \mathbb{R}^{1 \times 35}$.

$$ \tilde{\vec{v}} = \begin{bmatrix} \vec{v} \\ \vec{0} \end{bmatrix} \quad \tilde{{A}} = \begin{bmatrix} {A} & {C} \\ {D} & {E} \\ \end{bmatrix} \quad \tilde{\vec{u}} = \begin{bmatrix} \vec{u} \\ \vec{0} \end{bmatrix} \quad \tilde{\vec{b}} = \begin{bmatrix} \vec{b} \\ \vec{e} \end{bmatrix} $$ still represent the original Galerkin problem. In effect,

$$ \begin{aligned} \tilde{\vec{v}}^T \cdot \tilde{{A}} \cdot \tilde{\vec{u}} &= \tilde{\vec{v}}^T \cdot \tilde{\vec{b}} \\ \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} {A} & {C} \\ {D} & {E} \\ \end{bmatrix} \cdot \begin{bmatrix} \vec{u} \\ \vec{0} \end{bmatrix} & = \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \begin{bmatrix} \vec{b} \\ \vec{e} \end{bmatrix} \\ \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} {A} \cdot \vec{u} + {C} \cdot \vec{0} \\ {D} \cdot \vec{u} + {E} \cdot \vec{0} \\ \end{bmatrix} &= \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \begin{bmatrix} \vec{b} \\ \vec{e} \end{bmatrix} \\ \vec{v}^T \cdot {A} \cdot \vec{u} + \vec{0}^T \cdot {D} \cdot \vec{u} &= \vec{v}^T \cdot \vec{b} + \vec{0}^T \cdot \vec{e}\\ \vec{v}^T \cdot {A} \cdot \vec{u} &= \vec{v}^T \cdot \vec{b}\\ \end{aligned} $$

Since this identity must hold $\forall \vec{v}$, then ${A} \cdot \vec{u} - \vec{b} = 0$, which is the original homogeneous problem. We now need to prove that if we have

$$ {K} = \begin{bmatrix} {A} & {C} \\ {0} & {I} \\ \end{bmatrix} \quad \vec{f} = \begin{bmatrix} \vec{b} \\ \vec{0} \\ \end{bmatrix} $$ such that ${A} \cdot \vec{u} = \vec{b}$, where  ${I}$ is the identity matrix of size  $3 \times 3$, then the vector $\vec{\varphi}$ such that ${K} \cdot \vec{\varphi} = \vec{f}$ is equal to

$$ \vec{\varphi} = \begin{bmatrix} \vec{u} \\ \vec{0} \\ \end{bmatrix} $$

Indeed, let $\vec{\varphi} = \begin{bmatrix} \vec{\varphi}_1 & \vec{\varphi}_2 \end{bmatrix}^T$. Then ${K} \cdot \vec{\varphi}$ is

$$ \begin{bmatrix} {A} & {C} \\ {0} & {I} \end{bmatrix} \cdot \begin{bmatrix} \vec{\varphi}_1 \\ \vec{\varphi}_2 \end{bmatrix} = \begin{bmatrix} {A} \cdot \vec{\varphi}_1 + {C} \cdot \vec{\varphi}_2\\ {0} \cdot \vec{\varphi}_1 + {I} \cdot \vec{\varphi}_2 \end{bmatrix} = \begin{bmatrix} \vec{b}\\ \vec{0} \end{bmatrix} $$

From the second row,  $\vec{\varphi}_2 = \vec{0}$. Replacing this result in the first row, ${A} \cdot \vec{\varphi}_1 = \vec{b}$. Therefore $\vec{\varphi}_1 = {A}^{-1} \cdot \vec{b} = \vec{u}$.

where functions in $V$ vanish at  $\Gamma_D$, and  $u_g$ is a known function (i.e. a known lifting) that satisfies $u_g(\vec{x}) = g(\vec{x})$ for $\vec{x} \in \Gamma_D$. Note that the right-hand side of the formulation contains known functions only.

$$ u_h(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot u_j + \sum_{j=33}^{35} h_j(\vec{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\vec{u} \\ \vec{0}\end{bmatrix} $$

$$ v(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot v_j + \sum_{j=33}^{35} h_j(\vec{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\vec{v} \\ \vec{0}\end{bmatrix} $$ for another matrix  $\tilde{H}$ of size  $1 \times 35$.

$$ u_h(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot 0 + \sum_{j=33}^{35} h_j(\vec{x}) \cdot g(\vec{x}_j) = \tilde{H} \cdot \begin{bmatrix}\vec{0} \\ \vec{g}\end{bmatrix} $$ so the sum $u_h + u_g$ is

$$ u_h(\vec{x}) + u_g(\vec{x}) = \tilde{H} \cdot \begin{bmatrix}\vec{u} \\ \vec{g}\end{bmatrix} $$

$$ \begin{aligned} \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} A & C \\ D & E \end{bmatrix} \cdot \begin{bmatrix} \vec{u} \\ \vec{g} \end{bmatrix} &= \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} \vec{b} \\ \vec{e} \end{bmatrix} \\ \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} A \cdot \vec{u} + C \cdot \vec{g} \\ D \cdot \vec{u} + E \cdot \vec{g} \end{bmatrix} &= \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} \vec{b} \\ \vec{e} \end{bmatrix} \\ \vec{v}^T \cdot A \cdot \vec{u} + \vec{v}^T \cdot C \cdot \vec{g} &= \vec{v}^T \cdot \vec{b} \end{aligned} $$ for all $\vec{u} \in \mathbb{R}^J$. That is to say,

$$ A \cdot \vec{u} + C \cdot \vec{g} = \vec{b} $$

$$ K = \begin{bmatrix} A & C \\ 0 & I \\ \end{bmatrix} \quad \vec{f} = \begin{bmatrix} \vec{b} \\ \vec{g} \\ \end{bmatrix} $$ such that $A \cdot \vec{u} + C \cdot \vec{g} = \vec{b}$, then the vector $\vec{\varphi}$ such that $K \cdot \vec{\varphi} = \vec{f}$ is equal to

$$ \vec{\varphi} = \begin{bmatrix} \vec{u} \\ \vec{g} \\ \end{bmatrix} $$

Here's my justification for setting the nodal values to $g(\mathbf{x})$. There might be some details missing because this is an online answer and not a math paper.

Then the Galerkin approximation involves writing both $u$ and $v$ as linear combinations of 32 shape functions $h_j(\mathbf{x})$ nodal values

$$ u(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot u_j = H \cdot \mathbf{u} $$

$$ v(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot v_j = H \cdot \mathbf{v} $$

$$ A \cdot \mathbf{u} = \mathbf{b} $$ where the elements of the matrix  $A$ are

$$ a_{i,j} = a(h_i,h_j) $$ and the elements of vector $\mathbf{b}$ are

Note that since the $h_j(\mathbf{x})$ form a basis of $V$ they all vanish on  $\Gamma_D$ so the solution $u(\mathbf{x})$ also vanishes on  $\Gamma_D$. This procedure is right but it is inconvenient because it is not easy to find the right shape functions. We can simplify it by allowing nodes to exist on  $\Gamma_D$

$$ u_h(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot u_j + \sum_{j=33}^{35} h_j(\mathbf{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\mathbf{u} \\ \mathbf{0}\end{bmatrix} $$

$$ v(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot v_j + \sum_{j=33}^{35} h_j(\mathbf{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\mathbf{v} \\ \mathbf{0}\end{bmatrix} $$ for an extended matrix  $\tilde{H} \in \mathbb{R}^{1 \times 35}$.

$$ \tilde{\mathbf{v}} = \begin{bmatrix} \mathbf{v} \\ \mathbf{0} \end{bmatrix} \quad \tilde{{A}} = \begin{bmatrix} {A} & {C} \\ {D} & {E} \\ \end{bmatrix} \quad \tilde{\mathbf{u}} = \begin{bmatrix} \mathbf{u} \\ \mathbf{0} \end{bmatrix} \quad \tilde{\mathbf{b}} = \begin{bmatrix} \mathbf{b} \\ \mathbf{e} \end{bmatrix} $$ still represent the original Galerkin problem. In effect,

$$ \begin{aligned} \tilde{\mathbf{v}}^T \cdot \tilde{{A}} \cdot \tilde{\mathbf{u}} &= \tilde{\mathbf{v}}^T \cdot \tilde{\mathbf{b}} \\ \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} {A} & {C} \\ {D} & {E} \\ \end{bmatrix} \cdot \begin{bmatrix} \mathbf{u} \\ \mathbf{0} \end{bmatrix} & = \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \begin{bmatrix} \mathbf{b} \\ \mathbf{e} \end{bmatrix} \\ \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} {A} \cdot \mathbf{u} + {C} \cdot \mathbf{0} \\ {D} \cdot \mathbf{u} + {E} \cdot \mathbf{0} \\ \end{bmatrix} &= \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \begin{bmatrix} \mathbf{b} \\ \mathbf{e} \end{bmatrix} \\ \mathbf{v}^T \cdot {A} \cdot \mathbf{u} + \mathbf{0}^T \cdot {D} \cdot \mathbf{u} &= \mathbf{v}^T \cdot \mathbf{b} + \mathbf{0}^T \cdot \mathbf{e}\\ \mathbf{v}^T \cdot {A} \cdot \mathbf{u} &= \mathbf{v}^T \cdot \mathbf{b}\\ \end{aligned} $$

Since this identity must hold $\forall \mathbf{v}$, then ${A} \cdot \mathbf{u} - \mathbf{b} = 0$, which is the original homogeneous problem. We now need to prove that if we have

$$ {K} = \begin{bmatrix} {A} & {C} \\ {0} & {I} \\ \end{bmatrix} \quad \mathbf{f} = \begin{bmatrix} \mathbf{b} \\ \mathbf{0} \\ \end{bmatrix} $$ such that ${A} \cdot \mathbf{u} = \mathbf{b}$, where  ${I}$ is the identity matrix of size  $3 \times 3$, then the vector $\mathbf{\varphi}$ such that ${K} \cdot \mathbf{\varphi} = \mathbf{f}$ is equal to

$$ \mathbf{\varphi} = \begin{bmatrix} \mathbf{u} \\ \mathbf{0} \\ \end{bmatrix} $$

Indeed, let $\mathbf{\varphi} = \begin{bmatrix} \mathbf{\varphi}_1 & \mathbf{\varphi}_2 \end{bmatrix}^T$. Then ${K} \cdot \mathbf{\varphi}$ is

$$ \begin{bmatrix} {A} & {C} \\ {0} & {I} \end{bmatrix} \cdot \begin{bmatrix} \mathbf{\varphi}_1 \\ \mathbf{\varphi}_2 \end{bmatrix} = \begin{bmatrix} {A} \cdot \mathbf{\varphi}_1 + {C} \cdot \mathbf{\varphi}_2\\ {0} \cdot \mathbf{\varphi}_1 + {I} \cdot \mathbf{\varphi}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{b}\\ \mathbf{0} \end{bmatrix} $$

From the second row, $\mathbf{\varphi}_2 = \mathbf{0}$. Replacing this result in the first row, ${A} \cdot \mathbf{\varphi}_1 = \mathbf{b}$. Therefore $\mathbf{\varphi}_1 = {A}^{-1} \cdot \mathbf{b} = \mathbf{u}$.

where functions in $V$ vanish at  $\Gamma_D$, and  $u_g$ is a known function (i.e. a known lifting) that satisfies $u_g(\mathbf{x}) = g(\mathbf{x})$ for $\mathbf{x} \in \Gamma_D$. Note that the right-hand side of the formulation contains known functions only.

$$ u_h(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot u_j + \sum_{j=33}^{35} h_j(\mathbf{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\mathbf{u} \\ \mathbf{0}\end{bmatrix} $$

$$ v(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot v_j + \sum_{j=33}^{35} h_j(\mathbf{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\mathbf{v} \\ \mathbf{0}\end{bmatrix} $$ for another matrix  $\tilde{H}$ of size  $1 \times 35$.

$$ u_h(\mathbf{x}) = \sum_{j=1}^{32} h_j(\mathbf{x}) \cdot 0 + \sum_{j=33}^{35} h_j(\mathbf{x}) \cdot g(\mathbf{x}_j) = \tilde{H} \cdot \begin{bmatrix}\mathbf{0} \\ \mathbf{g}\end{bmatrix} $$ so the sum $u_h + u_g$ is

$$ u_h(\mathbf{x}) + u_g(\mathbf{x}) = \tilde{H} \cdot \begin{bmatrix}\mathbf{u} \\ \mathbf{g}\end{bmatrix} $$

$$ \begin{aligned} \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} A & C \\ D & E \end{bmatrix} \cdot \begin{bmatrix} \mathbf{u} \\ \mathbf{g} \end{bmatrix} &= \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} \mathbf{b} \\ \mathbf{e} \end{bmatrix} \\ \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} A \cdot \mathbf{u} + C \cdot \mathbf{g} \\ D \cdot \mathbf{u} + E \cdot \mathbf{g} \end{bmatrix} &= \begin{bmatrix} \mathbf{v}^T & \mathbf{0}^T \end{bmatrix} \cdot \begin{bmatrix} \mathbf{b} \\ \mathbf{e} \end{bmatrix} \\ \mathbf{v}^T \cdot A \cdot \mathbf{u} + \mathbf{v}^T \cdot C \cdot \mathbf{g} &= \mathbf{v}^T \cdot \mathbf{b} \end{aligned} $$ for all $\mathbf{u} \in \mathbb{R}^J$. That is to say,

$$ A \cdot \mathbf{u} + C \cdot \mathbf{g} = \mathbf{b} $$

$$ K = \begin{bmatrix} A & C \\ 0 & I \\ \end{bmatrix} \quad \mathbf{f} = \begin{bmatrix} \mathbf{b} \\ \mathbf{g} \\ \end{bmatrix} $$ such that $A \cdot \mathbf{u} + C \cdot \mathbf{g} = \mathbf{b}$, then the vector $\mathbf{\varphi}$ such that $K \cdot \mathbf{\varphi} = \mathbf{f}$ is equal to

$$ \mathbf{\varphi} = \begin{bmatrix} \mathbf{u} \\ \mathbf{g} \\ \end{bmatrix} $$

Here's my justification for setting the nodal values to $g(\vec{x})$. There might be some details missing because this is an online answer and not a math paper.

Let's start with the homogeneous variational problem of finding $u \in V$ such that

$$ a(u,v) = \mathcal{B}(v) \quad \forall \quad v \in V $$ where $V$ is the space of $L_2$ functions that vanish at $\Gamma_D$. Let's assume for simplicity the problem is 2D:

Then the Galerkin approximation involves writing both $u$ and $v$ as linear combinations of 32 shape functions $h_j(\vec{x})$ nodal values

$$ u(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot u_j = H \cdot \vec{u} $$

$$ v(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot v_j = H \cdot \vec{v} $$

for a matrix $H \in \mathbb{R}^{1 \times 32}$ such that the problem can be cast in matrix form as

$$ A \cdot \vec{u} = \vec{b} $$ where the elements of the matrix $A$ are

$$ a_{i,j} = a(h_i,h_j) $$ and the elements of vector $\vec{b}$ are

$$ b_i = \mathcal{B}(h_i) $$

Note that since the $h_j(\vec{x})$ form a basis of $V$ they all vanish on $\Gamma_D$ so the solution $u(\vec{x})$ also vanishes on $\Gamma_D$. This procedure is right but it is inconvenient because it is not easy to find the right shape functions. We can simplify it by allowing nodes to exist on $\Gamma_D$

and "extending" the expansion of $u$ and $v$ with zeros:

$$ u_h(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot u_j + \sum_{j=33}^{35} h_j(\vec{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\vec{u} \\ \vec{0}\end{bmatrix} $$

$$ v(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot v_j + \sum_{j=33}^{35} h_j(\vec{x} \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\vec{u} \\ \vec{0}\end{bmatrix} $$ for an extended matrix $\tilde{H} \in \mathbb{R}^{1 \times 35}$.

First, we note that these "extended" objects

$$ \tilde{\vec{v}} = \begin{bmatrix} \vec{v} \\ \vec{0} \end{bmatrix} \quad \tilde{{A}} = \begin{bmatrix} {A} & {C} \\ {D} & {E} \\ \end{bmatrix} \quad \tilde{\vec{u}} = \begin{bmatrix} \vec{u} \\ \vec{0} \end{bmatrix} \quad \tilde{\vec{b}} = \begin{bmatrix} \vec{b} \\ \vec{e} \end{bmatrix} $$ still represent the original Galerkin problem. In effect,

$$ \begin{aligned} \tilde{\vec{v}}^T \cdot \tilde{{A}} \cdot \tilde{\vec{u}} &= \tilde{\vec{v}}^T \cdot \tilde{\vec{b}} \\ \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} {A} & {C} \\ {D} & {E} \\ \end{bmatrix} \cdot \begin{bmatrix} \vec{u} \\ \vec{0} \end{bmatrix} & = \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \begin{bmatrix} \vec{b} \\ \vec{e} \end{bmatrix} \\ \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} {A} \cdot \vec{u} + {C} \cdot \vec{0} \\ {D} \cdot \vec{u} + {E} \cdot \vec{0} \\ \end{bmatrix} &= \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \begin{bmatrix} \vec{b} \\ \vec{e} \end{bmatrix} \\ \vec{v}^T \cdot {A} \cdot \vec{u} + \vec{0}^T \cdot {D} \cdot \vec{u} &= \vec{v}^T \cdot \vec{b} + \vec{0}^T \cdot \vec{e}\\ \vec{v}^T \cdot {A} \cdot \vec{u} &= \vec{v}^T \cdot \vec{b}\\ \end{aligned} $$

Since this identity must hold $\forall \vec{v}$, then ${A} \cdot \vec{u} - \vec{b} = 0$, which is the original homogeneous problem. We now need to prove that if we have

$$ {K} = \begin{bmatrix} {A} & {C} \\ {0} & {I} \\ \end{bmatrix} \quad \vec{f} = \begin{bmatrix} \vec{b} \\ \vec{0} \\ \end{bmatrix} $$ such that ${A} \cdot \vec{u} = \vec{b}$, where ${I}$ is the identity matrix of size $3 \times 3$, then the vector $\vec{\varphi}$ such that ${K} \cdot \vec{\varphi} = \vec{f}$ is equal to

$$ \vec{\varphi} = \begin{bmatrix} \vec{u} \\ \vec{0} \\ \end{bmatrix} $$

Indeed, let $\vec{\varphi} = \begin{bmatrix} \vec{\varphi}_1 & \vec{\varphi}_2 \end{bmatrix}^T$. Then ${K} \cdot \vec{\varphi}$ is

$$ \begin{bmatrix} {A} & {C} \\ {0} & {I} \end{bmatrix} \cdot \begin{bmatrix} \vec{\varphi}_1 \\ \vec{\varphi}_2 \end{bmatrix} = \begin{bmatrix} {A} \cdot \vec{\varphi}_1 + {C} \cdot \vec{\varphi}_2\\ {0} \cdot \vec{\varphi}_1 + {I} \cdot \vec{\varphi}_2 \end{bmatrix} = \begin{bmatrix} \vec{b}\\ \vec{0} \end{bmatrix} $$

From the second row,  $\vec{\varphi}_2 = \vec{0}$. Replacing this result in the first row, ${A} \cdot \vec{\varphi}_1 = \vec{b}$. Therefore $\vec{\varphi}_1 = {A}^{-1} \cdot \vec{b} = \vec{u}$.

This part proves that putting a one in the diagonal of the stiffness matrix and a zero in the RHS vector in the rows corresponding to the nodes at $\Gamma_D$ is correct.

Now let's investigate non-homogeneous Dirichlet BCs. The "lifting" procedure from textbooks translates into finding $u_h \in V$ such that

$$ a(u_h,v) = \mathcal{B}(v) - a(u_g,v) \quad \forall \quad v \in V $$

where functions in $V$ vanish at $\Gamma_D$, and $u_g$ is a known function (i.e. a known lifting) that satisfies $u_g(\vec{x}) = g(\vec{x})$ for $\vec{x} \in \Gamma_D$. Note that the right-hand side of the formulation contains known functions only.

Now, let's take back the $u_g$ into the left-hand side (we have already assumed $a$ was bi-linear) then we have to solve

$$ a(u_h+u_g,v) = \mathcal{B}(v) $$

Let's first write the functions $u_h$ and $v$ over the whole set of nodes

$$ u_h(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot u_j + \sum_{j=33}^{35} h_j(\vec{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\vec{u} \\ \vec{0}\end{bmatrix} $$

$$ v(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot v_j + \sum_{j=33}^{35} h_j(\vec{x} \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\vec{u} \\ \vec{0}\end{bmatrix} $$ for another matrix $\tilde{H}$ of size $1 \times 35$.

We then write the non-homogeneous part $u_g$ as

$$ u_h(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot 0 + \sum_{j=33}^{35} h_j(\vec{x}) \cdot g(\vec{x}_j) = \tilde{H} \cdot \begin{bmatrix}\vec{0} \\ \vec{g}\end{bmatrix} $$ so the sum $u_h + u_g$ is

$$ u_h(\vec{x}) + u_g(\vec{x}) = \tilde{H} \cdot \begin{bmatrix}\vec{u} \\ \vec{g}\end{bmatrix} $$

Now, the stiffness matrix of the first homogeneous problem was $A \in \mathbb{R}^{32 \times 32}$. We then extend it to a new matrix of size $35 \times 35$

$$ \begin{bmatrix} A & C \\ D & E \end{bmatrix} $$ such that the discretized problem is now

$$ \begin{aligned} \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} A & C \\ D & E \end{bmatrix} \cdot \begin{bmatrix} \vec{u} \\ \vec{g} \end{bmatrix} &= \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} \vec{b} \\ \vec{e} \end{bmatrix} \\ \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} A \cdot \vec{u} + C \cdot \vec{g} \\ D \cdot \vec{u} + E \cdot \vec{g} \end{bmatrix} &= \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} \vec{b} \\ \vec{e} \end{bmatrix} \\ \vec{v}^T \cdot A \cdot \vec{u} + \vec{v}^T \cdot C \cdot \vec{g} &= \vec{v}^T \cdot \vec{b} \end{aligned} $$ for all $\vec{u} \in \mathbb{R}^J$. That is to say,

$$ A \cdot \vec{u} + C \cdot \vec{g} = \vec{b} $$

Now you can prove as an exercise that if we have

$$ K = \begin{bmatrix} A & C \\ 0 & I \\ \end{bmatrix} \quad \vec{f} = \begin{bmatrix} \vec{b} \\ \vec{g} \\ \end{bmatrix} $$ such that $A \cdot \vec{u} + C \cdot \vec{g} = \vec{b}$, then the vector $\vec{\varphi}$ such that $K \cdot \vec{\varphi} = \vec{f}$ is equal to

$$ \vec{\varphi} = \begin{bmatrix} \vec{u} \\ \vec{g} \\ \end{bmatrix} $$

Here's my justification for setting the nodal values to $g(\vec{x})$. There might be some details missing because this is an online answer and not a math paper.

Let's start with the homogeneous variational problem of finding $u \in V$ such that

$$ a(u,v) = \mathcal{B}(v) \quad \forall \quad v \in V $$ where $V$ is the space of $L_2$ functions that vanish at $\Gamma_D$. Let's assume for simplicity the problem is 2D:

Then the Galerkin approximation involves writing both $u$ and $v$ as linear combinations of 32 shape functions $h_j(\vec{x})$ nodal values

$$ u(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot u_j = H \cdot \vec{u} $$

$$ v(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot v_j = H \cdot \vec{v} $$

for a matrix $H \in \mathbb{R}^{1 \times 32}$ such that the problem can be cast in matrix form as

$$ A \cdot \vec{u} = \vec{b} $$ where the elements of the matrix $A$ are

$$ a_{i,j} = a(h_i,h_j) $$ and the elements of vector $\vec{b}$ are

$$ b_i = \mathcal{B}(h_i) $$

Note that since the $h_j(\vec{x})$ form a basis of $V$ they all vanish on $\Gamma_D$ so the solution $u(\vec{x})$ also vanishes on $\Gamma_D$. This procedure is right but it is inconvenient because it is not easy to find the right shape functions. We can simplify it by allowing nodes to exist on $\Gamma_D$

and "extending" the expansion of $u$ and $v$ with zeros:

$$ u_h(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot u_j + \sum_{j=33}^{35} h_j(\vec{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\vec{u} \\ \vec{0}\end{bmatrix} $$

$$ v(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot v_j + \sum_{j=33}^{35} h_j(\vec{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\vec{v} \\ \vec{0}\end{bmatrix} $$ for an extended matrix $\tilde{H} \in \mathbb{R}^{1 \times 35}$.

First, we note that these "extended" objects

$$ \tilde{\vec{v}} = \begin{bmatrix} \vec{v} \\ \vec{0} \end{bmatrix} \quad \tilde{{A}} = \begin{bmatrix} {A} & {C} \\ {D} & {E} \\ \end{bmatrix} \quad \tilde{\vec{u}} = \begin{bmatrix} \vec{u} \\ \vec{0} \end{bmatrix} \quad \tilde{\vec{b}} = \begin{bmatrix} \vec{b} \\ \vec{e} \end{bmatrix} $$ still represent the original Galerkin problem. In effect,

$$ \begin{aligned} \tilde{\vec{v}}^T \cdot \tilde{{A}} \cdot \tilde{\vec{u}} &= \tilde{\vec{v}}^T \cdot \tilde{\vec{b}} \\ \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} {A} & {C} \\ {D} & {E} \\ \end{bmatrix} \cdot \begin{bmatrix} \vec{u} \\ \vec{0} \end{bmatrix} & = \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \begin{bmatrix} \vec{b} \\ \vec{e} \end{bmatrix} \\ \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} {A} \cdot \vec{u} + {C} \cdot \vec{0} \\ {D} \cdot \vec{u} + {E} \cdot \vec{0} \\ \end{bmatrix} &= \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \begin{bmatrix} \vec{b} \\ \vec{e} \end{bmatrix} \\ \vec{v}^T \cdot {A} \cdot \vec{u} + \vec{0}^T \cdot {D} \cdot \vec{u} &= \vec{v}^T \cdot \vec{b} + \vec{0}^T \cdot \vec{e}\\ \vec{v}^T \cdot {A} \cdot \vec{u} &= \vec{v}^T \cdot \vec{b}\\ \end{aligned} $$

Since this identity must hold $\forall \vec{v}$, then ${A} \cdot \vec{u} - \vec{b} = 0$, which is the original homogeneous problem. We now need to prove that if we have

$$ {K} = \begin{bmatrix} {A} & {C} \\ {0} & {I} \\ \end{bmatrix} \quad \vec{f} = \begin{bmatrix} \vec{b} \\ \vec{0} \\ \end{bmatrix} $$ such that ${A} \cdot \vec{u} = \vec{b}$, where ${I}$ is the identity matrix of size $3 \times 3$, then the vector $\vec{\varphi}$ such that ${K} \cdot \vec{\varphi} = \vec{f}$ is equal to

$$ \vec{\varphi} = \begin{bmatrix} \vec{u} \\ \vec{0} \\ \end{bmatrix} $$

Indeed, let $\vec{\varphi} = \begin{bmatrix} \vec{\varphi}_1 & \vec{\varphi}_2 \end{bmatrix}^T$. Then ${K} \cdot \vec{\varphi}$ is

$$ \begin{bmatrix} {A} & {C} \\ {0} & {I} \end{bmatrix} \cdot \begin{bmatrix} \vec{\varphi}_1 \\ \vec{\varphi}_2 \end{bmatrix} = \begin{bmatrix} {A} \cdot \vec{\varphi}_1 + {C} \cdot \vec{\varphi}_2\\ {0} \cdot \vec{\varphi}_1 + {I} \cdot \vec{\varphi}_2 \end{bmatrix} = \begin{bmatrix} \vec{b}\\ \vec{0} \end{bmatrix} $$

From the second row,  $\vec{\varphi}_2 = \vec{0}$. Replacing this result in the first row, ${A} \cdot \vec{\varphi}_1 = \vec{b}$. Therefore $\vec{\varphi}_1 = {A}^{-1} \cdot \vec{b} = \vec{u}$.

This part proves that putting a one in the diagonal of the stiffness matrix and a zero in the RHS vector in the rows corresponding to the nodes at $\Gamma_D$ is correct.

Now let's investigate non-homogeneous Dirichlet BCs. The "lifting" procedure from textbooks translates into finding $u_h \in V$ such that

$$ a(u_h,v) = \mathcal{B}(v) - a(u_g,v) \quad \forall \quad v \in V $$

where functions in $V$ vanish at $\Gamma_D$, and $u_g$ is a known function (i.e. a known lifting) that satisfies $u_g(\vec{x}) = g(\vec{x})$ for $\vec{x} \in \Gamma_D$. Note that the right-hand side of the formulation contains known functions only.

Now, let's take back the $u_g$ into the left-hand side (we have already assumed $a$ was bi-linear) then we have to solve

$$ a(u_h+u_g,v) = \mathcal{B}(v) $$

Let's first write the functions $u_h$ and $v$ over the whole set of nodes

$$ u_h(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot u_j + \sum_{j=33}^{35} h_j(\vec{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\vec{u} \\ \vec{0}\end{bmatrix} $$

$$ v(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot v_j + \sum_{j=33}^{35} h_j(\vec{x}) \cdot 0 = \tilde{H} \cdot \begin{bmatrix}\vec{v} \\ \vec{0}\end{bmatrix} $$ for another matrix $\tilde{H}$ of size $1 \times 35$.

We then write the non-homogeneous part $u_g$ as

$$ u_h(\vec{x}) = \sum_{j=1}^{32} h_j(\vec{x}) \cdot 0 + \sum_{j=33}^{35} h_j(\vec{x}) \cdot g(\vec{x}_j) = \tilde{H} \cdot \begin{bmatrix}\vec{0} \\ \vec{g}\end{bmatrix} $$ so the sum $u_h + u_g$ is

$$ u_h(\vec{x}) + u_g(\vec{x}) = \tilde{H} \cdot \begin{bmatrix}\vec{u} \\ \vec{g}\end{bmatrix} $$

Now, the stiffness matrix of the first homogeneous problem was $A \in \mathbb{R}^{32 \times 32}$. We then extend it to a new matrix of size $35 \times 35$

$$ \begin{bmatrix} A & C \\ D & E \end{bmatrix} $$ such that the discretized problem is now

$$ \begin{aligned} \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} A & C \\ D & E \end{bmatrix} \cdot \begin{bmatrix} \vec{u} \\ \vec{g} \end{bmatrix} &= \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} \vec{b} \\ \vec{e} \end{bmatrix} \\ \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} A \cdot \vec{u} + C \cdot \vec{g} \\ D \cdot \vec{u} + E \cdot \vec{g} \end{bmatrix} &= \begin{bmatrix} \vec{v}^T & \vec{0}^T \end{bmatrix} \cdot \begin{bmatrix} \vec{b} \\ \vec{e} \end{bmatrix} \\ \vec{v}^T \cdot A \cdot \vec{u} + \vec{v}^T \cdot C \cdot \vec{g} &= \vec{v}^T \cdot \vec{b} \end{aligned} $$ for all $\vec{u} \in \mathbb{R}^J$. That is to say,

$$ A \cdot \vec{u} + C \cdot \vec{g} = \vec{b} $$

Now you can prove as an exercise that if we have

$$ K = \begin{bmatrix} A & C \\ 0 & I \\ \end{bmatrix} \quad \vec{f} = \begin{bmatrix} \vec{b} \\ \vec{g} \\ \end{bmatrix} $$ such that $A \cdot \vec{u} + C \cdot \vec{g} = \vec{b}$, then the vector $\vec{\varphi}$ such that $K \cdot \vec{\varphi} = \vec{f}$ is equal to

$$ \vec{\varphi} = \begin{bmatrix} \vec{u} \\ \vec{g} \\ \end{bmatrix} $$