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).


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