How to increase unbounding and dominating numbers for $(\kappa^\lambda,\leq^*)$

Let $$\kappa$$ be regular and $$\lambda\geq\kappa$$. For $$f, g\in\kappa^\lambda$$ say that $$f\le^* g$$ if the set $$\{\gamma<\lambda:f(\gamma)>g(\gamma)\}$$ has size less than $$\kappa$$. Set

$$\mathfrak{b}_\kappa^\lambda:=\min\{|F|:F\subseteq \kappa^\lambda\text{ and }\neg\exists y\in \kappa^\lambda\forall x\in F(x\leq^* y)\}$$,

$$\mathfrak{d}_\kappa^\lambda:=\min\{|D|:D\subseteq \kappa^\lambda\text{ and }\forall x\in \kappa^\lambda\exists y\in D(x\leq^* y)\}$$.

I'm studying the cardinals $$\mathfrak{b}_\kappa^\lambda$$ and $$\mathfrak{d}_\kappa^\lambda$$. My question what forcing should I use to use to increase $$\mathfrak{b}_\kappa^\lambda$$ and $$\mathfrak{d}_\kappa^\lambda$$?

For example: If $$\lambda\geq\kappa$$, $$\mu>\lambda$$ and $$\mathrm{cf}(\mu)>\lambda$$. What forcing can I use for $$\mathfrak{d}_\kappa^\lambda\geq\mu$$?

Short version: an appropriate version of Cohen forcing will increase $$\mathfrak d^\lambda_\kappa$$.

With more details: Consider the forcing $$Q=Q(\mu,\kappa,\mathord<\kappa)$$, the set of all partial functions from $$\mu$$ (equivalently, from $$\mu\times \lambda$$) into $$\kappa$$ of size $$<\kappa$$. The generic function $$g:\mu\times\lambda\to \kappa$$ induces a family $$(g_i:i<\mu)$$, each $$g_i$$ a function from $$\lambda$$ to $$\kappa$$.

Assuming $$\kappa^{<\kappa} = \kappa$$, the forcing notion $$Q$$ has the $$\kappa^+$$-cc, and adds no bounded sets to $$\kappa$$, hence preserves all cardinals.

I claim that $$Q$$ forces $${\mathfrak d^\kappa_\lambda \ge \mu}$$. Indeed, any function $$f\colon \lambda\to \kappa$$ in the extension already lives in a $$Q(A, \kappa, \mathord<\kappa)$$-extension, for some $$A$$ of size $$\lambda$$, as the values of $$f$$ are decided by a family of $$\lambda$$ many (labelled) antichains. Such a function $$f$$ cannot be forced to dominate any $$g_i$$ for $$i\notin A$$. Hence if you have fewer than $$\mu$$ many functions, you will find an index $$i<\mu$$ such that these functions cannot dominate $$g_i$$.

A partial answer concerning $$\mathfrak{b}_\kappa^\lambda$$:

Assume $$\lambda > \kappa$$, regular. I claim $$\mathfrak{b}_\kappa^\lambda =\kappa$$:

It is obvious, why $$\mathfrak{b}_\kappa^\lambda <\kappa$$ is impossible.

The family $$(f_\alpha \equiv \alpha)_{\alpha < \kappa}$$ is unbounded. Assume $$g$$ dominates all $$f_\alpha$$. As $$\lambda$$ is regular, there exists $$\beta < \kappa$$ such that $$\{i < \lambda \colon g(i) = \beta\}$$ has size $$\lambda$$. So $$g$$ does not dominate $$f_{\beta +1}$$.