# How to show this weak formulation has unique solution?

Suppose $a$ is a bounded and coercive bilinear form on a Hilbert space $H$ and that $b$ is a bounded bilinear form on $H$ and $\ell$ is a bounded linear function also on $H$.

How do I show that: For sufficiently small $\varepsilon > 0$, the equation $$a(u^\epsilon, v) + \varepsilon b(u^\varepsilon, v) = \ell(v), \quad \mathit{ for\,\,all}\,\, v \in H,$$ has a unique solution $u^\varepsilon$.

I can't apply Lax-Milgram because $b$ is not necessarily coercive. What's the general technique for these kinds of problems?

If $$\epsilon$$ is small enough, $$a+\epsilon b$$ will be coercive.
You can define $$D(w,v):=a(w,v)+\epsilon b(w,v)$$. Since coercivity implies that the induced norm is bounded from below by an arbitrary vector up to the value of some constant $$C\in(0,\infty)$$, i.e. $$a(w,w)\geq C||w||_H$$ It is easy to show that $$D$$ is a bilinear form and that for small enough $$\epsilon$$ it must also be a coercive since if $$b(w,v)$$ is negative, you can reduce $$\epsilon$$ to arbitrarily small value ($$b$$ is bounded).