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A Remark on the Continuous Subsolution Problem for the Complex Monge-Ampère Equation

  • Sławomir Kołodziej
  • Ngoc Cuong NguyenEmail author
Open Access
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Abstract

We prove that if the modulus of continuity of a plurisubharmonic subsolution satisfies a Dini-type condition then the Dirichlet problem for the complex Monge-Ampère equation has the continuous solution. The modulus of continuity of the solution also given if the right hand side is locally dominated by capacity.

Keywords

Dirichlet problem Complex Monge-Ampère equation Weak solutions Subsolution problem 

Mathematics Subject Classification (2010)

32W20 32U40 

1 Introduction

In this note, we consider the Dirichlet problem for the complex Monge-Ampère equation in a strictly pseudoconvex domain \({\varOmega }\subset \mathbb {C}^{n}\). Let ψ be a continuous function on the boundary of Ω. We look for the solution to the equation:
$$ \begin{array}{@{}rcl@{}} && u\in PSH({\varOmega}) \cap C^{0}(\bar{\varOmega}), \\ && (dd^{c} u)^{n} = d\mu, \\ && u = \psi \quad \text{on }\partial\varOmega. \end{array} $$
(1.1)
Here, PSH stands for plurisubharmonic functions, and \(d^{c} = i (\overline {\partial } -\partial )\). It was shown in [9] and [10] that for the measures satisfying certain bound in terms of the Bedford-Taylor capacity [4], the Dirichlet problem has a (unique) solution. The precise statement is as follows.
Let \(h :\mathbb R_{+} \rightarrow (0, \infty ) \) be an increasing function such that
$$ {\int}_{1}^{\infty} \frac{1}{x [h(x) ]^{\frac{1}{n}} } dx < +\infty. $$
We call such a function admissible. If h is admissible, then so is Ah for any number A > 0. Define
$$ F_{h}(x) = \frac{x}{h(x^{-\frac{1}{n}})}. $$
Suppose that for such a function Fh(x) a Borel measure μ satisfies
$$ {\int}_{E} d\mu \leq F_{h}(\text{cap}(E)) $$
(1.2)
for any Borel set EΩ. Then, by [9] the Dirichlet problem (1.1) has a solution.
This statement is useful as long as we can verify the condition (1.2). In particular, if μ has density with respect to the Lebesgue measure in Lp, p > 1 then this bound is satisfied [9]. By the recent results in [12, 13] if μ is bounded by the Monge-Ampère measure of a Hölder continuous plurisubharmonic function φ
$$ \mu \leq (dd^{c}\varphi)^{n} \quad \text{in }{\varOmega} , $$
then (1.2) holds for a specific h, and consequently, the Dirichlet problem (1.1) is solvable with Hölder continuous solution. The main result in this paper says that we can considerably weaken the assumption on φ and still get a continuous solution of the equation.
Let \(\varpi (t):= \varpi (t;\varphi ,\bar {\varOmega })\) denote the modulus of continuity of φ on \(\bar {\varOmega }\), i.e.,
$$ \varpi(t)= \sup \left\{|\varphi(z) -\varphi(w)| : z,w \in \bar{\varOmega}, |z-w| \leq t\right\}. $$
Thus |φ(z) − φ(w)|≤ ϖ(|zw|) for every \( z,w\in \bar {\varOmega }\). Let us state the first result.

Theorem 1.1

Let\(\varphi \in PSH({\varOmega }) \cap C^{0}(\bar {\varOmega }), \)φ = 0 onΩ.Assume that its modulus of continuity satisfies the Dini type condition
$$ {{\int}_{0}^{1}} \frac{[\varpi(t)]^{\frac{1}{n}}}{t |\log t|} dt <+\infty. $$
(1.3)
If the measure μ satisfies μ ≤ (ddcφ)n in Ω, then the Dirichlet problem (1.1) admits a unique solution.

Let us mention in this context that it is still an open problem if a continuous subsolution φ implies the solvability of (1.1).

The modulus of continuity of the solution to the Dirichlet problem (1.1) was obtained in [3] for μ = fdV2n with f(x) being continuous on \(\bar {\varOmega }\). We also wish to study this problem for the measures which satisfy the inequality (1.2). For simplicity, we restrict ourselves to measures belonging to \(\mathcal H(\alpha ,{\varOmega })\). In other words, we take the function h(x) = Cxnα for positive constants C,α > 0 in the inequality (1.2).

We introduce the following notion, which generalizes the one in [8]. Consider a continuous increasing function \(F_{0}:[0,\infty ) \to [0,\infty )\) with F(0) = 0.

Definition 1.2

The measure μ is called uniformly locally dominated by capacity with respect to F0 if for every cube I(z,r) =: IBI := B(z,2r) ⊂⊂Ω and for every set EI,
$$ \mu(E) \leq \mu(I) F_{0}\left( \text{cap} (E, B_{I}) \right). $$
(1.4)

According to [1], the Lebesgue measure dV2n satisfies this property with \(F_{0} = C_{\alpha } \exp (-\alpha / x^{-1/n})\) for every 0 < α < 2n. The case F0(x) = Cx was considered in [8]. We refer the reader to [5] for more examples of measures satisfying this property. Here is our second result.

Theorem 1.3

Assume\(\mu \in \mathcal H(\alpha ,{\varOmega })\)withcompact support and satisfying the condition (1.4) for someF0. Then,the modulus of continuity of the solution u of the Dirichlet problem(1.1) satisfies for0 < δ < R0and2R0 = dist(suppμ,Ω) > 0,
$$ \varpi(\delta;u,{\varOmega}) \leq \varpi(\delta;\psi,\partial{\varOmega}) + C \left[ \left( \log \frac{R_{0}}{\delta}\right)^{-\frac{1}{2}} + F_{0} \left( \frac{C_{0}}{[\log (R_{0}/\delta)]^{\frac{1}{2}}}\right)\right]^{\alpha_{1}}, $$
where the constants C,α1 depend only on α,μ,Ω.

2 Preliminaries

Here, we gather some basic facts from pluripotential theory taken from [4], and used in the sequel. Given a compact set K in a domain \({\varOmega }\subset \mathbb {C}^{n}, \) its relative extremal function uK is given by
$$ u_{K} = \sup \{ u\in PSH ({\varOmega} ) : u<0, u\leq -1 \text{ on } K\}. $$
Its upper semicontinuous regularization \(u_{K}^{\ast }\) is plurisubharmonic. When uK is continuous, we call K a regular set. It is easy to see that the 𝜖-envelope
$$ K_{\epsilon } =\{ z : \operatorname{dist} (z, K) \leq \epsilon \} $$
of a compact set K is regular, and thus any compact set can be approximated from above by regular compact sets.
The relative capacity of a compact set K with respect to Ω (now usually called the Bedford-Taylor capacity) is defined by the formula
$$ \text{cap} (K, {\varOmega} ) = \sup \left\{ {\int}_{K} (dd^{c} u )^{n} : u\in PSH ({\varOmega} ), -1\leq u\leq 0 \right\}, $$
and by [4], can be expressed as
$$ \text{cap} (E, {\varOmega} ) = {\int}_{K} (dd^{c} u_{K}^{\ast } )^{n} . $$
We say that a positive Borel measure μ belongs to \(\mathcal H(\alpha ,{\varOmega })\), α > 0, if there exists a uniform constant C > 0 such that for every compact set EΩ,
$$ \mu(E) \leq C \left[\text{cap} (E,{\varOmega}) \right]^{1+ \alpha}. $$

3 Proof of Theorem 1.1

In this section, we shall prove Theorem 1.1. We need the following lemma. The proof of this lemma is based on a similar idea as the one in [11, Lemma 3.1] where the complex Hessian equation is considered. The difference is that we have much stronger volume-capacity inequality for the Monge-Ampère equation.

Lemma 3.1

Assume the measureμiscompactly supported. Fix 0 < α < 2nandτ = α/(2n + 1).There exists a uniform constant C such that for every compact setKΩ,
$$ \mu(K) \leq C \left\{ \varpi\left( \exp \left( \frac{-\tau}{2[\text{cap}(K)]^{\frac{1}{n}}}\right) \right) + \exp\left( \frac{2n\tau -\alpha}{2[\text{cap}(K)]^{\frac{1}{n}}} \right)\right\} \cdot \text{cap}(K), $$
(3.1)
where cap(K) := cap(K,Ω).

Proof

Fix a compact subset K ⊂⊂Ω. Without loss of generality, we may assume that K is regular. Denote by φε the standard regularization of φ in the terminology of [10]. We choose ε > 0 so small that
$$ \operatorname{supp} \mu \subset {\varOmega}^{\prime\prime} \subset\subset {\varOmega}^{\prime}\subset {\varOmega}_{\varepsilon} \subset \varOmega,$$
where Ωε = {zΩ : dist(z,Ω) > ε}. Since for every \(K \subset {\varOmega }^{\prime \prime }\) we have
$$ C_{0} \text{cap}(K,{\varOmega}) \leq \text{cap}(K, {\varOmega}^{\prime}) \leq C_{0}^{-1} \text{cap}(K,{\varOmega})$$
(for a constant C0 depending only on \(\varOmega , {\varOmega }^{\prime }\)) in what follows we shall write cap(K) for either one of these capacities. We have
$$ 0\leq \varphi_{\varepsilon} - \varphi \leq \varpi(\varepsilon) := \delta \quad \text{on } {\varOmega}^{\prime}.$$
Let uK be the relative extremal function of K with respect to \({\varOmega }^{\prime }\). Consider the set \(K^{\prime } = \{ 3\delta u_{K} + \varphi _{\varepsilon } < \varphi - 2\delta \}\). Then,
$$ K \subset K^{\prime} \subset \left\{u_{K} < -\frac{1}{2} \right\} \subset {\varOmega}^{\prime}. $$
(3.2)
Hence, by the comparison principle [4],
$$ \text{cap}(K^{\prime}) \leq 2^{n} \text{cap}(K). $$
(3.3)
Note that
$$ dd^{c} \varphi_{\varepsilon} \leq \frac{C}{\varepsilon^{2}} dd^{c} |z|^{2}, \quad \|\varphi_{\varepsilon} + u_{K}\|_{\infty}=:M \leq \|\varphi\|_{\infty} +1. $$
(3.4)
The comparison principle, the bounds (3.4), and the volume-capacity inequality from [1, Theorem A] (in the last inequality below) give the following:
$$ \begin{array}{@{}rcl@{}} {\int}_{K^{\prime}} (dd^{c} \varphi )^{n} &\leq& {\int}_{K^{\prime}} (dd^{c} (3\delta u_{K} + \varphi_{\varepsilon}) )^{n} \\ &\leq& 3\delta{\int}_{K^{\prime}} \left[dd^{c} (u_{K} + \varphi_{\varepsilon})\right]^{n} + {\int}_{K^{\prime}} (dd^{c} \varphi_{\varepsilon} )^{n} \\ &\leq& 3\delta M^{n} \text{cap}(K^{\prime}) + C(\alpha) \varepsilon^{-2n} \exp\left( \frac{-\alpha}{[\text{cap}(K^{\prime})]^{\frac{1}{n}}} \right) \text{cap}(K^{\prime}). \end{array} $$
Choose
$$ \varepsilon = \exp\left( \frac{-\tau}{[\text{cap}(K^{\prime})]^{\frac{1}{n}}} \right) $$
(we assume that ε is so small that it satisfies (3.2), otherwise the inequality (3.1) holds true by increasing the constant) and plug in the formula for δ to get that
$$ \begin{array}{@{}rcl@{}} \mu(K) &&\leq {\int}_{K^{\prime}} (dd^{c} (\varphi) )^{n} \\ &&\leq 3 M^{n} \varpi\left( \exp \left( \frac{-\tau}{[\text{cap}(K^{\prime})]^{\frac{1}{n}}}\right) \right) \cdot \text{cap}(K^{\prime}) + C \exp\left( \frac{2n\tau -\alpha}{[\text{cap}(K^{\prime})]^{\frac{1}{n}}} \right). \end{array} $$
This combined with (3.3) gives the desired inequality. □
We are ready to finish the proof of the theorem. It follows from Lemma 3.1 that
$$ h (x)= \frac{1}{C \varpi(\exp (-\tau x))} $$
is a function which satisfies (1.2) for the measure μ once we have
$$ {\int}_{1}^{\infty} \frac{1}{x [h(x) ]^{\frac{1}{n}} } dx < +\infty. $$
By changing the variable s = 1/x, and then t = eτ/s, this is equivalent to
$$ {\int}_{0}^{e^{-\tau}} \frac{\left[\varpi(t) \right]^{\frac{1}{n}}}{t |\log t|} dt <+\infty. $$
The last inequality is guaranteed by (1.3). Thus, our assumption on the modulus of continuity ϖ(t) implies that h is admissible in the case of μ with compact support. Then, by [10, Theorem 5.9] the Dirichlet problem (1.1) has a unique solution.
To deal with the general case, consider the exhaustion of Ω by compact sets
$$ E_{j} =\{ \varphi \leq -1/j \} $$
and define μj to be the restriction of μ to Ej. Denote by uj the solution of (1.1) with μ replaced by μj. By the comparison principle
$$ u_{j} + \max (\varphi , -1/j ) \leq u \leq u_{j} , $$
and so the sequence uj tends to \(u=\lim u_{j}\) uniformly and the continuity of u follows. The proof is complete.

4 The Modulus of Continuity of Solutions

In this section, we study the modulus of continuity of the solution of the Dirichlet problem with the right hand side in the class \(\mathcal H(\alpha ,{\varOmega })\) under the additional condition that a given measure is locally dominated by capacity.

In what follows we need [8, Lemma 2] whose proof is based on the lemma due to Alexander and Taylor [2, Lemma 3.3]. For the reader’s convenience, we give the proofs. The latter can be simplified by using the Błocki inequality [6].

Lemma 4.1

Let\(B^{\prime } = \{|z-z_{0}| <r \} \subset \subset B= \{|z-z_{0}| <R\}\)be two concentric balls centered atz0in\(\mathbb C^{n}\).Let\(u \in PSH(B) \cap L^{\infty }(B)\)withu < 0. There is aconstant\(C = C(n, \frac {R}{r})\)independent of u suchthat
$$ {\int}_{B^{\prime}} (dd^{c} u)^{n} \leq C |u(z_{0})| \sup_{z\in B} |u(z)|^{n-1}. $$
In particular, if R/r = 3 then the constant C depends only on n.

Proof

Without loss of generality, we may assume z0 = 0. Set ρ := (r + R)/2 and B(ρ) = {|zz0| < ρ}. We use the Błocki inequality [6] for v(z) = |z|2ρ2 and β := ddcv = ddc|z|2, to get
$$ \begin{array}{@{}rcl@{}} {\int}_{B^{\prime}} (dd^{c} u)^{n} &&\leq \frac{1}{(\rho^{2} -r^{2})^{n-1}}{\int}_{B(\rho)} |v|^{n-1} (dd^{c} u)^{n} \\ &&\leq \frac{(n-1)! \|u\|_{B_{\rho}}^{n-1}}{(\rho^{2} -r^{2})^{n-1}} {\int}_{B(\rho)} dd^{c} u \wedge \beta^{n-1}. \end{array} $$
By Jensen’s formula
$$ u(0) + N(\rho) = \frac{1}{\sigma_{2n-1}} {\int}_{\{|\zeta|=1\}} u(\rho \zeta) d\sigma(\zeta), $$
where σ2n− 1 is the area of the unit sphere,
$$ N(\rho) = {\int}_{0}^{\rho} \frac{n(t)}{t^{2n-1}} dt $$
and
$$ n(t) = \frac{1}{\sigma_{2n-1}} {\int}_{\{|z| \leq t\}} {\varDelta} u(z) dV_{2n}(z) = a_{n} {\int}_{\{|z| \leq t\}} dd^{c} u \wedge \beta^{n-1}. $$
Since n(t)/t2n− 2 is increasing, we have
$$ N(R) \geq {\int}_{\rho}^{R} \frac{n(t)}{t^{2n-1}} dt \geq \frac{n(\rho)}{\rho^{2n-2}} \log (R/\rho). $$
From u < 0, it follows that N(R) < −u(0). Hence,
$$ {\int}_{B_{\rho}} dd^{c}u \wedge \beta^{n-1} \leq \frac{n(\rho)}{a_{n}} \leq \frac{N(R)\rho^{n-2}}{\log(R/\rho)} \leq \frac{\rho^{2n-2} |u(0)|}{\log(R/\rho)}. $$
Combining the above inequalities, we get the desired estimate with the constant
$$ C = \frac{(n-1)!\rho^{2n-2}}{(\rho^{2}-r^{2})^{n-1} \log (R/\rho)}. $$
If R = 3r, then C is also independent of r. □

Lemma 4.2

Denote forρ ≥ 0,Bρ = {|zz0| < eρR0}. Givenz0Ωand twonumbersR > 1,R0 > 0 such thatBM ⊂⊂Ω, and givenvPSH(Ω) such that− 1 < v < 0, denote by E theset
$$ E=E(\delta) = \left\{z \in B_{0} : (1-\delta)v \leq \sup_{B_{0}} v\right\}, $$
where δ ∈ (0,1). Then, there exists C0 depending only on n such that
$$ \text{cap}(E, B_{2}) \leq \frac{C_{0}}{R\delta}. $$

Proof

From the logarithmic convexity of the function \(r \mapsto \sup _{|z-z_{0}|<r} v(z)\) it follows that for zBRB0 and \(a_{0}:= \sup _{B_{0}} v\) we have
$$ v(z) \leq a_{0} \left( 1 - \frac{1}{R} \log\frac{|z-z_{0}|}{R_{0}}\right). $$
Hence,
$$ a := \sup_{B_{2}} v \leq a_{0}\left( 1 - \frac{2}{R}\right). $$
Let \(u = u_{E,B_{2}}\) the relative extremal function of E with respect to B2. One has
$$ \frac{v-a}{a - a_{0}/(1-\delta)} \leq u. $$
So, for some z1Ḅ0, we have
$$ u(z_{1}) \geq \frac{a_{0} - a}{a - a_{0}/(1-\delta)} \geq \frac{2(\delta -1)}{(M-2)\delta + 2}. $$
Note that E ⊂{|zz1| < 2R0}⊂ B2. Therefore, Lemma 4.1 gives
$$ \text{cap}(E,B_{2}) = {\int}_{\{|z-z_{1}| < 6R_{0}\}} (dd^{c} u)^{n} \leq C_{0} \|u\|_{B_{2}}^{n-1} |u(z_{1})| \leq \frac{C_{0}}{R \delta}. $$
This is the desired inequality. □
Let us proceed with the proof of Theorem 1.3. Since \(\mu \in \mathcal H(\alpha ,{\varOmega })\), according to [9] and [10, Theorem 5.9] we can solve the Dirichlet problem (1.1) to obtain a unique continuous solution u. Define for δ > 0 small
$$ {\varOmega}_{\delta} := \left\{z \in {\varOmega} : \operatorname{dist}(z, \partial {\varOmega}) > \delta\right\}; $$
and for zΩδ set
$$ u_{\delta}(z) := \sup_{|\zeta| \leq \delta} u(z+ \zeta). $$
Thanks to the arguments in [12, Lemma 2.11] it is easy to see that there exists δ0 > 0 such that
$$ u_{\delta}(z) \leq u(z) + \varpi(\delta;\psi,\partial{\varOmega}) $$
(4.1)
for every zΩδ and 0 < δ < δ0. Here, we used the result of Bedford and Taylor [3, Theorem 6.2] (with minor modifications) to extend ψ plurisubharmonically onto Ω so that its modulus of continuity on \(\bar {\varOmega }\) is controlled by the one on the boundary. Therefore, for a suitable extension of uδ to Ω, using the stability estimate for measure in \(\mathcal H(\alpha ,{\varOmega })\) as in [7, Theorem 1.1] (see also [12, Proposition 2.10]), we get

Lemma 4.3

There are uniform constantsC,α1dependingonly onΩ,α,μsuchthat
$$ \sup_{{\varOmega}_{\delta}} (u_{\delta} - u) \leq \varpi(\delta; \psi, \partial{\varOmega}) + C \left( {\int}_{{\varOmega}_{\delta}} (u_{\delta} -u) d\mu \right)^{\alpha_{1}} $$
for every 0 < δ < δ0.

Thanks to this lemma, we know that the right hand side tends to zero as δ decreases to zero. We shall use the property “locally dominated by capacity” to obtain a quantitative bound via Lemma 4.2.

Let us denote the support of μ by K. Since \(\|u\|_{\infty }\) is controlled by a constant C = C(α,Ω,μ), without loss of generality, we may assume that
$$ -1 \leq u \leq 0. $$
Then for every 0 < ε < 1
$$ {\int}_{{\varOmega}_{\delta}} (u_{\delta} -u) d\mu \leq \varepsilon \mu({\varOmega}) + {\int}_{\{u < u_{\delta} - \varepsilon\} \cap K} d\mu. $$
(4.2)
We shall now estimate the second term on the right hand side. We may assume that Ω ⊂⊂ [0,1]2n. Let us write \(z = (x^{1}, \dots ,x^{2n}) \in \mathbb R^{2n}\) and denote the semi-open cube centered at a point z0, of diameter 2r by
$$ I(z_{0},r):= \{z = (x^{1}, \dots, x^{2n})\in \mathbb C^{n} : -r \leq x^{i} - {x_{0}^{i}}< r, \forall i = 1,\dots,2n\}. $$
Then, by the assumption, μ satisfies for every cube
$$I(z,r)=:I \subset B_{I} := B(z,2r) \subset\subset {\varOmega}$$
and for every set EI, the inequality
$$ \mu (E) \leq \mu (I(z,r)) F_{0}\left( \text{cap} (E, B_{I}) \right), $$
(4.3)
where \(F_{0}: [0,\infty ] \to [0,\infty ]\) is an increasing continuous function and F0(0) = 0.
Consider the semi-open cube decomposition of \({\varOmega } \subset \subset I_{0}:=[0,1)^{2n} \subset \mathbb R^{2n}\) into 32ns congruent cubes of diameter 3s = 2δ, where \( s \in \mathbb N\). Then
$$ \{u < u_{\delta} -\varepsilon\} \cap I_{s} \subset \left\{z\in B_{I_{s}} : u< \sup_{B_{I_{s}}} u -\varepsilon\right\}, $$
(4.4)
where Is = I(zs,δ) and \(B_{I_{s}} = B(z_{s}, 2\delta )\) for some zsI0. Hence,
$$ {\int}_{\{u < u_{\delta} - \varepsilon\}} d\mu \leq {\sum}_{I_{s} \cap K \neq \emptyset }{\int}_{\{u< u_{\delta} -\varepsilon \} \cap I_{s}} d\mu. $$
Using (4.3), (4.4), and then applying Lemma 4.2 for r = 2δ and R = 2R0, we have for Bs := B(zs,4δ) corresponding to each cube Is
$$ \begin{array}{@{}rcl@{}} {\int}_{\{u< u_{\delta} - \varepsilon\} \cap I_{s}} d\mu \leq \mu(I_{s}) F_{0}(\text{cap}(E(\varepsilon, u, B_{I_{s}}), B_{s})) \leq \mu(I_{s}) F_{0} \left( \frac{C_{0}}{\varepsilon \log (R_{0}/\delta)}\right), \end{array} $$
(4.5)
where 2R0 = dist(K,Ω). Therefore, combining the above inequalities, we get that
$$ {\int}_{\{u < u_{\delta} - \varepsilon\}} d\mu \leq \mu({\varOmega}) F_{0} \left( \frac{C_{0}}{\varepsilon \log (R_{0}/\delta)}\right). $$
We conclude from this and Lemma 4.3 that
$$ \omega(\delta;u, \bar{\varOmega}) \leq \sup_{{\varOmega}_{\delta}} (u_{\delta} - u) \leq \varpi(\delta;\psi,\partial{\varOmega}) + C \left[\varepsilon + F_{0} \left( \frac{C_{0}}{\varepsilon \log (R_{0}/\delta)}\right)\right]^{\alpha_{1}}. $$
If we choose \(\varepsilon = (\log R_{0}/\delta )^{-1/2}\), then Theorem 1.3 follows.

Notes

Acknowledgements

The first author was partially supported by NCN grant 2017/27/B/ST1/01145. The second author was supported by the NRF Grant 2011-0030044 (SRC-GAIA) of The Republic of Korea. He also would like to thank Kang-Tae Kim for encouragement and support.

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Authors and Affiliations

  1. 1.Faculty of Mathematics and Computer ScienceJagiellonian UniversityKrakówPoland
  2. 2.Department of Mathematics, Center for Geometry and its ApplicationsPohang University of Science and TechnologyPohangRepublic of Korea

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