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As explained by Noam Elkies, the fibers you want are of Kodaira type:

• $I_2$ : two rational curves intersecting at two distinct points
• $III$ : two rational curves meeting at a double point

Since these are the only fibers you want, I will recommand not using a Weierstrass equation but rather a Jacobi quartic form for your elliptic fibration. Indeed, a smooth Weierstrass model admits only singular fibers of type $I_1$ (nodal curves) and type $II$ (cuspidial curve). So if you want to have singular fibers of type $I_2$ and $III$, you will have to introduce singularities in the total space of your fibration and resolve them following Tate's algorithm.

You can avoid dealing with the resolution of singularities if you start with the Jacobi quartic form. Consider a weighted projective plane $\mathbb{P}^2_{1,2,1}$ bundle, a smooth quartic equation will define an elliptic curve as you can see by using the adjunction formula. The canonical form of such a curve is

$$\text{Jacobi form}: \quad y^2=x^4+ e x^2 z^2+ f x z^3+ g z^4$$

where $[x:y:z]$ are the projective coordinates of weight $1$, $2$ and $1$ respectively.

For a fibration $Y\rightarrow B$, the weighted projective plane is replaced by a weighted projective bundle $\mathbb{P}_{1,2,1}[\mathscr{O}\oplus\mathscr{L}\oplus\mathscr{L}^2]$ with weight $1,2,1$, where $\mathscr{L}\$ is a line bundle over the base $B$ of your fibration. In that equation $e,f,g$ are now sections of $\mathscr{L}^2$, $\mathscr{L}^3$ and $\mathscr{L}^4$ respectively. The projective coordinates $x$, $z$, $y$ are respectively sections of $\mathscr{L}\otimes \mathscr{O}(1)$, $\mathscr{O}(1)$ and $\mathscr{L}^2\otimes\mathscr{O}(2)$, where $\mathscr{O}(1)$ is the tautological line bundle of the weighted projective bundle.

Using the adjuction, you can see that you have a $K3$ elliptic fibration iff $c_1(\mathscr{L})=c_1(B)$.

In case you actually want a Weierstrass model, the Jacobian fibration will give you a birationally equivalent Weierstrass model: $y^2 =x^3+ Fx + G$ with $F=-4 g +e^2/3$ and $G=f^2-8/3 eg +2e^3/27$. You can compute the birational transformation using Maple.

Going back to the Jacobi form, it is not difficult to see that a smooth elliptic fibration will have fibers of type $I_1$, $II$, $I_2$ and $III$.

In particular, the fibers are:

• type $I_1$ at a general point of $4 F^3+27 G^2=0$
• type $II$ at $F=G=0$ and $e\neq 0$
• type $I_2$ iff $f= e^2-4 g=0$ and $g\neq 0$
• type $III$ iff $f=e=g=0$.

With a $\mathbb{P}^1$ base and $\mathscr{L}=O(2H)$, you have a $K3$ surface with fibers $I_1,II, I_2, III$.

You can avoid the fiber $I_1$ and $II$ by manipulating the equation. For example if you take the equation with a $\mathbb{P}^1$ base and $\mathscr{L}=O(2H)$, you can write the following K3 elliptic surface

$$y^2=x^4+ g z^4\Longrightarrow \ \text{8 fibers of type}\ III$$

You need $g$ to be a smooth polynomial of degree 8 in $P^1$. For example $g=x_1^8+x_0^8$ with $[x_0:x_1]$ the projective coordinates of $\mathbb{P}^1$. You will have $8$ fibers of type $III$ at the zero of $g$.Note that with that choice, the singular fibers are at the vertices of an octahedron octagon on $\mathbb{P}^1$ defined by $x_1^8+x_0^8=0$.

These types of fibrations based on the Jacobi quartic form are known as $E_7$ elliptic fibration in the string theory literature. They have some nice topological properties and transitions. You can read more about them in this paper.

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As explained by Noam Elkies, the fibers you want are of Kodaira type:

• $I_2$ : two rational curves intersecting at two distinct points
• $III$ : two rational curves meeting at a double point

Since these are the only fibers you want, I will recommand not using a Weierstrass equation but rather a Jacobi quartic form for your elliptic fibration. Indeed, a smooth Weierstrass model admits only singular fibers of type $I_1$ (nodal curves) and type $II$ (cuspidial curve). So if you want to have singular fibers of type $I_2$ and $III$, you will have to introduce singularities in the total space of your fibration and resolve them following Tate's algorithm.

You can avoid dealing with the resolution of singularities if you start with the Jacobi quartic form. Consider a weighted projective plane $\mathbb{P}^2_{1,2,1}$ bundle, a smooth quartic equation will define an elliptic curve as you can see by using the adjunction formula. The canonical form of such a curve is

$$\text{Jacobi form}: \quad y^2=x^4+ e x^2 z^2+ f x z^3+ g z^4$$

where $[x:y:z]$ are the projective coordinates of weight $1$, $2$ and $1$ respectively.

For a fibration $Y\rightarrow B$, the weighted projective plane is replaced by a weighted projective bundle $\mathbb{P}_{1,2,1}[\mathscr{O}\oplus\mathscr{L}\oplus\mathscr{L}^2]$ with weight $1,2,1$, where $\mathscr{L}\$ is a line bundle over the base $B$ of your fibration. In that equation $e,f,g$ are now sections of $\mathscr{L}^2$, $\mathscr{L}^3$ and $\mathscr{L}^4$ respectively. The projective coordinates $x$, $z$, $y$ are respectively sections of $\mathscr{L}\otimes \mathscr{O}(1)$, $\mathscr{O}(1)$ and $\mathscr{L}^2\otimes\mathscr{O}(2)$, where $\mathscr{O}(1)$ is the tautological line bundle of the weighted projective bundle.

Using the adjuction, you can see that you have a $K3$ elliptic fibration iff $c_1(\mathscr{L})=c_1(B)$.

In case you actually want a Weierstrass model, the Jacobian fibration will give you a birationally equivalent Weierstrass model: $y^2 =x^3+ Fx + G$ with $F=-4 g +e^2/3$ and $G=f^2-8/3 eg +2e^3/27$. You can compute the birational transformation using Maple.

Going back to the Jacobi form, it is not difficult to see that a smooth elliptic fibration will have fibers of type $I_1$, $II$, $I_2$ and $III$.

In particular, the fibers are:

• type $I_1$ at a general point of $4 F^3+27 G^2=0$
• type $II$ at $F=G=0$ and $e\neq 0$
• type $I_2$ iff $f= e^2-4 g=0$ and $g\neq 0$
• type $III$ iff $f=e=g=0$.

With a $\mathbb{P}^1$ base and $\mathscr{L}=O(2H)$, you have a $K3$ surface with fibers $I_1,II, I_2, III$.

You can avoid the fiber $I_1$ and $II$ by manipulating the equation. For example if you take the equation with a $\mathbb{P}^1$ base and $\mathscr{L}=O(2H)$, you can write the following K3 elliptic surface

$$y^2=x^4+ g z^4\Longrightarrow \ \text{8 fibers of type}\ III$$

You need $g$ to be a smooth polynomial of degree 8 in $P^1$. For example $g=x_1^8+x_0^8$ with $[x_0:x_1]$ the projective coordinates of $\mathbb{P}^1$. You will have $8$ fibers of type $III$ at the zero of $g$.Note that with that choice, the singular fibers are at the vertices of an octahedron on $\mathbb{P}^1$ defined by $x_1^8+x_0^8=0$.

These types of fibrations based on the Jacobi quartic form are known as $E_7$ elliptic fibration in the string theory literature. They have some nice topological properties and transitions. You can read more about them in this paper.

2 added 258 characters in body

As explained by Noam Elkies, the fibers you want are of Kodaira type:

• $I_2$ : two rational curves intersecting at two distinct points
• $III$ : two rational curves meeting at a double point

Since these are the only fibers you want, I will recommand not using a Weierstrass equation but rather a Jacobi quartic form for your elliptic fibration. Indeed, a smooth Weierstrass model admits only singular fibers of type $I_1$ (nodal curves) and type $II$ (cuspidial curve). So if you want to have singular fibers of type $I_2$ and $III$, you will have to introduce singularities in the total space of your fibration and resolve them following Tate's algorithm.

You can avoid dealing with the resolution of singularities if you start with the Jacobi quartic form. Consider a weighted projective plane $\mathbb{P}^2_{1,2,1}$ bundle, a smooth quartic equation will define an elliptic curve as you can see by using the adjunction formula. The canonical form of such a curve is

$$\text{Jacobi form}: \quad y^2=x^4+ e x^2 z^2+ f x z^3+ g z^4$$

where $[x:y:z]$ are the projective coordinates of weight $1$, $2$ and $1$ respectively.

For a fibration $Y\rightarrow B$, the weighted projective plane is replaced by a weighted projective bundle $\mathbb{P}_{1,2,1}[\mathscr{O}\oplus\mathscr{L}\oplus\mathscr{L}^2]$ with weight $1,2,1$, where $\mathscr{L}\$ is a line bundle over the base $B$ of your fibration. In that equation $e,f,g$ are now sections of $\mathscr{L}^2$, $\mathscr{L}^3$ and $\mathscr{L}^4$ respectively. The projective coordinates $x$, $z$, $y$ are respectively sections of $\mathscr{L}\otimes \mathscr{O}(1)$, $\mathscr{O}(1)$ and $\mathscr{L}^2\otimes\mathscr{O}(2)$, where $\mathscr{O}(1)$ is the tautological line bundle of the weighted projective bundle.

Using the adjuction, you can see that you have a $K3$ elliptic fibration iff $c_1(\mathscr{L})=c_1(B)$.

In case you actually want a Weierstrass model, the Jacobian fibration will give you a birationally equivalent Weierstrass model: $y^2 =x^3+ Fx + G$ with $F=-4 g +e^2/3$ and $G=f^2-8/3 eg +2e^3/27$. You can compute the birational transformation using Maple.

Going back to the Jacobi form, it is not difficult to see that a smooth elliptic fibration will have fibers of type $I_1$, $II$, $I_2$ and $III$.

In particular, the fibers are:

• type $I_1$ at a general point of $4 F^3+27 G^2=0$
• type $II$ at $F=G=0$ and $e\neq 0$
• type $I_2$ iff $f= e^2-4 g=0$ and $g\neq 0$
• type $III$ iff $f=e=g=0$.

With a $\mathbb{P}^1$ base and $\mathscr{L}=O(2H)$, you have a $K3$ surface with fibers $I_1,II, I_2, III$.

You can avoid the fiber $I_1$ and $II$ by manipulating the equation. For example if you take the equation with a $\mathbb{P}^1$ base and $\mathscr{L}=O(2H)$, you can write the following K3 elliptic surface

$$y^2=x^4+ g z^4\Longrightarrow \ \text{8 fibers of type}\ III$$

You need $g$ to be a smooth polynomial of degree 8 in $P^1$. For example $g=x_1^8+x_0^8$ with $[x_0:x_1]$ the projective coordinates of $\mathbb{P}^1$. You will have $8$ fibers of type $III$ at the zero of $g$.

These types of fibrations based on the Jacobi quartic form are known as $E_7$ elliptic fibration in the string theory literature. They have some nice topological properties and transitions. You can read more about them in this paper.

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