Let $X$ be a projective, smooth variety over $\mathbb{C}$, and let $D$ be a irreducible, **big** Cartier divisor(notice, I do not assume nefness). Then is it true that $${\rm{H}}^1(X, K\_X + D) = 0\quad?$$


On the one hand, I doubt it because the result seems to be written nowhere; on the other hand, I feel the conditions here are much nicer than many vanishing theorem, and I was wondering if some modification would reduce it to one of the vanishing theorem. 


Besides, any references/results/comments on vanishing theorem for big divisors are great welcome!



Vanishing theorem for big divisors
Let $X\to \mathbb P ^1$ be a family of smooth quadric surfaces degenerating to $X\_0=\mathbb F \_2=\mathbb P (\mathcal O \_{\mathbb P ^1}\oplus \mathcal O \_{\mathbb P ^1} (2))$ where $0\in \mathbb P ^1$. We denote by $E$ the $-2$ curve in $X\_0$, $F$ the fiber of $X\_0\to \mathbb P ^1$ and by $f:X\to Z$ the corresponding small contraction (following the notation of arXiv:0901.0389 remark 4.4 and remark 4.9). For $b>>0$ consider $L$ a line bundle whose restriction to the general fiber is $\mathcal O \_{\mathbb P ^1} (b+1)\boxtimes \mathcal O \_{\mathbb P ^1} (b-1)$ and whose restriction to the special fiber is $\mathcal O \_{X\_0}(b(E+2F)+E)$ and $D\in |L|$ a general member. 


We claim that $R^1f\_\*\omega \_X(D)\ne 0$. To see this consider the short exact sequence $0\to \omega \_X(D-X\_0)\to \omega \_X(D)\to \omega \_{X}(D)|\_{X\_0}\to 0$. Since $R^if\_\*\omega \_X(D-X\_0)\cong R^if\_\*\omega \_X(D)\otimes \mathcal O\_{Z}(-Z\_0)$ (by the projection formula), then it follows that $R^1f\_\* \omega \_X(D)\to R^1f\_\*\omega \_{X}(D)|\_{X\_0}\cong R^1f\_\* \omega \_{X\_0}(D|\_{X\_0})$ is surjective. Consider now the short exact sequence $0\to \omega \_{X\_0}\to \omega \_{X\_0}(D|\_{X\_0})\to \omega \_{D|\_{X\_0}}\to 0$. Since $R^1f\_\*\omega \_{X\_0}=0$ (by Grauert-Riemanschneider), it follows that $R^1f\_\* \omega \_{X\_0}(D|\_{X\_0})\to R^1f\_\*\omega \_{D|\_{X\_0}}$ is surjective. 
Since $|L|\_{X\_0}|=|b(E+2F)|+E$ where $E+2F$ is the pull back of an ample divisor $H$ on $X\_0$, we have $R^1f\_\*\omega \_{D|\_{X\_0}}\cong R^1f\_\*\omega \_E\otimes \mathcal O \_{Z\_0}(bH)$. Since $H^1(\omega \_E)\cong H^0(\mathcal O \_E)^\vee\ne 0$, we have $R^1f\_\*\omega \_{D|\_{X\_0}}\ne 0$ and the claim follows.


It should be easy to see that $D$ is big, it is not nef because $D\cdot E =E^2=-2$, moreover we can arrange that $H^1(f\_\*\omega \_X(D))=0$ (if necessary replace $D$ by $D$ plus the pull-back of a sufficiently ample divisor on $Z$ and apply Serre vanishing).
Then, by the Leray spectral sequence $H^1(\omega \_X (D))\cong H^0(R^1f\_\*\omega \_X (D))\ne 0$.


Maybe there are easier examples, but this is the first one that came to mind. I do not think there can be surface examples, because if $D$ is irreducible and not nef, then $D^2<0$, but as $D$ is big, $D\sim \_{\mathbb Q}D'\neq D$ subtracting common components of $D$ we obtain $\lambda D\sim \_{\mathbb Q}D''$ with $0<\lambda$ and the support of $D''$ does not contain $D$. But then $D^2=\frac 1 {\lambda ^2} D\cdot D''\geq 0$ a contradiction.


