Let Σ_{2} be a translation surface of type 2 defined by
$$\begin{array}{}\mathbf{x}(x,y)=(x+y,g(y),f(x)),\end{array}$$
where *f* and *g* are smooth functions.

If Σ_{2} is a surface in *G*_{3} with a linear density *e*^{ϕ}, where *ϕ* = *ax* + *by* + *cz*, *a*, *b*, *c* not all zero, then the weighted minimality condition *H*_{ϕ} = 0 becomes
$$\begin{array}{}H-{\displaystyle \frac{1}{2\omega}\u3008(0,{f}^{\prime}(x),{g}^{\prime}(y)),(a,b,c)\u3009=0.}\end{array}$$(20)

Let us distinguish two cases according to the value of *a*.

Case 1. *a* ≠ 0.

In this case, *H* = 0 and from (12) we have
$$\begin{array}{}{f}^{\u2033}(x){g}^{\prime}(y)+{f}^{\prime}(x){g}^{\u2033}(y)=0.\end{array}$$

If *f* or *g* is constant, then Σ_{2} is a plane. Now, we assume that *f*′*g*′ ≠ 0. Then, the above equation writes as
$$\begin{array}{}{\displaystyle \frac{{f}^{\u2033}(x)}{{f}^{\prime}(x)}=-\frac{{g}^{\u2033}(y)}{{g}^{\prime}(y)}}\end{array}$$
which implies there exists a real number *m* ∈ ℝ such that
$$\begin{array}{}{\displaystyle \frac{{f}^{\u2033}(x)}{{f}^{\prime}(x)}=-\frac{{g}^{\u2033}(y)}{{g}^{\prime}(y)}=m.}\end{array}$$(21)

If *m* = 0, then the functions *f* and *g* are linear functions, which generates an isotropic plane.

If *m* ≠ 0, then the general solutions of (21) are given by
$$\begin{array}{}f(x)={\displaystyle \frac{1}{m}{e}^{mx}+{d}_{1},\phantom{\rule{1em}{0ex}}g(y)=-\frac{1}{m}{e}^{-my}+{d}_{2},}\end{array}$$
where *d*_{1}, *d*_{2} are constant.

Case 2. *a* = 0.

From (12) and (20) the weighted minimality condition *H*_{ϕ} = 0 becomes
$$\begin{array}{}{f}^{\u2033}(x){g}^{\prime}(y)+{f}^{\prime}(x){g}^{\u2033}(y)=({f}^{\mathrm{\prime}2}(x)+{g}^{\mathrm{\prime}2}(y))(b{f}^{\prime}(x)+c{g}^{\prime}(y)).\end{array}$$(22)

Taking the derivative with respect to *x* and next with respect to *y*, we have the following ordinary differential equation:
$$\begin{array}{}{f}^{\u2034}(x){g}^{\u2033}(y)+{f}^{\u2033}(x){g}^{\u2034}(y)=2b{f}^{\u2033}(x){g}^{\prime}(y){g}^{\u2033}(y)+2c{f}^{\prime}(x){f}^{\u2033}(x){g}^{\u2033}(y).\end{array}$$(23)

If *f*″(*x*) = 0 or *g*″(*y*) = 0, then *f* or *g* is a linear function. Now we assume that *f*″(*x*)*g*″(*y*) ≠ 0.

Subcase 2.1. *b* = 0. Dividing (23) by *f*″(*x*)*g*″(*y*), we have
$$\begin{array}{}{\displaystyle \frac{{f}^{\u2034}(x)}{{f}^{\u2033}(x)}-2c{f}^{\prime}(x)=-\frac{{g}^{\u2034}(y)}{{g}^{\u2033}(y)}.}\end{array}$$

Then there exists a real number *m* ∈ ℝ such that
$$\begin{array}{}{\displaystyle \frac{{f}^{\u2034}(x)}{{f}^{\u2033}(x)}-2c{f}^{\prime}(x)=-\frac{{g}^{\u2034}(y)}{{g}^{\u2033}(y)}=m.}\end{array}$$

If *m* = 0,
$\begin{array}{}g(y)={\displaystyle \frac{1}{2}{d}_{1}{y}^{2}+{d}_{2}y+{d}_{3}}\end{array}$
with *d*_{1}, *d*_{2}, *d*_{3} ∈ ℝ. Substituting the function *g*(*y*) into (22), equation (22) is a polynomial in *y* with functions of *x* as coefficients. Thus, all the coefficients must be zero. The coefficient of the highest degree *y*^{3} in (22) is
$\begin{array}{}c{d}_{1}^{3},\end{array}$
thus it follows *d*_{1} = 0 and *g*″(*y*) = 0, a contradiction.

Suppose that *m* ≠ 0. The solution of the ODE *g*‴(*y*) + *m**g*″(*y*) = 0 is
$$\begin{array}{}g(y)={m}^{2}{e}^{-my+{d}_{1}}+{d}_{2}y+{d}_{3}.\end{array}$$

Substituting the function *g*(*y*) into (22), we get a polynomial on *e*^{−my + d1} with functions of *x* as coefficients. In the polynomial we can obtain the coefficient of *e*^{3(−my + d1)} and it is −*m*^{2}*c*, a contradiction.

Subcase 2.2. *c* = 0. This subcase is similar to the previous one. Thus, there are no weighted minimal translation surfaces of type 2 in *G*_{3} with a linear density *e*^{by}.

Subcase 2.3. *bc* ≠ 0. Dividing (23) by *f*″(*x*)*g*″(*y*), we get
$$\begin{array}{}{\displaystyle \frac{{f}^{\u2034}(x)}{{f}^{\u2033}(x)}-2c{f}^{\prime}(x)=2b{g}^{\prime}(y)-\frac{{g}^{\u2034}(y)}{{g}^{\u2033}(y)}.}\end{array}$$

Hence, we deduce the existence of a real number *m* ∈ ℝ such that
$$\begin{array}{}{\displaystyle \frac{{f}^{\u2034}(x)}{{f}^{\u2033}(x)}-2c{f}^{\prime}(x)=2b{g}^{\prime}(y)-\frac{{g}^{\u2034}(y)}{{g}^{\u2033}(y)}=m.}\end{array}$$(24)

If *m* = 0, the general solution of (24) is given by
$$\begin{array}{}f(x)=-{\displaystyle \frac{1}{c}\mathrm{ln}|-c(x+{d}_{2})|-\frac{{d}_{1}}{c},}\\ g(y)=-{\displaystyle \frac{1}{b}\mathrm{ln}|-b(y+{d}_{4})|-\frac{{d}_{3}}{b},}\end{array}$$(25)
where *d*_{i}(*i* = 1, ⋯, 4) are constant. Substituting the function (25) into (22) we can obtain the equation:
$$\begin{array}{}{b}^{4}{\displaystyle \frac{1}{(x+{d}_{2}{)}^{3}}+{c}^{4}\frac{1}{(y+{d}_{4}{)}^{3}}=0,}\end{array}$$
which is impossible.

Suppose *m* ≠ 0. Equation (24) can be written as
$$\begin{array}{}{f}^{\u2034}(x)-c({f}^{\mathrm{\prime}2}(x){)}^{\prime}=m{f}^{\u2033}(x),\\ {g}^{\u2034}(y)-b({g}^{\mathrm{\prime}2}(y){)}^{\prime}=-m{g}^{\u2033}(y).\end{array}$$(26)

First, in order to solve the first equation of (26) we put *p* = *f*′(*x*). Then we have
$$\begin{array}{}{\displaystyle \frac{dp}{dx}=p(cp+m)}\end{array}$$
and its solution is
$$\begin{array}{}p={\displaystyle \frac{m}{{e}^{-m(x+{d}_{1})}-c}.}\end{array}$$

Thus a first integration implies
$$\begin{array}{}f(x)={m}^{2}{\displaystyle \int \frac{1}{{e}^{-m(x+{d}_{1})}-c}.}\end{array}$$(27)

By using the similar method, the function *g* satisfying the second equation of (26) is given by
$$\begin{array}{}g(y)=-{m}^{2}{\displaystyle \int \frac{1}{{e}^{m(y+{d}_{1})}-b}.}\end{array}$$(28)

When substituting (27) and (28) into (22), equation (22) appears a polynomial in *e*^{−m(x + d1)} and *e*^{m(y+d1)}. The coefficient of *e*^{− 3m(x + d1)} in the polynomial is −*cm*^{3}, which is impossible because *m* and *c* are non-zero constant.

Thus we have:

#### Theorem 5.1

*Let* Σ_{2} *be a translation surface of type 2 in the Galilean 3*-*space* *G*_{3} *with a log*-*linear density* *e*^{ax + by + cz}. *If* Σ_{2} *is weighted minimal, then* Σ_{2} *is an isotropic plane or parameterized as* **x**(*x*, *y*) = (*x* + *y*,
$\begin{array}{}-\frac{1}{m}{e}^{-my}+{d}_{1},\frac{1}{m}{e}^{mx}+{d}_{2})\end{array}$
*with* *m* ≠ 0, *d*_{1}, *d*_{2} ∈ ℝ.

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